This is not a substitute for R documentation, just a list of things I'm using to remember what I read in the email list.
Heed the words of Brian D. Ripley, "One enquiry came to me yesterday which suggested that some users of binary distributions do not know that R comes with two Guides in the doc/manual directory plus an FAQ and the help pages in book form. I hope those are distributed with all the binary distributions (they are not made nor installed by default). Windows-specific versions are available in PDF in rw1001d1.zip and rw1001d2.zip"
Credit for the FaqManager software goes to Stas Bekman, its author.
Table of contents
This is truly easy. Suppose you've got numbers in a space-separated file "myData", with column names in the first row. R's
myDataFrame<-read.table("myData",header=TRUE)
command works great.
If you type "?read.table" it tells about importing files with other delimiters.
Suppose you have tab delimited data with blank spaces to indicate "missing" values. Do this:
> myDataFrame<-read.table("myData",sep="\t",na.strings=" ",header=TRUE)
(new R1.3 feature) Suppose your data is in a gzipped file, myData.gz. Do this:
myDataFrame <- read.table(gzfile("myData.gz"), header=T)
If you read columns in from separate files, combine into a data frame as:
variable1 <- scan("file1")
variable2 <- scan("file2")
mydata <- cbind(variable1, variable2)
#or use the equivalent:
#mydata <- data.frame(variable1, variable2)
#Optionally save dataframe in R object file with:
write.table(mydata, file="filename3")
[top]
To access the columns of a data frame "x" with the column number, say x[,1], to get the first column. If you know the column name, say "pjVar1", it is the same as x$pjVar1.
If you attach that data frame with attach(x) then you can just refer to the variable pjVar1 and R will find it in the data frame. I have had some trouble with this.
In a regression, include the option data=x and then you can use pjVar1 as a variable name. That makes the cleanest code.
Grab an element in a list as x[[1]]
For instance, if a data frame is "nebdata" then
nebdata[1, 2]or
nebdata[nebdata$Volt=="ABal", 2]or by attaching the data frame
attach(nebdata) nebdata[Volt=="ABal", 2] detach(nebdata)(from Diego Kuonen) [top]
With R-1.2, the R team released a manual about how to get data into R and out of R. That's the first place to look if you need help. Right now it is at http://lib.stat.cmu.edu/R/CRAN/doc/manuals/R-data.pdf
[top]Patience is the key. If you understand the format in which your data is currently held, chances are good you can translate it into R with more or less ease.
Most commonly, people seem to want to import Microsoft Excel spreadsheets. I believe there is an ODBC approach to this, but I think it is only for MS Windows.
In the gdata package, which was formerly part of gregmisc, there is a function that can use Perl to import an Excel spreadsheet. If you install gdata, the function to use is called "read.xls". You have to specify which sheet you want. That works well enough if your data is set in a numeric format inside Excel. If it is set with the GENERAL type, I've seen the imported numbers turn up as all asterixes.
In the R-help list last week, I saw mention of another Excel importer program that is in the xlsReadWrite package. I've not tried it. If you install xlsReadWrite, the syntax is the same as the function in the gdata package.
mydata <- read.xls("mydata.xls", sheet = 'Sheet1")
In the R-help list, I also see reference to an Excel program addon called RExcel that can install an option in the Excel menus called "Put R Dataframe". http://sunsite.univie.ac.at/rcom/download/RAndFriends.distro/RAndFriendsLightSetup2081V3.0-8-1.exe THe home for that project is http://sunsite.univie.ac.at/rcom/
I usually proceed like this.
Step 1. Manually massage the Excel sheet so that it has a row of names at the top, and it has numbers or NA values filled in for all cells. Be sure that the columns are declared with the proper Excel format. Numerical information should have a numerical type, while text should have text as the type. Avoid "General".
Step 2. First, try the easy route. In the try a "read.xls" method. As long as you tell it which sheet you want to read out of your data set, and you add header=T and whatever other options you'd like in an ordinary read.table usage, then it has worked well for us.
Step 3. Suppose step 2 did not work. Then you have something irregular in your Excel sheet and you should proceed as follows. Open the Excel sheet in a spreadsheet (we like gnumeric best, but all are probably OK) and check it over, and then choose File/Save As and look for an option like "text configurable" or such. Choose the delimiter as the tab key, and if you have options for numerical output, try to turn off the usage of commas or other unnecessary visual delimiters in numbers. Sometimes for the delimiter between columns, I use the symbol "|" because there's little chance it is in use inside a column of data. If your columns have addresses or such, usage of the COMMA as a delimiter is very dangerous.
After you save that file in text mode, open it in Emacs and look through to make sure that 1) your variable names in the first row do not have spaces or quotation marks or underscores or special characters, and 2) look to the bottom to make sure your spreadsheet program did not insert any crud. If so, delete it. If you see commas in numeric variables, just use the Edit/Search/replace option to delete them.
Then read in your data with read.table ("filename",header=T,sep="\t")
I have tested the "foreign" package by importing an SPSS file and it worked great. I've had great results importing Stata Data sets too.
Here's a big caution for you, however. If your SPSS or Stata numeric variables have some "value lables", say 98=No Answer and 99=Missing, then R will think that the variable is a factor, and it will convert it into a factor with 99 possible values. The foreign library commands for reading spss and dta files have options to stop R from trying to help with factors and I'd urge you to read the manual and use them.
If you use read.spss, for example, setting use.value.labels=F will stop R from creating factors at all. If you don't want to go that far, there's an option max.value.labels that you can set to 5 or 10, and stop it from seeing 98=Missing and then creating a factor with 98 values. It will only convert variables that have fewer than 5 or 10 values. If you use read.dta (for Stata), you can use the option convert.factors=F.
Also, if you are using read.table, you may have trouble if your numeric variables have any "illegal" values, such as letters. Then R will assume you really intend them to be factors and it will sometimes be tough to fix. If you add the option as.is=T, it will stop that cleanup effort by R.
At one time, the SPSS import support in foreign did not work for me, and so I worked out a routine of copying the SPSS data into a text file, just as described for Excel.
SPSS: Use |File, Export| menu in the data editor and export as a tab-separated file (for simplicity, say |mydata.txt| in the top directory on the |C:| drive). That creates tab-separated columns. Make sure that it contains a header line with the variable names and use
read.csv2("C:/mydata.txt",sep="\t")
to read it in. This refers to locales where the comma is used as decimal separator, otherwise use read.csv().
I have a notoriously difficult time with SAS XPORT files and don't even try anymore. I've seen several email posts by Frank Harrel in r-help and he has some pretty strong words about it. I do have one working example of importing the Annenberg National Election Study into R from SAS and you can review that at http://lark.cc.ku.edu/~pauljohn/ANES/2002. I wrote a long boring explanation. Honestly, I think the best thing to do is to find a bridge between SAS and R, say use some program to convert the SAS into Excel, and go from there. Or just write the SAS data set to a file and then read into R with read.table() or such.
[top]update:Merge is confusing! But if you study this, you will see everything in perfect clarity:
x1 <- rnorm(100)
x2 <- rnorm(100)
x3 <- rnorm(100)
x4 <- rnorm(100)
ds1 <- data.frame(city=rep(1,100),x1=x1,x2=x2)
ds2 <- data.frame(city=rep(2,100),x1=x1,x3=x3,x4=x4)
merge(ds1,ds2,by.x=c("city","x1"),by.y=c("city","x1"),all=T)
The trick is to make sure R understands which are the "common variables" in the two datasets so it lines them up, and then all=T is needed to say that you don't want to throw away the variables that are only in one set or the other. Read the help page over and over, you eventually get it.
More examples:
experiment<-data.frame(times=c(0,0,10,10,20,20,30,30),expval=c(1,1,2,2,3,3,4,4))simul<-data.frame(times=c(0,10,20,30),simul=c(3,4,5,6))
I want a merged datatset like:
> times expval simul
> 1 0 1 3
> 2 0 1 3
> 3 10 2 4
> 4 10 2 4
> 5 20 3 5
> 6 20 3 5
> 7 30 4 6
> 8 30 4 6
Suggestions
merge(experiment, simul)
does all the work for you. (from Brian D. Ripley)
Or consider:
exp.sim<-data.frame(experiment,simul=simul$simul[match(experiment$times,simul$times)])(from Jim Lemon)
I have dataframes like this:
state count1 percent1
CA 19 0.34
TX 22 0.35
FL 11 0.24
OR 34 0.42
GA 52 0.62
MN 12 0.17
NC 19 0.34
state count2 percent2
FL 22 0.35
MN 22 0.35
CA 11 0.24
TX 52 0.62
And I want
state count1 percent1 count2 percent2
CA 19 0.34 11 0.24
TX 22 0.35 52 0.62
FL 11 0.24 22 0.35
OR 34 0.42 0 0
GA 52 0.62 0 0
MN 12 0.17 22 0.35
NC 19 0.34 0 0
( from Yu-Ling Wu )
In response, Ben Bolker said
state1 <- c("CA","TX","FL","OR","GA","MN","NC")
count1 <- c(19,22,11,34,52,12,19)
percent1 <- c(0.34,0.35,0.24,0.42,0.62,0.17,0.34)
state2 <- c("FL","MN","CA","TX")
count2 <- c(22,22,11,52)
percent2 <- c(0.35,0.35,0.24,0.62)
data1 <- data.frame(state1,count1,percent1)
data2 <- data.frame(state2,count2,percent2)
datac <- data1m <- match(data1$state1,data2$state2,0)
datac$count2 <- ifelse(m==0,0,data2$count2[m])
datac$percent2 <- ifelse(m==0,0,data2$percent2[m])
If you didn't want to keep all the rows in both data sets (but just theshared rows) you could use
merge(data1,data2,by=1)
Question: I would like to create an (empty) data frame with "headings" for every column (column titles) and then put data row-by-row into this data frame (one row for every computation I will be doing), i.e.
no. time temp pressure <---the headings 1 0 100 80 <---first result 2 10 110 87 <---2nd result .....
Answer: Depends if the cols are all numeric: if they are a matrix would be better.
If you know the number of results in advance, say, N, do this
df <- data.frame(time=numeric(N), temp=numeric(N), pressure=numeric(N)) df[1, ] <- c(0, 100, 80) df[2, ] <- c(10, 110, 87) ...or
m <- matrix(nrow=N, ncol=3)
colnames(m) <- c("time", "temp", "pressure")
m[1, ] <- c(0, 100, 80)
m[2, ] <- c(10, 110, 87)
The matrix form is better size it only needs to access one vector
(a matrix is a vector with attributes), not three.
If you don't know the final size you can use rbind to add a row at a time, but that is substantially less efficient as lots of re-allocation is needed. It's better to guess the size, fill in and then rbind on a lot more rows if the guess was too small.(from Brian Ripley)
[top](from Stephen Arthur)
Example Data:File1 C A T File2 1 2 34 56 2 3 45 67 3 4 56 78
To Yield:
> C A T 1 2 34 56 > C A T 2 3 45 67 > C A T 3 4 56 78
This works:
repcbind <- function(x,y) {
nx <- nrow(x)
ny <- nrow(y)
if (nx<ny)
x <- apply(x,2,rep,length=ny)
else if (ny<nx)
y <- apply(y,2,rep,length=nx)
cbind(x,y)
}
(from Ben Bolker)
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If you load things, R does not warn you when they are not found, it records them as NULL. You have the responsibility of checking them. Use
is.null(list$component)
to check a thing named component in a thing named list.
Accessing non-existent dataframe columns with "[" does give an error, so you could do that instead.
>data(trees) >trees$aardvark NULL >trees[,"aardvark"]Error in [.data.frame(trees, , "aardvark") : subscript out of bounds (from Thomas Lumley) [top]
runif() --uniform random on [0,1) rnorm() --normal random
multivariate normal: Given covar matrix:
0.25, 0.20 0.20, 0.25
Brian Ripley said: "mvrnorm in package MASS (part of the VR bundle).
mvrnorm(2, c(0,0), matrix(c(0.25, 0.20, 0.20, 0.25), 2,2))
and Peter Dalgaard observed "a less general solution for this particular case would be
rnorm(1,sd=sqrt(0.20)) + rnorm(2,sd=sqrt(0.05))
Wait. You want randomly drawn integers? Here:
# If you mean sampling without replacement: >sample(1:10,3,replace=FALSE) #If you mean with replacement: >sample(1:10,3,replace=TRUE)(from Bill Simpson) [top]
If you generate random numbers with a given tool, you don't get a sample with the exact mean you specify. A generator with a mean of 0 will create samples with varying means, right?
So, how to force the sample to have a mean of 0? Take a 2 step approach:
R> x <- rnorm(100, mean = 5, sd = 2) R> x <- (x - mean(x))/sqrt(var(x)) R> mean(x) [1] 1.385177e-16 R> var(x) [1] 1and now create your sample with mean 5 and sd 2:
R> x <- x*2 + 5 R> mean(x) [1] 5 R> var(x) [1] 4(from Torsten.Hothorn) [top]
x<-c(10,40,50,100) # income vector for instance w<-c(2,1,3,2) # the weight for each observation in x with the same rep(x,w) [1] 10 10 40 50 50 50 100 100
(from P. Malewski)
[top]Also, most of the functions that have weights have frequency weights rather than probability weights: that is, setting a weight equal to 2 has exactly the same effect as replicating the observation.
If you have frequency weights you may have to expand your dataset.
expanded.data<-as.data.frame(lapply(compressed.data,
function(x) rep(x,compressed.data$weights)))
should do it.
(from Thomas Lumley)
[top]Given a 8 dimensional crosstab, you want a data frame with 8 factors and 1 column for frequencies of the cells in the table.
R1.2 introduces a function as.data.frame.table() to handle this.
This can also be done manually. Here's a function(it's a simple wrapper around expand.grid):
dfify <- function(arr, value.name = "value", dn.names =
names(dimnames(arr))) {
Version <- "$Id: dfify.sfun,v 1.1 1995/10/09 16:06:12 d3a061 Exp $"
dn <- dimnames(arr <- as.array(arr))
if(is.null(dn))
stop("Can't data-frame-ify an array without dimnames")
names(dn) <- dn.names
ans <- cbind(expand.grid(dn), as.vector(arr))
names(ans)[ncol(ans)] <- value.name
ans
}
The name is short for "data-frame-i-fy".
For your example, assuming your multi-way array has proper dimnames, you'd just do:
my.data.frame <- dfify(my.array, value.name="frequency")
(from Todd Taylor)
[top]Say I have a command that produced a 28 x 28 data matrix. How can I output the matrix into a txt file (rather than copy/paste the matrix)?
write.table(mat,file="filename")
Pesl Thomas said, "I want to output a file to my disc with the contents of a data frame I called "tab"
> tabor 1 2 3 4 5 6 1 57 80 25 23 46 23 2 106 35 59 8 51 4
You can use
write.table(unclass(tab))
You could also do
write(t(tab),ncol=NCOL(tab))since write() transposes rows and columns. (from Thomas Lumley)
Note MASS library has a function write.matrix which is faster if you need to write a numerical matix, not a data frame. Good for big jobs.
[top]In r-help, I asked " I keep finding myself in a situation where I want to calculate a variable name and then use it on the left hand side of an assignment. For example
> iteration <- 1
> varName <- paste("run",iteration,sep="")
> myList$parse(text=varName) <- aColumn
I got many useful answers that opened my eyes!
Brian Ripley said:
For a data frame you could usemydf[paste("run",iteration,sep="")] <- aColumnand for a list or a data frameRobject[[paste("run",iteration,sep="")]] <- aColumn
And Thomas Lumley added: " If you wanted to do something of this sort for which the above didn't workyou could also learn about substitute():
eval(substitute(myList$newColumn<-aColumn),
list(newColumn=as.name(varName)))
as an alternative to parse().
-(Thomas Lumley)
[top]Paste together a variable name, set it to a value. Use assign. As in
> assign(paste("file", 1, "max", sep=""), 1)
> ls()
[1] "file1max"
(Brian Ripley, June 18, 2001)
[top]
Please read the documentation for transform() and replace() and also learn how to use the magic of R vectors.
The transform() function works only for data frames. Suppose a data frame is called "mdf" and you want to add a new variable "newV" that is a function of var1 and var2:
mdf <- transform(mdf, newV=log(var1) + var2))
I'm inclined to take the easy road when I can. Proper use of indexes in R will help a lot, especially for recoding discrete valued variables. Some cases are particularly simple because of the way arrays are processed.
Suppose you create a variable, and then want to reset some values to missing. Go like this:
x <- rnorm(10000)And if you don't want to replace the original variable, create a new one first ( xnew <- x ) and then do that same thing to xnew.
x[x > 1.5] <- NA
You can put other variables inside the brackets, so if you want x to equal 234 if y equals 1, then
x[y == 1] <- 234
Suppose you have v1, and you want to add another variable v2 so that there is a translation. If v1=1, you want v2=4. If v1=2, you want v2=4. If v1=3, you want v2=5. This reminds me of the old days using SPSS and SAS. I think it is clearest to do:
> v1 <- c(1,2,3)
# now initialize v2
> v2 <- rep( -9, length(v1))
# now recode v2
> v2[v1==1] <- 4
> v2[v1==2]<-4
> v2[v1==3]<-5
> v2[1] 4 4 5
Note that R's "ifelse" command can work too:
x<-ifelse(x>1.5,NA,x)
One user recently asked how to take data like a vector of names and convert it to numbers, and 2 good solutions appeared:
y <- c( "OLDa", "ALL", "OLDc", "OLDa", "OLDb", "NEW", "OLDb", "OLDa", "ALL","...")or
> el <- c("OLDa", "OLDb", "OLDc", "NEW", "ALL")
> match(y,el)[1] 1 5 3 1 2 4 2 1 5 NA
> f <- factor(x,levels=c("OLDa", "OLDb", "OLDc", "NEW", "ALL") )
> as.integer(f)
> [1] 1 5 3 1 2 4 2 1 5
I asked Mark Myatt (author of the ReX guide mentioned below) for more examples:
For example, suppose I get a bunch of variables coded on a scale
1 = noRecode that into a new variable name with 0=no, 1=yes, and all else NA.
6 = yes
8 = tied
9 = missing
10 = not applicable.
It seems like the replace() function would do it for single values but you end up with empty levels in factors but that can be fixed by re-factoring the variable. Here is a basic recode() function:
recode <- function(var, old, new){
x <- replace(var, var == old, new)
if(is.factor(x)) factor(x) else x}For the above example:test <- c(1,1,2,1,1,8,1,2,1,10,1,8,2,1,9,1,2,9,10,1)Although it is probably easier to use replace():
testtest <- recode(test, 1, 0)
test <- recode(test, 2, 1)
test <- recode(test, 8, NA)
test <- recode(test, 9, NA)
test <- recode(test, 10, NA)
test
test <- c(1,1,2,1,1,8,1,2,1,10,1,8,2,1,9,1,2,9,10,1)
testtest <- replace(test, test == 8 | test == 9 | test == 10, NA)
test <- replace(test, test == 1, 0)
test <- replace(test, test == 2, 1)
test
I suppose a better function would take from and to lists as arguments:
For the example:
recode <- function(var, from, to){
x <- as.vector(var)
for (i in 1:length(from)){
x <- replace(x, x == from[i], to[i])}
if(is.factor(var)) factor(x) else x}
test <- c(1,1,2,1,1,8,1,2,1,10,1,8,2,1,9,1,2,9,10,1)and it still works with single values.
testtest <- recode(test, c(1,2,8:10), c(0,1))
test
Suppose somebody gives me ascale from 1 to 100, and I want to collapse it into 10 groups, how do I go about it?
Mark says: Use cut() for this. This cuts into 10 groups:
To get ten groups without knowing the minimum and maximum value you canuse pretty():
test <- trunc(runif(1000,1,100))
groups <-cut(test,seq(0,100,10))
table(test, groups)
You can specify the cut-points:
groups <- cut(test,pretty(test,10))
table(test, groups)
And they don't need to be even groups:
groups <- cut(test,c(0,20,40,60,80,100))
table(test, groups)
Mark added, "I think I will add this sort of thing to the REX pack."
groups <- cut(test,c(0,30,50,75,100))
table(test, groups)
2003-12-01, someone asked how to convert a vector of numbers to characters, such as
>if x[i] < 250 then col[i] = "red"and so forth. Many interesing answers appeared in R-help. A big long nested ifelse would work, as in:
>else if x[i] < 500 then col[i] = "blue"
x.col <- ifelse(x < 250, "red", ifelse(x<500, "blue", ifelse(x<750, "green", "black")))
There were some nice suggestions to use cut, such as Gabor Grothendeick's advice:
The following results in a character vector:
>colours <- c("red", "blue", "green","back")
>colours[cut(x, c(-Inf,250,500,700,Inf),right=F,lab=F)]
While this generates a factor variable:
colours <- c("red", "blue", "green","black")
cut(x, c(-Inf,250,500,700,Inf),right=F,lab=colours)
[top]
2 examples:
c is a column, you want dummy variable, one for each valid value. First, make it a factor, then use model.matrix():
> x <- c(2,2,5,3,6,5,NA) > xf <- factor(x,levels=2:6) > model.matrix(~xf-1) xf2 xf3 xf4 xf5 xf6 1 1 0 0 0 0 2 1 0 0 0 0 3 0 0 0 1 0 4 0 1 0 0 0 5 0 0 0 0 1 6 0 0 0 1 0 attr(,"assign") [1] 1 1 1 1 1
(from Peter Dalgaard )
Question: I have a variable with 5 categories and I want to create dummy variables for each category.
Answer: Use row indexing or model.matrix.
ff <- factor(sample(letters[1:5], 25, replace=TRUE))
diag(nlevels(ff))[ff,]
or
model.matrix(~ ff - 1)
(from Brian D. Ripley)
[top]Peter Dalgaard explained, "the simple way is to create a new variable which shifts the response, i.e.
yshft <- c(y[-1], NA) # pad with missing summary(lm(yshft ~ x + y))Alternatively, lag the regressors:
N <- length(x) xlag <- c(NA, x[1:(N-1)]) ylag <- c(NA, y[1:(N-1)]) summary(lm(y ~ xlag + ylag))[top]
The best way to drop levels, BTW, is
problem.factor <- problem.factor[,drop=TRUE] (from Brian D. Ripley)
[top]Want to take observations for which variable Y is greater than A and less than B: do
X[Y>=A & Y<=B]
Suppose you want observations with c=1 in df1. This makes a new data frame you want.
df2 <- df1[df1$c == 1,]
and note that indexing is pretty central to using S (the language), so it is worth learning all the ways to use it. (from Brian Ripley)
Or use "match" select values from the column "d" by taking the ones that match the values of another column, as in
> d <- t(array(1:20,dim=c(2,10)))(from Peter Wolf)
> i <- c(13,5,19)
> d[match(i,d[,1]), 2][1] 14 6 20
Till Baumgaertel wanted to select observations for men over age 40, and sex was coded either m or M. Here are two working commands:
1.) maleOver40 <- subset(d, sex %in% c("m","M") & age > 40)
2.) maleOver40 <- d[(d$sex == "m" | d$sex == "M") & d$age >40,]To decipher that, do How to deal with values that are already marked as missing? John Fox says:
x2 <- na.omit(x)produces a copy of the data frame x with all rows that contain missing data removed. The function na.exclude could be used also. For more information on missings, check help : ?na.exclude.
For exclusion of missing, Peter Dalgaard likes
subset(x,complete.cases(x)) or x[complete.cases(x),]adding "is.na(x) is preferable to x != "NA" [top]
Given data like:
1 ABK 19910711 11.1867461 0.0000000 108 2 ABK 19910712 11.5298979 11.1867461 111 6 CSCO 19910102 0.1553819 0.0000000 106 7 CSCO 19910103 0.1527778 0.1458333 166remove the first observation for each value of the "sym" variable (the one coded ABK,CSCO, etc). . If you just need to remove rows 1,5, and 13, do:
newhilodata <- hilodata[-c(1,6,13),]To solve the more general problem of omitting the first in each group, assuming "sym" is a factor, try something like
newhilodata <- subset(hilodata, diff(c(0,as.integer(sym))) != 0)(actually, the as.integer is unnecessary because the c() will unclass the factor automagically)
(from Peter Dalgaard)
Alternatively, you could use the match function because it returns the first match. Suppose jm is the data set. Then:
> match(unique(jm$sym), jm$sym) [1] 1 6 13 > jm <- jm[ -match(unique(jm$sym), jm$sym), ](from Douglas Bates)
As Robert pointed out to me privately: duplicated() does the trick
subset(hilodata, duplicated(sym))has got to be the simplest variant.
(from Peter Dalgaard)
[top]sample(N, n, replace=F)
and
seq(N)[rank(runif(N)) <= n]
is another general solution.
(from Brian D. Ripley)
[top]You can manage data directly by deleting lines or so forth, but subset() can be used to achieve the same effect without editing the data at all. Do ?select to find out more. Subset is also an option in many statistical functions like lm.
Peter Dalgaard gave this example, using the "builtin" dataset airquality.
data(airquality) names(airquality) lm(Ozone~.,data=subset(airquality,select=Ozone:Month)) lm(Ozone~.,data=subset(airquality,select=c(Ozone:Wind,Month))) lm(Ozone~.-Temp,data=subset(airquality,select=Ozone:Month))
The "." on the RHS of lm means "all variables" and the subset command on the rhs picks out different variables from the dataset. "x1:x2" means variables between x1 and x2, inclusive.
[top]data<-read.table("lastline.txt",header=T,as.is = TRUE) indices<-1:dim(data)[2] indices<-na.omit(ifelse(indices*sapply(data,is.numeric),indices,NA)) mean<-sapply(data[,indices],mean) sd<-sapply(data[,indices],sd)
[top]Can someone tell me how can I sort a list that contains duplicates (name) but keeping the duplicates together when sorting the values.
name M 1234 8 1234 8.3 4321 9 4321 8.1
I also would like to set a cut-off, so that anything below a certain values will not be sorted.(from Kong, Chuang Fong)
I take it that the cutoff is on the value of M. OK, suppose it is the value of `coff'.
sort.ind <- order(name, pmax(coff, M)) # sorting index name <- name[sort.ind] M <- M[sort.ind]Notice how using pmax() for "parallel maximum" you can implement the cutoff by raising all values below the mark up to the mark thus putting them all into the same bin as far as sorting is concerned.
If your two variables are in a data frame you can combine the last two steps into one, of course.
sort.ind <- order(dat$name, pmax(coff, dat$M)) dat <- dat[sort.ind, ]In fact it's not long before you are doing it all in one step:
dat <- dat[order(dat$name, pmax(coff, dat$M)), ]
(from Bill Venables)
I want the ability to sort a data frame lexicographically according to several variables
Here's how:
spsheet[order(name,age,zip),] (from Peter Dalgaard)
[top]Read the help for by()
> by(x[2], x$group, rank) x$group: A [1] 4.0 1.5 1.5 3.0 ------------------------------------------------------------ x$group: B [1] 3 2 1(from Brian D. Ripley) [top]> c(by(x[2], x$group, rank), recursive=T) A1 A2 A3 A4 B1 B2 B3 4.0 1.5 1.5 3.0 3.0 2.0 1.0
To make sure missings are omitted from a dataset called "insure":
thisdf <- na.omit(insure[,c(1, 19:39)]) body.m <- lm(BI.PPrem ~ ., data=thisdf, na.action=na.fail)
is.na() returns a T for missings, and that can be used to spot and change them.
To convert missing values to a numerical value, Matt Wiener (2005-01-05) says " The commands below construct a matrix, insert some NA's, and then convert them all to 0.
> temp1 <- matrix(runif(25), 5, 5) > temp1[temp1 < 0.1] <- NA > temp1[is.na(temp1)] <- 0[top]
Question: I want to read values from a text file - 200 lines, 32 floats per line - and calculate a mean for each of the 32 values, so I would end up with an "x" vector of 1-200 and a "y" vector of the 200 means.
Peter Dalgaard says do thi:
y <- apply(as.matrix(read.table("myfile")), 1, mean) x <- seq(along=y)
(possibly adding a couple of options to read.table, depending on the file format)
[top]How about this: Z<-aggregate(X, f, sum)
(assumes all X variables can be summed)
Or: [If] X contains also factors. I have to select variables for which summary statistics have to be computed. So I used:
Z <- data.frame(f=levels(f),x1=as.vector(tapply(x1,f,sum)))
(from Wolfgang Koller)
[top]Given
10 20 23 44 33 10 20 33 23 67and you want
10 20 56 67 100try this:
dat<-read.table("data.dat",header=TRUE)
aggregate(dat[,-(1:2)], by=list(std=dat$std, cf=dat$cf), sum)
note the first two columns are excluded by [,(1:2)] and the by optoin preserves those values in the output.
[top]
I want to "take out" the first decimal place of each output, plot them based on their appearence frequencies. Then take the second decimal place, do the same thing.
a<- log(1:1000) d1<-floor(10*(a-floor(a))) # first decimal par(mfrow=c(2,2)) hist(d1,breaks=c(-1:9)) table(d1) d2<-floor(10*(10*a-floor(10*a))) # second decimal hist(d2,breaks=c(-1:9)) table(d2)(from Yudi Pawitan)
x <- 1:1000(from Martin Maechler) [top]ndig <- 6
(ii <- as.integer(10^(ndig-1) * log(x)))[1:7] (ci <- formatC(ii, flag="0", wid= ndig))[1:7] cm <- t(sapply(ci, function(cc) strsplit(cc,NULL)[[1]])) cm [1:7,]
apply(cm, 2, table) #--> Nice tables
# The plots : par(mfrow= c(3,2), lab = c(10,10,7)) for(i in 1:ndig) hist(as.integer(cm[,i]), breaks = -.5 + 0:10, main = paste("Distribution of ", i,"-th digit"))
Let Z denote the list of matrices. All matrices have the same order. Suppose you need to take element [1,2] from each.
lapply(Z, function(x) x[1,2])
should do this, giving a list. Use sapply if you want a vector. (Brian Ripley)
[top]xlevels<-sort(unique(x))
Which you can use in a contour plot, as in
contour(xlevels,ylevels,zvals)
[top]Use the eval(parse()) pairing:
String2Eval <- "A valid R statement"
eval(parse(text = String2Eval))
(from Mark Myatt)
Also check out substitute(), as.name() et al. for other methods of manipulating expressions and function calls (from Peter Dalgaard)
Or
eval(parse(text="ls()"))
Or
eval(parse(text = "x[3] <- 5"))
[top]To analyse large vectors of data using boxplot to find outliers, try:
which(x==boxplot(x,range=1)$out)
[top]Jason Liao "I have 10000 integer triplets stored in A[1:10000, 1:3]. I would like
to find the unique triplets among the 10000 ones with possible
duplications."
Peter Dalgaard answers, "As of 1.4.0 (!):
unique(as.data.frame(A))"
2.24 Select only numerical variables
(12/11/2002 Paul Johnson <pauljohn@ku.edu>)
sapply(dataframe, is.numeric)
Thomas Lumley 11/11/2002
or
which(sapply(data.frame, is.numeric))
Some things in R are so startlingly simple and different from other programs that the experts have a difficult time understanding our misunderstanding.
Mathematically, I believe a vector is a column in which all elements are of the same type (e.g., real numbers). (1,2,3) is a vector of integers as well as a column in the data frame sense. If you put dissimilar-typed items into an R column, it really creates a "list" object. (In R, a list is a collection of dissimilar types).
To create a mathematical vector, you can just initialize a column of numbers with c(). As long as all of your input is of the same type, you don't hit any snags. Note the is.vector() function will say this is a vector:
> v <- c(1, 2, 3) > is.vector(v)
Caution: If your input is not all of the same type, then R converts it to characters, which you may not intend.
> gg <- c("1",3,4)
> is.vector(gg)
[1] TRUE
> gg
[1] "1" "3" "4"
> is.character(gg)
[1] TRUE
The worst case scenario is that it manufactures a list. The double brackets [[]] are the big signal that you really have a list:
> bb <- c(1,c,4) > is.vector(bb) [1] TRUE > bb [[1]] [1] 1[[2]] .Primitive("c")
[[3]] [1] 4
> is.character(bb) [1] FALSE > is.integer(bb) [1] FALSE > is.list(bb) [1] TRUE
To make sure you get what you want, you can use the vector() function, and then tell it what type of elements you want in your vector. Please note it puts a "false" value in each position, not a missing.
> vector(mode="integer",10) [1] 0 0 0 0 0 0 0 0 0 0 > vector(mode="double",10) [1] 0 0 0 0 0 0 0 0 0 0 > vector(mode="logical",10) [1] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSEThere are also methods to directly create the particular kind of vector you want (factor, character, integer, etc). For example, if you want a real-valued vector, there is a shortcut method numeric()
> x <- numeric(10) > x [1] 0 0 0 0 0 0 0 0 0 0
If you want to add values onto the end of a vector, you have options!
If you want to add a single value onto the vector, try:
> r <- rnorm(1) > v <- append(v, r)
The length() command can not only tell you how long a vector is, but it can be used to RESIZE it. This command says if the length of v has reached the value vlen, then expand the length by doubling it.
if (vlen == length(v)) length(v) <- 2*length(v)
These were mentioned in an email from Gabor Grothendieck to the list on 2005-01-04.
[top]diag(n)
Or, go the long way:
n<-c(5) I <- matrix(0,nrow=n,ncol=n) I[row(I)==col(I)] <- 1E.D.Isaia [top]
dim(m)<-NULL, or c(m) (from Peter Dalgaard )
[top]I don't know why this is useful, but it shows some row and matrix things:
I ended up using Brian Ripley's method, as I got it first and it worked, ie.
A <- matrix(1, n-1, n-1) rA <- row(A) rA[rA + col(A) <= n]However, thanks to Andy Royle I have since discovered that there is a much more simple and sublime solution:
sequence(n-1:1) [1] 1 2 3 4 1 2 3 1 2 1Karen Kotschy [top]
extract every nth element from a very long vector, vec?
vec[seq(n, length(vec), n)](from Brian Ripley)
seq(1,11628,length=1000) will give 1000 evenly spaced numbers from 1:11628 that you can then index with. (from Thomas Lumley)
My example:
vec <- rnorm(1999) newvec <- vec[1, length(vec), 200] > newvec [1] 0.2685562 1.8336569 0.1371113 0.2204333 -1.2798172 0.3337282 [7] -0.2366385 0.5060078 0.9680530 1.2725744This shows the items in vec at indexes (1, 201, 401, ..., 1801) [top]
n <- 1000 x <- sort(rnorm(n)) x0 <- 1.5 dx <- abs(x-x0) which(dx==min(dx))(from Jan Schelling)
which(abs(x - 1.5) == min(abs(x - 1.5)))(from Uwe Ligges) [top]
which(x!=0) (from Uwe Ligges)
rfind <- function(x) seq(along=x)[x != 0] (from Brian D. Ripley)
Concerning speed and memory efficiency I find
as.logical(x)
is better than
x!=0
and
seq(along=x)[as.logical(x)]
is better than
which(as.logical(x))
thus
which(x!=0)
is shortest and
rfind <- function(x)seq(along=x)[as.logical(x)]
seems to be computationally most efficient
(from Jens Oehlschlägel-Akiyoshi)
[top]Suppose the vector "Pes" has 600 observations. Don't do this:
(1:600)[is.na(Pes)]
The `approved' method is
seq(along=Pes)[is.na(Pes)]
In this case it does not matter as the subscript is of length 0, but it has floored enough library/package writers to be worth thinking about.
(from Brian Ripley)
However, the solution I gave
which(is.na(Pes))
is the one I still really recommend; it does deal with 0-length objects, and it keeps names when there are some, and it has an `arr.ind = FALSE' argument to return array indices instead of vector indices when so desired. (from Martin Maechler)
[top]T[which(A==max(A,na.rm=TRUE))]
(pj: note na.rm deprecated? I think it is now na.omit)
[top]> tmat <- matrix(rep(0,3*3),ncol=3) > tmat [,1] [,2] [,3] [1,] 0 0 0 [2,] 0 0 0 [3,] 0 0 0 > tmat[tmat==0]<-1 > tmat [,1] [,2] [,3] [1,] 1 1 1 [2,] 1 1 1 [3,] 1 1 1 (from Jan Goebel)
If la is a data frame, you have to coerce the data into matrix form, with:
la <- as.matrix(la) la[la==0] <- 1
Try this:
la <- ifelse(la == 0, 1, la)(from Martyn Plummer) [top]
> x <- matrix(1:10,,2) > x[x[,1]%in%c(2,3),] > x[!x[,1]%in%c(2,3),](from Peter Malewski)
mat[!(mat$first %in% 713:715),](from Peter Dalgaard ) [top]
Apply "length()" to results of which() described in previous question, as in
length(which(v<7)) or sum( v<7,na.rm=TRUE)
If you apply sum() to a matrix, it will scan over all columns. To focus on a particular column, use subscripts like sum(v[,1]>0) or such. If you want separate counts for columns, there's a method colSums() that will count separately for each column.
[top]I need to compute partial correlation coefficients between multiple variables (correlation between two paired samples with the "effects of all other variables partialled out")? (from Kaspar Pflugshaupt)
Actually, this is quite straightforward. Suppose that R is the correlation matrix among the variables. Then,
Rinv<-solve(R)
D<-diag(1/sqrt(diag(Rinv)))
P<- -D %*% Rinv %*% D
The off-diagonal elements of P are then the partial correlations between each pair of
variables "partialed" for the others. (Why one would want to do this is another
question.)
(from John Fox )
In general you invert the variance-covariance matrix and then rescale it so the diagonal is one. The off-diagonal elements are the negative partial correlation coefficients given all other variables.
pcor2 <- function(x){
conc <- solve(var(x))
resid.sd <- 1/sqrt(diag(conc))
pcc <- - sweep(sweep(conc, 1, resid.sd, "*"), 2, resid.sd, "*")
return(pcc)
}
pcor2(cbind(x1,x2,x3))
see J. Whittaker's book "Graphical models in applied multivariate statistics" from Martyn Plummer)
This is the version I'm using now, together with a test for significance of each coefficient (H0: coeff=0):
f.parcor <-
function (x, test = F, p = 0.05)
{
nvar <- ncol(x)
ndata <- nrow(x)
conc <- solve(cor(x))
resid.sd <- 1/sqrt(diag(conc))
pcc <- -sweep(sweep(conc, 1, resid.sd, "*"), 2, resid.sd,
"*")
colnames(pcc) <- rownames(pcc) <- colnames(x)
if (test) {
t.df <- ndata - nvar
t <- pcc/sqrt((1 - pcc^2)/t.df)
pcc <- list(coefs = pcc, significant = t > qt(1 - (p/2),
df = t.df))
}
return(pcc)
}
(from Kaspar Pflugshaupt)
[top]
Suppose you have data matrices A and B on the same observations. Then
cancor <- function(A, B)
{
Ap <- prcomp(scale(A, T, F), retx=T)
Apc <- Ap$x %*% diag(1/Ap$sdev)
Bp <- prcomp(scale(B, T, F), retx=T)
Bpc <- Bp$x %*% diag(1/Bp$sdev)
Sigma <- crossprod(Apc, Bpc)/(nrow(A) - 1)
s <- svd(Sigma, ncol(A), ncol(B))
return(list(cor=s$d, canvar.x=Apc %*% s$u, canvar.y=Bpc %*% s$v))
}
should do the trick. The canonical variates are zero-mean, unit-variance
(unlike S-PLUS). (from Brian D. Ripley)
[top]
Brian Ripley said:
my.array<-array(0,dim=c(10,5,6,8))
will give you a 4-dimensional 10 x 5 x 6 x 8 array.
Or
array.test <- array(1:64,c(4,4,4)) array.test[1,1,1] 1 array.test[4,4,4] 64[top]
Given a list with 100s of (nx2) matrices, how do you combine them into a giant (Nx2) matrix?
list is the list of matrices.
nr <- sapply(list, nrow) cs <- cumsum(nr) st <- c(0, cs[-length(cs)]) + 1 res <- matrix(, sum(nr), 2) for(i in seq(along=nr)) res[st[i]:cs[i],] <- list[[i]](from Brian Ripley) [top]
Want a matrix of 0s and 1s indicating whether a cell has a neighbor at a location:
N <- 3
x <- matrix(1:(N^2),nrow=N,ncol=N)
rowdiff <- function(y,z,mat)abs(row(mat)[y]-row(mat)[z])
coldiff <- function(y,z,mat)abs(col(mat)[y]-col(mat)[z])
rook.case <- function(y,z,mat){coldiff(y,z,mat)+rowdiff(y,z,mat)==1}
bishop.case <- function(y,z,mat){coldiff(y,z,mat)==1 & rowdiff(y,z,mat)==1}
queen.case <- function(y,z,mat){rook.case(y,z,mat) | bishop.case(y,z,mat)}
matrix(as.numeric(sapply(x,function(y)sapply(x,rook.case,y,mat=x))),ncol=N^2,nrow=N^2)
matrix(as.numeric(sapply(x,function(y)sapply(x,bishop.case,y,mat=x))),ncol=N^2,nrow=N^2)
matrix(as.numeric(sapply(x,function(y)sapply(x,queen.case,y,mat=x))),ncol=N^2,nrow=N^2)
(from Ben Bolker)
[top]
The question was:
I have a matrix of predictions from an proportional odds model (using the polr function in MASS), so the columns are the probabilities of the responses, and the rows are the data points. I have another column with the observed responses, and I want to extract the probabilities for the observed responses.
As a toy example, if I have
> x <- matrix(c(1,2,3,4,5,6),2,3) > y <- c(1,3)and I want to extract the numbers in x[1,1] and x[2,3] (the columns being indexed from y), what do I do?
Is
x[cbind(seq(along=y), y)]
what you had in mind? The key is definitely matrix indexing. (from Brian Ripley)
[top]There are many ways to do this. Whenever it comes up in r-help, there is a different answer.
Suppose you want a matrix like
a^0 0 0 a^1 a^0 0 a^2 a^1 a^0
I don't know why you'd want that, but look at the differences among answers this elicited.
ma <- matrix(rep(c(a^(0:n), 0), n+1), nrow=n+1, ncol=n+1) ma[upper.tri(ma)] <- 0(from Uwe Ligges)
> n <- 3 > a <- 2 > out <- diag(n+1) > out[lower.tri(out)] <- a^apply(matrix(1:n,ncol=1),1,function(x)c(rep(0,x),1:(n-x+1)))[lower.tri(out)] > out(from Woodrow Setzer)
tmpmat <- function(a,n) {
x <- matrix(a,nrow=n,ncol=n)
x <- x^pmax(0,row(x)-col(x))
x[row(x)
(from Ben Bolker)
If you want an upper triangular matrix, this use of "ifelse" looks great to me:
fun <- function(x,y) { ifelse(y>x, x+y, 0) }
(then use this with outer())
or
m <- outer(x,y,FUN="+")
m[lower.tri(m, diag=T)] <- 0
(from Kaspar Pflugshaupt)
[top]
3.20 Calculate inverse of X
(20/06/2001 ? <?>)
solve(A) yields the inverse of A.
[top]
3.21 Interesting use of Matrix Indices
(20/06/2001 ? <?>)
Here's a similar problem.
Mehdi Ghafariyan said "I have two vectors A=c(5,2,2,3,3,2)
and B=c(2,3,4,5,6,1,3,2,4,3,1,5,1,4,6,1,4)
and I want to make the following matrix
using the information I have from the above
vectors.
0 1 1 1 1 1
1 0 1 0 0 0
0 1 0 1 0 0
1 0 1 0 1 0
1 0 0 1 0 1
1 0 0 1 0 0
so the first vector says that I have 6
elements therefore I have to make a
6 by 6 matrix and then I have to read
5 elements from the second
vector , and put 1 in [1,j] where j=2,3,4,5,6
and put zero elsewhere( i.e. in [1,1])
and so on. Any idea how this can be done in R ?
Use matrix indices:
a<-c(5,2,2,3,3,2)
b<-c(2,3,4,5,6,1,3,2,4,3,1,5,1,4,6,1,4)
n<-length(a)
M<-matrix(0,n,n)
M[cbind(rep(1:n,a),b)]<-1
(from Peter Dalgaard )
[top]
3.22 Eigenvalues example
(20/06/2001 ? <?>)
Tapio Nummi asked about double-checking results of spectral decomposition
In what follows, m is this matrix:
0.4015427 0.08903581 -0.2304132
0.08903581 1.60844812 0.9061157
-0.23041322 0.9061157 2.9692562
Brian Ripley posted:
> sm <- eigen(m, sym=TRUE)
> sm
$values
[1] 3.4311626 1.1970027 0.3510817
$vectors
[,1] [,2] [,3]
[1,] -0.05508142 -0.2204659 0.9738382
[2,] 0.44231784 -0.8797867 -0.1741557
[3,] 0.89516533 0.4211533 0.1459759
> V <- sm$vectors
> t(V) %*% V
[,1] [,2] [,3]
[1,] 1.000000e+00 -1.665335e-16 -5.551115e-17
[2,] -1.665335e-16 1.000000e+00 2.428613e-16
[3,] -5.551115e-17 2.428613e-16 1.000000e+00
> V %*% diag(sm$values) %*% t(V)
[,1] [,2] [,3]
[1,] 0.40154270 0.08903581 -0.2304132
[2,] 0.08903581 1.60844812 0.9061157
[3,] -0.23041320 0.90611570 2.9692562
[top]
4 Applying functions, tapply, etc
4.1 Return multiple values from a function
(30/01/2001 Paul Johnson <pauljohn@ku.edu>)
Below is a summary on returning variable from subfunction, Thanks to
Brian Ripley and Douglas Bates:
To access the variables within a function, return a list or data structure
assigned to a variable in the calling function. For more than one varaible,
assign them to a list variable and separate the components in the calling
function using listname$variablename.
"test" <-
function()
{
"hello" <-
function()
{
x <- 1:3
y <- 23:34
z <- cbind(x,y)
return(x,y,z)
}
c <- hello()
return(c$z,c$x, c$y)
}
Sam McClatchie
[top]
4.2 Grab "p" values out of a list of significance tests
(22/08/2000 Paul Johnson <pauljohn@ku.edu>)
Suppose chisq1M11 is a list of htest objects, and you want a vector of p values. Kjetil Kjernsmo observed this works:
> apply(cbind(1:1000), 1, function(i) chisq1M11[[i]]$p.value)
And Peter Dalgaard showed a more elegant approach:
sapply(chisq1M11,function(x)x$p.value)
In each of these, a simple R function called "function" is created and applied to each item
[top]
4.3 ifelse usage
(06/04/2001 Paul Johnson <pauljohn@ku.edu>)
If you want to calculate something, but only if y is *between* 0 and 1 then
ifelse(y==0,0,y*log(y))
(from Ben Bolker)
[top]
4.4 Apply to create matrix of probabilities, one for each "cell"
(14/08/2000 Paul Johnson <pauljohn@ku.edu>)
Suppose you want to calculate the gamma density for various values of "scale" and "shape". So you create vectors "scales" and "shapes", then create a grid if them with expand.grid, then write a function, then apply it (this example courtesy of Jim Robison-Cox) :
gridS <- expand.grid(scales, shapes)
survLilel <- function(ss) sum(dgamma(Survival,ss[1],ss[2]))
Likel <- apply(gridS,1,survLilel)
Actually, I would use
sc <- 8:12; sh <- 7:12
args <- expand.grid(scale=sc, shape=sh)
matrix(apply(args, 1, function(x) sum(dgamma(Survival, scale=x[1],
shape=x[2], log=T))), length(sc), dimnames=list(scale=sc, shape=sh))
(from Brian Ripley)
[top]
4.5 Outer.
(15/08/2000 Paul Johnson <pauljohn@ku.edu>)
outer(x,y,f) does just one call to f with arguments created by stacking x
and y together in the right way, so f has to be vectorised. (from Thomas Lumley)
[top]
4.6 Check if something is a formula/function
(11/08/2000 Paul Johnson <pauljohn@ku.edu>)
Formulae have class "formula", so inherits(obj, "formula") looks best.
(from Prof Brian D Ripley)
And if you want to ensure that it is ~X+Z+W rather than Y~X+Z+W you can
use
inherits(obj,"formula") && (length(obj)==2)
(from Thomas Lumley)
[top]
4.7 Optimize with a vector of variables
(11/08/2000 Paul Johnson <pauljohn@ku.edu>)
The function being optimized has to be a function of
a single argument. If alf and bet are both scalars you can combine
them into a vector and use
opt.func <- function(arg)
t(Y-(X[,1] * arg[1] + X[,2] * arg[2])^delta) %*% covariance.matrix.inverse %*%
(Y-(X[,1] * arg[1] + X[,2] * arg[2])^delta)
(from Douglas Bates)
[top]
4.8 slice.index, like in S+
(14/08/2000 Paul E Johnson <pauljohn@ku.edu>)
slice.index<-function(x,i){k<-dim(x)[i];sweep(x,i,1:k,function(x,y)y)}
(from Peter Dalgaard)
[top]
5 Graphing
5.1 Adjust features with par before graphing
(18/06/2001 Paul Johnson <pauljohn@ku.edu>)
par() is a multipurpose command to affect qualities of graphs. Examples include:
1. put the next plots into a single picture:
par(mfrow=c(3,2))
creates a 3x2 matrix of plots.
2. Adjust margins, as in
par(mfrow=c(2,1))
par(mar=c(10, 5, 4, 2))
x <- 1:10
plot(x)
box("figure", lty="dashed")
par(mar=c(5, 5, 10, 2))
plot(x)
box("figure", lty="dashed")
[top]
5.2 Save graph output
(09/29/2005 Paul Johnson <pauljohn@ku.edu>)
There are two ways to save and/or print graphs.
Before you create your graphs, you can decide on a format
for output. A graph on the screen is using the default device, either x11 on Unix or windows on MS Windows.
Saving output is a complicated topic, and let me start by just telling you some examples that work "pretty well."
If you want to save a graph on the screen as an encapsulated post script file for inclusion in a document, do this
dev.copy(postscript, file="myfile.eps", height=6, width=6, horizontal=F, onefile=F)
dev.off()
Height and width are in inches, and the onefile=F option is absolutely vital if you want the result to be truly useful in your output document. The onefile=F makes sure the EPS bounding box is set, so programs that use the figure can find the boundaries in it.
If you want to save the same as a picture image, say a png file with white background, do
dev.copy(png,filename="myfile.png",height=600, width=800,bg="white")
dev.off()
png format required height and width in pixels, not inches like postscript.
Note you can also set the pointsize for text. Peter Dalgaard told
me that on postscript devices, a point is 1/72nd of an inch. If you expect to shrink an image for final presentation, set a big pointsize, like 20 or so, in order
to avoid having really small type.
The dev.copy approach works well ALMOST ALL OF THE TIME. Sometimes, however, you get in trouble because the size on the computer screen does not translate into the size you specify in your output. So you can either re-do your screen output or revise your copy command.
The R experts recommend that instead of copying a result from screen into a device file, we should instead create the device and then run the plot commands. Then R knows it is supposed to create the graph for the indicated device and everything is supposed to be internally consistent.
R uses the term "device" to refer to the various graphical output formats. The command dev.off() is vital because it lets the device know you are finished drawing in there.
There are many devices representing kinds of file output.
You can choose among device types, postscript, png, jpeg,
bitmap, etc. (windows users can choose windows metafile).
Type
>?device
to see a list of devices R finds on your system.
For each device, you can find out more about it.
?x11
or
?png
or, in MS Windows
?windows
(and so forth)
You start a device with a configuration command, like
png(filename="myfile.png",width=800, height=600,bg="blue")
for the default png (a gif-like) format.
Suppose we want png output. We create an instance of a png
device and adjust features for that
device, as in:
> png(filename="mypicture.png", width=480, height=640, pointsize=12)
Note, if no filename is here, it will print to the default printer.
You can then run your graph, and when it is finished you have to
close the device:
> dev.off()Some print devices can accept multiple plots, and dev.off() is needed to tell them when to stop recording. On a single-plot device, if you try to run another graph before closing this one, you'll get an error.
The postscript device has options to output multiple graphs, one after the other. If you want to set options to the postscript device, but don't actually want to open it at the moment, use ps.options.
Here's an example of how to make a jpeg:
jpeg(filename="plot.jpg") plot(sin, 2*pi) dev.off()
After you create a device, check what you have on:
> dev.list()And, if you have more than one device active, you tell which one to use with dev.set(n), for the n device. Suppose we want the third device:
> dev.set(3)
For me, the problem with this approach is that I don't usually know what I want to print until I see it. If I am using the jpeg device, there is no screen output, no picture to look at. So I have to make the plot, see what I want, turn on an output device and run the plot all over in order to save a file. It seems complicated, anyway.
So what is the alternative?
If you already have a graph on the screen, and did not prepare a device, use dev.copy() or dev.print(). This may not work great if the size of your display does not match the target device.
dev.copy() opens the device you specify.
It is
dev.copy(device=png, file="foo", width=500, height=300) dev.off()
The dev.copy() function takes a device type as the first argument, and then it can take any arguments that would ordinarily be intended for that device, and then it can take other options as well. Be careful, some devices want measurements in inches, while others want them in pixels. Note that, if you do not specify a device, dev.copy() expects a which= option to be specified telling it which pre-existing device to use.
If you forget the last line, it won't work. It's like a write command.
If you use dev.copy or dev.print, you may run into the problem that your graph elements have to be resized and they don't fit together the way you expect. The default x11 size is 7in x 7in, but postscript size is 1/2 inch smaller than the usable papersize. That mismatch means that either you should set the size of the graphics window on the monitor display device to match the eventual output device, or you fiddle the dev.copy() or dev.print() command to make sure the sizes are correct. Recently I was puzzled over this and Peter Dalgaard said you can force the sizes to match, as in
dev.copy(pdf, file="foo", height=7, width=7)or
x11(height=6, width=6)
There are some specialized functions that can do this in a single step, as in dev.copy2eps() or dev2bitmap(). dev.copy2eps creates an eps file, and I'm not sure how it is different from the eps-compatible output created by the postscript() device. dev.copy2eps *is* an indirect way of calling dev.copy(device=postscript,...). It takes the dimensions from the current plot, and sets a couple of options. The same is true of dev.print().
The dev.print() function, as far as I can see, is basically the same as dev.copy(), except it has two appealing features. First, the graph is rescaled according to paper dimensions, and the fonts are rescaled accordingly. Due to the problem mentioned above, not everything gets rescaled perfectly, however, so take care. Second, it automatically turns off the device after it has printed/saved its result.
dev.print(device=postscript,file="yourFileName.ps")
If you just need to test your computer setup, Bill Simpson offered this. Here is a plot printing example
x<-1:10 y<-x plot(x,y) dev.print(width=5,height=5, horizontal=FALSE)
A default jpeg is 480x480, but you can change that:
jpeg(filename="plot.jpg" width = 460, height = 480, pointsize = 12, quality = 85)The format of png use is the same.
As of R1.1, the dev.print() and dev.copy2eps() will work when called from a function, for example:
> ps <- function(file="Rplot.eps", width=7, height=7, ...) {
dev.copy2eps(file=file, width=width, height=height, ...)
}
> data(cars)
> plot(cars)
> ps()
The mismatch of "size" between devices even comes up when you want to print out an plot. This command will print to a printer:
> dev.print(height=6, width=6, horizontal=FALSE)You might want to include pointsize=20 or whatever so the text is in proper proportion to the rest of your plot.
One user observed, "Unfortunately this will also make the hash marks too big and put a big gap between the axis labels and the axis title...", and in response Brian Ripley observed: "The problem is your use of dev.print here: the ticks change but not the text size. dev.copy does not use the new pointsize: try
x11(width=3, height=3, pointsize=8) x11(width=6, height=6, pointsize=16) dev.set(2) plot(1:10) dev.copy()Re-scaling works as expected for new plots but not re-played plots. Plotting directly on a bigger device is that answer: plots then scale exactly, except for perhaps default line widths and other things where rasterization effects come into play. In short, if you want postscript, use postscript() directly.
The rescaling of a windows() device works differently, and does rescale the fonts. (I don't have anything more to say on that now)
[top]postscript(file="test3.%d.ps",onefile=FALSE) gives files called test3.1.ps, test3.2.ps and so on. (from Thomas Lumley)
[top]options(papersize="letter")
whereas ps.options is only for the postscript device
[top]Advice seems to be to output R graphs in a scalable vector format, eps on Linux or eps or metafile on windows. If you are using Tex for document prep, on linux use "tetex" and on windows it has an implementation called http://www.tile.net/listserv/fptex.html and another called Miktex www.miktex.org I've installed Miktex and its pretty good.
Keith Reckdahl's Using EPS in LaTeX documents is freely available on the CTAN sites. (Brian Ripley)
For a description of EPSF and various other vector and bitmap formats, see the Encyclopedia of Graphics File Formats from O'Reilly at http://www.oreilly.com/catalog/gffcd/ (from Joel West)
I personally swear by the LaTeX Companion and the LaTeX Graphics companion, both published by Addison Wesley. (from James Marca)
Thanks to Prof. Ripley for the hint, maybe someone other is interested, so I will shortly outline what "to use pstricks with LaTeX" means:
* get PSfrag (e.g. www.dante.de)
* get graphics (same url)
* create your ps-graphic in R and define "tags" e.g.
postscript("myfile.ps")
plot(..., xlab="rho", ylab="wrho")
dev.off()
system("ps2epsi myfile.ps")
* add
\usepackage{graphics}
\usepackage{psfrag}
to your LaTeX-File
* and now we reach the climax:
\begin{figure}
\begin{center}
\psfrag{rho}{$\varrho$} <-- this replaces rho with \varrho
\psfrag{wrho}{$W(\varrho)$} <-- this replaces wrho with W(\varrho)
\includegraphics{myfile.epsi}
\caption{wonderful R plotting \label{fig}}
\end{center}
\end{figure}
* run latex and dvips and enjoy :-) (from Torsten Hothorn)
If you need to create a preview inside an eps file, you can do it with GSView (available free).
[top]Ordinarily, the graphics device throws away the old graph to make room for the new one.
Have a look to ?recordPlot to see how to keep snapshots. (Doo myGraph<-recordPlot() to save snap, then replayPlot(myGraph) to see it again);
On Windows you can choose "history"-"recording" in the menu and scroll through the graphs using the PageUp/Down keys. (from Uwe Ligges)
[top]You only need to specify the points to be plotted:
## upper triangle of the correct size: temp <- upper.tri(mmtop94.2, diag = TRUE) z <- mmtop94.2[temp] # values y <- nrow(mmtop94.2) + 1 - row(mmtop94.2)[temp] # row indices x <- col(mmtop94.2)[temp] # col indices scatterplot3d(x, y, z, type="h")(from Uwe Ligges) [top]
x<-seq(-3,3,.1) plot(x,dnorm(x,0,1), type="l")
(Bill Simpson)
[top]Here is how to do it.
x<-c(1,2,3,4,5) y<-c(1.1, 2.3, 3.0, 3.9, 5.1) ucl<-c(1.3, 2.4, 3.5, 4.1, 5.3) lcl<-c(.9, 1.8, 2.7, 3.8, 5.0) plot(x,y, ylim=range(c(lcl,ucl))) arrows(x,ucl,x,lcl,length=.05,angle=90,code=3) #or segments(x,ucl,x,lcl)(from Bill Simpson) [top]
If you want a density estimate and a histogram on the same scale, I suggest you try something like this:
> IQR <- diff(summary(data)[c(5,2)]) > dest <- density(data, width = 2*IQR) # or some smaller width, maybe, > hist(data, xlim = range(dest$x), xlab = "x", ylab = "density", + probability = TRUE) # <<<--- this is the vital argument > lines(dest, lty=2)(from Bill Venables) [top]
This is a question where terminology is important.
If you mean overlay as in overlay "slides" on a projector, it can be done. This literally super-imposes 2 graphs.
Use par("new"=TRUE) to stop the previous output from being erased, as in
tmp1<-plot(acquaint~T,type='l', ylim=c(0,1),ylab="average
proportion",xlab="PERIOD",lty=1,pch=1,main="")
par("new"=TRUE)
tmp2<-plot(harmony~T,type='l', ylim=c(0,1),ylab="average
proportion",xlab="PERIOD",lty=2,pch=1,main="")
par("new"=TRUE)
tmp3<-plot(identical~T,type='l', ylim=c(0,1),ylab="average
proportion",xlab="PERIOD",lty=3,pch=1,main="")
Note the par() is used to overlay "high" level plotting commands.
Here's an example for histograms:
> hist(rnorm(100, mean = 20, sd =12), xlim=range(0,100), ylim=range(0,50)) > par(new = TRUE) > hist(rnorm(100, mean = 88, sd = 2), xlim=range(0,100), ylim=range(0,50))The label at the bottom is all messed up. You can put the option xlab="" to get rid of them.
There are very few times when you actually want an overlay in that sense. Instead, you want to draw one graph, and then add points, lines, text, an so forth. That is what the R authors actually intend.
First you create the baseline plot, the one that sets the axes values. Then add points, lines, etc. Here are some self-contained examples to give you the idea.
Here's a regression example
x <- rnorm(100) e <- rnorm (100) y <- 12 + 4*x +13 *e mylm <- lm(y ~ x ) plot(x,y,main="My regression") abline(mylm)
From Roland Rau in r-help:
hist(rnorm(10000), freq=FALSE) xvals <- seq(-5,5,length=100) lines(x=xvals, y=dnorm(xvals))
Here's another example that works great for scatterplots
> x1<-1:10 x2<-2:12 > y1<-x1+rnorm(length(x1)) > y2<-.1*x2+rnorm(length(x2)) > plot(x1,y1,xlim=range(x1,x2), ylim=range(y1,y2), pch=1) > points(x2,y2,pch=2)(from Bill Simpson)
If you just want to plot several lines on a single plot, the easiest thing is to use "matplot" which is intended for that kind of thing. But plot can do it too.
Here's another idea: form a new data.frame and pass it through as y:
plot(cbind(ts1.ts,ts2.ts),xlab="time", ylab="", plot.type="single")or better something like:
plot(cbind(ts1.ts,ts2.ts),plot.type="single",
+ col=c("yellow","red"),lty=c("solid","dotted")) #colours and patterns
It can also be helpful to contrast 'c(ts1.ts,ts2.ts)' with
'cbind(ts1.ts,ts2.ts)'. (from Guido Masarotto)
For time series data, there is a special function for this:
ts.plot(ts1.ts, ts2.ts) # same as above
See help on plot.ts and ts.plot (from Brian D. Ripley)
Often, I find myself using the plot command to plot nothing, and then fill in beautiful stuff with text, lines, points, and so forth. If you plot 2 variables with type="n", then the inside of the graph is empty.
[top]The standard way to array several images across a page is the par(mfrow=) or par(mfcol=) option, as in:
par(mfrow=c(2,3))And then the next 6 plot/image commands will be laid out in a 2 row x 3 column arrangement.
The layout option can give a powerful set of controls for that kind of thing as well, e.g. (from Paul Murrell):
layout(matrix(c(1, 0, 2), ncol=1),
heights=c(1, lcm(5), 1))
plot(x)
box("figure", lty="dashed")
plot(x)
box("figure", lty="dashed")
I think you have a data with a like:(from Peter Malewski. Per Peter Dalgaard's advice, I've replaced usage of codes with as.integer. pj)xx <- data.frame(y=rnorm(100), x1=as.factor(round(runif(100,1,4))), x2=as.factor(round(runif(100,1,4))) ) attach(xx) by(y,list(x1,x2),plot) by(y,list(x1,x2),function(x)print(x)) xn <- as.integer(x1) ## either use coplot or par(mfrow=) approach: coplot(y~xn|x2) ##or you use a loop: par(mfrow=c(2,2)) for( i in unique(as.integer(x1))) plot.default(x2 [as.integer(x1)== i] , y[as.integer(x1)==i ] ,main=paste("Code:",i) ) ##of course there might be ways to use tapply, lapply etc.
And to label the whole page, use the "mtext" function
[top]David James was kind enough to help me out and enlighten me to the fact that the x-scales used by barplot are independent of those from my data. He also sent me a function which I include below (with a minor modification). This does exactly what I was looking for. Thanks!
xbarplot <- function(y, col=par("col"), border = par("fg"), gap=gap, ...)
{
ny <- length(y)
x <- seq(0,ny)+0.5
n <- length(x)
space <- gap * mean(diff(x))
old <- par(xaxt = "n")
on.exit(par(old))
plot(range(x, na.rm=T), range(y, na.rm=T), bty="n",
xlab="", ylab = "", type = "n", ...)
rect(x[-n]+space/2, rep(0,length(y)),
x[-1]-space/2, y, col = col, border = border)
}
(from Dirk Eddelbuettel)
[top]
suppose you have
reg <-lm( y ~ x) plot(x,y)
then do
abline(reg)
[top]Plots show dots:
plot(rnorm(100),rnorm(100),pch=".") (from Peter Malewski)
Plots with numbers indicating values of a third variable:
text(tx, ty, label = ch, col = "red", bg = "yellow", cex = 3)
If you specify pch, only the first character is taken as the symbol for your point.
Matt Wiener suggested creating a color vector and then using the value of another value to control both the lable and the color:
> col.vec <- c("black", "red", "blue")
> text(chem.predict[,1:2], labels = km$cluster, color =
col.vec[km$clusters])
[top]
Remember pch uses integers to select plotting characters.
x <- 1:10
y <- 1:10
res <- -4:5
plot(x, y, pch = ifelse(res < 0, 20, 1))
If true, use character 20. Else use character 1. Experiment with different numbers there.
Then note you can create a vector of integers to control symbols, as in
g <- 15:24plot (x,y, pch= g)(From post by Uwe Ligges (11/7/2002)
Here's another cool example from Roger Bivand (11/7/2002),
> x <- rnorm(25)
> y <- rnorm(25)
> z <- rnorm(25)
> pchs <- c(1, 20)
> pchs[(z < 0)+1][1] 20 20 20 20 1 1 20 20 1 20 20 20 20 20 1 20 1 ...
> plot(x, y, pch=pchs[(z < 0)+1])
> text(x, y, labels=round(z, 2), pos=1, offset=0.5)
Roger adds, "This "cuts" the z vector at zero, using the convenient slight-of-hand that TRUE and FALSE map to integers 1 and 0, and thus gives the pch argument in plot() or points() a vector of values of indices of the pchs vector. More generally, use cut() to break a numeric vector into class intervals (possibly within ordered())."
[top]test<-c(2,6,4,7,5,6,8,3,7,2)
plot(test,type="n",main="Color/size test plot",ylab="Size scale",xlab="Colorscale")
colsiz<-function (yvec) {
ymin <- min(yvec)
cexdiv <- max(yvec)/2 for (i in 1:length(yvec)) {
nextcex <- (yvec[i] - ymin)/cexdiv + 1
par(cex = nextcex)
points(i, yvec[i], type = "p", col = i)
}
}
colsiz(test)
Paul Murrell posted this nice example (11/7/2002):
x <- rnorm(50)[top]
y <- rnorm(50)
z <- rnorm(50)
pch <- rep(1, 50)
pch[z < 0] <- 20
cex <- (abs(z)/max(abs(z))) * 4.9 + 0.1
plot(x, y, pch=pch, cex=cex)
Have a look at:
?symbols
e.g.:
symbols(1:10,1:10,circles=1:10, inches=0.1)(from Uwe Ligges) [top]
Rado Bonk asked how to smooth a line connecting some points. One answer was, "You may be interested in spline(). For example:
x <- 1:5 y <- c(1,3,4, 2.5,2) plot(x, y) sp <- spline(x, y, n = 50) lines(sp)Roger Peng [top]
I don't know how to get the estimate of the line printed, but I have seen one example of how to get the R-square:
I used this to indicate the R-squared value on a regression plot:
text(5, 0.6, as.expression(substitute(R^2 == r, list(r=round(R2.the,3)))))
where R2.the was computed a few lines above. (from Emmanuel Paradis)
Or
that <- 1 plot(1:10) title(substitute(hat(theta) == that, list(that=that)))(from Brian Ripley) [top]
Ticks, but no numbers:
plot(1:10, xaxt="n") axis(1, labels=FALSE)
This leaves off the x and y titles. You still have the numerical labels next to each tick.
plot(x,y,ann=FALSE)
If you want a lot of control you can do
plot(x,y,ann=FALSE,axes=FALSE)
box()
axis(side=1,labels=TRUE)
axis(side=2,labels=TRUE)
mtext("Y", side=2, line=2)
mtext("X", side=1, line=2)
do ?axis (from Bill Simpson)
Control range of axes through the graph command itself. This shows the "image" function:
> x <- 1:50 > y <- 1:50 > z <- outer(x,y) > image(z) > image(z, xlim=c(0.2,0.5)) > image(z, xlim=c(0.2,0.5), ylim=c(0,0.5))The same works with contour(). (from Kaspar Pflugshaupt) [top]
At one time, apparently par(las=) would do this, as in
par(las=2) or par(las=3)to make the 90 degrees to axes labels.
More recently, Paul Murrell has observed:
"The drawing of xlab and ylab ignores the par(las) setting.
...You can override this "sensible" default behaviour if it annoys you. For example, you can stop plot() drawing xlab and ylab by something like plot(ann=F) or plot(xlab="", ylab="") AND you can draw them yourself using mtext(), which does listen to par(las).
A couple of examples of what I mean ...
par(mfrow=c(2,2), mar=c(5.1, 4.1, 4.1, 2.1), las=0)
plot(0:10, 0:10, type="n")
text(5, 5, "The default axes")
box("figure", lty="dashed")
par(las=1)
plot(0:10, 0:10, type="n")
text(5, 5, "The user says horizontal text please\n\nbut R knows
better
!")
box("figure", lty="dashed")
par(las=1, mar=c(5.1, 6.1, 4.1, 2.1))
plot(0:10, 0:10, type="n", ann=F)
mtext("0:10", side=2, line=3)
mtext("0:10", side=1, line=3)
text(5, 5, "The user fights back !")
(posted 6/11/2001)
[top]
Try package "date", available at CRAN.
Example for plotting:
library(date) plot.date(.....) ## if x-values are dates, you can also use plot(...) after library(date)(from Uwe Ligges) [top]
In R1.1.1, there is a bug, but in later versions Ross Ihaka says do the following:
x <- rnorm(100) y <- rnorm(100) plot(x, y, xlim=rev(range(x)), ylim=rev(range(y)))
and get the result you expect.
Ben Bolker said, at this time (R1.1.1), you have to do something like
x <- rnorm(100) y <- rnorm(100) plot(max(x)-x,max(y)-y,axes=FALSE) xlabvals <- pretty(x,5) ylabvals <- pretty(y,5) axis(side=1,at=max(x)-xlabvals,labels=xlabvals) axis(side=2,at=max(y)-ylabvals,labels=ylabvals)
I think it's time to put this one in the FAQ ... (although, again, I don't know if I would have found it just searching the archives if I hadn't remembered that it was there in the first place).
Another person said try (if you don't want axes):
plot(x, -y, axes=FALSE) axis(1) axis(2)[top]
I have to plot the data set like the following. > x <- c("05/27/00","06/03/00","08/22/00",...) > y <- c(10,20,15,...)You should try the package "date", available at CRAN. Example:
library(date)
x <- as.date(c("05/27/2000","06/03/2000","08/22/2000"))
y <- c(10,20,15)
plot(x,y)
Uwe Ligges
Here is one using chron.
library(chron)
x <- dates(c("05/27/00","06/03/00","08/22/00"))
y <- c(10,20,15)
plot(x, y)
which will work if you were in the USA and your first string means 2000
May
27, 2000-05-27 in ISO 8601 date format. (R 1.2.0 will have facilties to
read those as well as US and European shorthands.)
(from Brian D. Ripley)
[top]
plot(c(1:10), ylab= expression(x^"++")) (from Peter Dalgaard)
[top]Question: When I plot them default x axis takes place on the bottom of the chart and the y axis takes place on the left of the chart. Is it possible to make axes pass from the origin.
Maybe this helps:
plot(...., xaxt="n", yaxt="n") axis(1, pos=0) axis(2, pos=0)Uwe Ligges [top]
I think using arrows(..., code=3, angle=90, ...) is quite simple, e.g.:
x <- rnorm(10)
y <- rnorm(10)
se.x <- runif(10)
se.y <- runif(10)
plot(x, y, pch=22)
arrows(x, y-se.y, x, y+se.y, code=3, angle=90, length=0.1)
arrows(x-se.x, y, x+se.x, y, code=3, angle=90, length=0.1)
The first arrows() draws the error bars for y, and the second one for x,
'code=3' draws a head at both ends of the arrow, 'angle=' is the angle of
the head with the main axis of the arrow, and 'length=' is the length of
the head. You can also add usual graphic parameters (col, lwd, ...).
Emmanuel Paradis
[top]To do the two in the same graph:
plot(x) lines(y)or
points(y)or
matplot(cbind(x,y),type="l")To do separate graphs one above the other:
par(mfrow=c(2,1)) plot(x) plot(y)
You can do other things like set colors, ranges, axis labels, etc.
(from Jon Baron 12/29/2001)
[top]> library(ts) data(LakeHuron) md <- arima0(LakeHuron, order=c(2,0,0), xreg=1:98) plot(LakeHuron) lines(LakeHuron-md$resid,col="red")
(from Adrian Trapletti)
[top]I want this in a plot: "The average temperature is 23.1 degrees."
text(x,y,paste("The average temperature is ", variable, "degrees"))
(from Thomas Lumley)
[top]Graphs blow up if a variable is unbounded, so if you are plotting tan(x), for example:
You probably want to plot something like pmax(-10,pmin(10,tan(x))) or ifelse(abs(tan(x))>10,NA,tan(x))
(from Thomas Lumley)
[top]Don Wingate wrote "More generally, what I need to do is dynamically create a vector of math expressions (using the powers of string manipulation), which can then be used as the textual values for a plot function."
Uve Ligges answered: What about something like:
numbers <- 1:2 # the dynamical part
my.names <- NULL
for(i in numbers)
my.names <- c(my.names,
eval(as.expression(substitute(expression(x[i]^2), list(i=i)))))
dotplot(1:2, labels = my.names)
[top]
Use substitute(). Something like
e<-substitute(expression(paste("Power plot of ", x ^ alpha, " for ",
alpha == ch.a)), list(ch.a=formatC(alpha,wid=1)))
plot(x, x^alpha, main=e)
(from Peter Dalgaard)
Suppose you have
> that <- 1 > plot(1:10)
and you need to specify "that" in the title.
expression(paste(hat(theta),'= ',that)))
Generally, to get an expression either use parse on your pasted character string or substitute on an expression. The neatest is
title(substitute(hat(theta) == that, list(that=that)))
(note it is == not =)
(from Roger Kroenker, from Brian Ripley)
Want a variable value in the title, as in "Data at the 45o North:
lat<-45
plot(1,main=substitute("Data at "*lat*degree*" North",list(lat=lat)))
(from Thomas Lumley)
You can use different way: First export plot for R as an (e)ps file and than use LaTeX psfrag package to add text and formulae. (from Jan Krupa)
I want a Greek letter subscripted in the title, as in
[Greek] gamma [subscript]1 = [Roman] Threshold 1
Try:
plot(1,1,main=expression(gamma[1]=="Threshold 1"))(from Thomas Lumley)
Try help(plotmath)
[top]plot(x,y,color=ifelse(is.na(z),"red","black")) (from Peter Dalgaard)
plot(x[!is.na(z)],y[!is.na(z)],xlim=range(x),ylim=range(y)) points(x[is.na(z)],y[is.na(z)],col=2) (from Ross Darnell)
[top]The lattice library came along recently, I've not explored it much. But here is a usage example, emaphsizing the aspect setting from Paul Murrell:
library(lattice)
# 10 rows
x <- matrix(rnorm(100), ncol=10)
lp1 <- levelplot(x, colorkey=list(space="bottom"), aspect=10/10)
print(lp1)
# 5 rows -- each row taller than in previous plot
x <- matrix(rnorm(50), ncol=5)
lp2 <- levelplot(x, colorkey=list(space="bottom"),aspect=5/10)
print(lp2)
This one from Deepayan Sarkar (deepayan@stat.wisc.edu) is especially neat.
library(lattice)[top]
xyplot(rnorm(100) ~ rnorm(100) | gl(2, 50),
strip = function(factor.levels, ...) {
strip.default(factor.levels = expression(1 * degree, 2 * degree),
...)
})
Type
?persp
?contour
to see about features in the base package.
There is a scatterplot3d package on CRAN.
See also:
http://www.statistik.uni-dortmund.de/leute/ligges.htm (from Uwe Ligges)
See also:
Get Xgobi (an X windows graphics library) and the R contrib package "xgobi". The web page for Xgobi is http://lib.stat.cmu.edu/general/XGobi/ This does 3-D graphs, rotations, etc. Very nice. Mainly for Unix/X users.
[top]Suppose you want a contour plot solid contour lines for positive values and dotted contour lines for negative values.
The trick is to specify the levels and to use the add= argument.
x <- seq(-1, 1, length=21)
f <- function(x,y) (x^2 + y^2) / 2 - 0.5
z <- outer(x, x, f)
contour(z, levels = seq(-0.5, -0.1, by = 0.1), lty = "dotted")
contour(z, levels = 0, lty = "dashed", add = TRUE)
contour(z, levels = seq(0.1, 1, by = 0.1), add = TRUE)
(from Ross Ihaka)
[top]
It is quite easy to do with ImageMagick (www.imagemagick.org), which can be installed on most OSes. I tried this simple sequence and it worked beautifully.
In R, create 15 histograms:
> for(i in 1:5) {
+ jpeg(paste("fig", i, ".jpg", sep = ""))
+ hist(rnorm(100))
+ dev.off()
+ }
Then from the command line (I tried it using Linux, though it should be the same on any platform):
% convert -delay 50 -loop 50 fig*.jpg animated.gifThis created animated.gif, a nice animation of my sequence of files.You can control the timing of the animation by playing with -delay and -loop. (from Matthew R. Nelson, Ph.D. ) [top]
Use "par" to find out where the boundaries/axes are, and "rect" to draw the picture ... the 6th plot in demo(graphics) does the same thing.
For example:
## code for coloring backgroundx <- runif(10) y <- runif(10)
plot(x,y,type="n") u <- par("usr") rect(u[1],u[3],u[2],u[4],col="magenta") points(x,y)
(from Ben Bolker)
(pj: see next because it includes a usage of do.call)
I think one is stuck with things like
plot(2:10,type="n")
do.call("rect",c(as.list(par("usr")[c(1,3,2,4)]),list(col="pink")))
points(2:10)
(or for the middle line, somewhat simpler
bb <- par("usr")
rect(bb[1],bb[3],bb[2],bb[4], col="pink")
)
(from Peter Dalgaard )
[top]
1. example(rug). 'nuff said.
2. jitter
[top]Tito de Morais Luis emailed this example to me. It shows several useful and widely-used "tricks" for customizing plots
# truncated y-axis plot # from Wolfgang Viechtbauer R-help archive available at # http://tolstoy.newcastle.edu.au/~rking/R/help/03a/7415.html y <- c(140, 120, 110, 108, 104) plot(x,y, yaxt="n") axis(side=2, at=c(100, 105, 110, 115, 120, 140), labels=c(100, 105, 110, 115, 120, 1000)) rect(0, 130, 1, 131, col="white", border="white") par(xpd=T) lines(x=c(0.7,1), y=c(130, 130)) lines(x=c(0.7,1), y=c(131, 131))# My use is based on the above. # It allows the plot on the same graph of water level curves from 2 African man-made lakes level <- data.frame(Selengue=c(343.9, 347.5, 348.5, 348.7, 348.4, 347.8, 347.2, 346.3, 345.2, 343.9, 342.3, 341.5), Manatali=c(203.0, 207.0, 208.0, 208.5, 208.0, 207.7, 207.0, 206.0, 204.5, 203.0, 202.0, 201.0)) rownames(level) <- c("Aug", "Sep", "Oct", "Nov", "Dec", "Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul")
lev <- data.frame(Selengue=level[,1]-130,Manantali=level[,2]) plot(lev[,1],ylim=c(200,220),xlab="",ylab="", yaxt="n", xaxt="n", type='b', pch=19) par(new=T) plot(lev[,2],ylim=c(200,220),xlab="Month",ylab="Water level (m)", yaxt="n", xaxt="n", type='b', pch=21) axis(side=2, at=c(200, 205, 210, 215, 220), labels=c(200, 205, 210, 245, 350)) axis(side=1, at=c(1,2,3,4,5,6,7,8,9,10,11,12), labels=rownames(lev)) rect(-0.5, 210.8, 1, 211.8, col="white", border="white") par(xpd=T) lines(x=c(0.4,0.7), y=c(210.8, 210.8)) lines(x=c(0.4,0.7), y=c(211.8, 211.8))
----------------------------------
x <- c(1,2,3,4); y1 <- c(1,2,3,4); y2 <- c(2,2,2,2) Fred <- c(1,2)(from Anon.)postscript(file="d:/Bob/Papers/IFM/try2.ps") plot(x,y1, type="l") lines(x,y2,lty="33") legend(1,4, c("y1","y2"), lty=Fred) graphics.off()
-----------------------------------------
Plot sin and cos
layout.my <- function (m, n) {
par(mfrow = c(m, n))
}
x <- 0:12566 / 1000 # Range from 0 to 4*pi
layout.my( 1, 2 )
plot( sin(x), type = "l",
xaxs = "i", yaxs = "i", axes = F,
xlab = "x", ylab = "sin(x)",
main = "Y = sin(x), x = [ 0, 720 ]"
)
axis( 2, at = seq( -1, 1, by=1 ),las = 2 )
box(lty="dotted")
abline( h = 0, lwd = 1 )
plot( cos(x), type = "l",
xaxs = "i", yaxs = "i", axes = F,
xlab = "x", ylab = "cos(x)",
main = "Y = cos(x), x = [ 0, 720 ]"
)
axis( 2, at = seq( -1, 1, by=1 ),las = 2 )
box(lty="dotted")
abline( h = 0, lwd = 1 )
(from Ko-Kang Wang)
-------------------------------------------
Plot a regular polygon
Below is a function that generates regular polygons, filled or borders, of n sides (n>8 => circle), with "diameter" prescribed in mm, for use alone or with apply.(from Denis White)############################################################
ngon <- function (xydc, n=4, type=1) # draw or fill regular polygon # xydc a four element vector with # x and y of center, d diameter in mm, and c color # n number of sides of polygon, n>8 => circle # if n odd, vertex at (0,y), else midpoint of side # type=1 => interior filled, type=2 => edge # type=3 => both { u <- par("usr") p <- par("pin") d <- as.numeric(xydc[3]) inch <- d/25.4 rad <- inch*((u[2]-u[1])/p[1])/2 ys <- inch*((u[4]-u[3])/p[2])/2/rad if (n > 8) n <- d*4 + 1 th <- pi*2/n costh <- cos (th) sinth <- sin (th) x <- y <- rep (0,n+1) if (n %% 2) { x[1] <- 0 y[1] <- rad } else { x[1] <- -rad*sin(th/2) y[1] <- rad*cos(th/2) } for (i in 2:(n+1)) { xl <- x[i-1] yl <- y[i-1] x[i] <- xl*costh - yl*sinth y[i] <- xl*sinth + yl*costh } x <- x + as.numeric(xydc[1]) y <- y*ys + as.numeric(xydc[2]) if (type %% 2) polygon (x,y,col=xydc[4],border=0) if (type %/% 2) lines (x,y,col=xydc[4]) invisible() }
---------------------------------------------
plot(1:1000, rnorm(1000), axes=FALSE) # plot with no axes axis(1, at = seq(0, 1000, by = 100)) # custom x axis axis(2) # default y axis box() # box around the plot
(from Ross Ihaka)
-------------------------------------------------
Here is an example of drawing a cubic with axes in the middle of the plot.
# Draw the basic curve (limits were established by trial and error).
# The axes are turned off blank axis labels used.
curve(x * (2 * x - 3) * (2 * x + 3), from = -1.85, to = 1.85,
xlim = c(-2, 2), ylim = c(-10, 10),
axes = FALSE, xlab = "", ylab = "")
# Draw the axes at the specified positions (crossing at (0, 0)).
# The ticks positions are specified (those at (0,0) are omitted).
# Note the use of las=1 to produce horizontal tick labels on the
# vertical axis.
axis(1, pos = 0, at = c(-2, -1, 1, 2))
axis(2, pos = 0, at = c(-10, -5, 5, 10), las = 1)
# Use the mathematical annotation facility to label the plot.
title(main = expression(italic(y==x * (2 * x - 3) * (2 * x + 3))))
(from Ross Ihaka)
---------------------------------- Question: is it possible to shade the 3d surface like a contour plot? i.e. black for large z, white for small z, say
Answer:
# Create a simple surface f(x,y) = x^2 - y^2
nx <- 21
ny <- 21
x <- seq(-1, 1, length = nx)
y <- seq(-1, 1, length = ny)
z <- outer(x, y, function(x,y) x^2 - y^2)
# Average the values at the corner of each facet
# and scale to a value in [0, 1]. We will use this
# to select a gray for colouring the facet.
hgt <- 0.25 * (z[-nx,-ny] + z[-1,-ny] + z[-nx,-1] + z[-1,-1])
hgt <- (hgt - min(hgt))/ (max(hgt) - min(hgt))
# Plot the surface with the specified facet colours.
persp(x, y, z, col = gray(1 - hgt), theta = 35)
persp(x, y, z, col = cm.colors(10)[floor(9 * hgt + 1)], theta = 35)
(from Ross Ihaka)
--------------------------------------------------------
Polygon!
y <- c(1.84147098480790, 1.90929742682568, 1.14112000805987, 0.243197504692072, -0.958924274663133,0.720584501801074, 1.65698659871879, 1.98935824662338, 1.41211848524176,0.45597888911063)plot(y,ylim=c(0.0000000001,2),type="l",log="y")
x <- 1:10 temp <- which(y<0) polygon(x[-temp], y[-temp], col="red")
Uwe Ligges
---------------------------------------------------
A use of matplot side-by-side output (Peter Dalgaard, June 11, 2001)
[top]p <- 1:3 u <- matrix(c(1,1,1,2,2,2,3,3,3),3,) r <- p*u X11(height=3.5,width=7) par(mfcol=c(1,3),cex=1.5,mex=0.6, mar=c(5,4,4,1)+.1) matplot(p,r,type="b",main="A",col="black") matplot(log(p),log(r),type="b",main="B",col="black") r <- p+u matplot(p,r,type="b",main="C",col="black")
A crosstab facility xtabs() was introduced in R1.2. Read all about it in help. table is there too.
These give counts, not percentages. There are special functions for that. Here's code that creates a table and then calls "prop.table" to get percents on the columns:
> x<- c(1,3,1,3,1,3,1,3,4,4)
> y <- c(2,4,1,4,2,4,1,4,2,4)
> hmm <- table(x,y) > prop.table(hmm,2) * 100
apply(hmm, 2, sum)
which gives the counts of the columns.
If you have an old R, here is a "do it yourself" crosstab
>count <- scan()
65 130 67 54 76 48 100 111 62 34 141 130 47 116 105 100 191 104
>s <- c("Lo", "Med", "Hi")
>satisfaction <- factor( rep(c(s,s), rep(3,6)), levels = >c("Lo","Med","Hi"))
>contact <- factor( rep(c("Low","High"), rep(9,2)),levels = >c("Low","High"))
>r <- c("Tower", "Flat", "House")
>residence <-factor(rep(r,6))
>tapply(count,list(satisfaction,contact,residence),sum)
On 2005-01-05, Brian Ripley proposed one way to make a table that has the "value" being counted in the first column (for tables of counts that works nicely for non-factor variables)
>xx <- unique(x) >rbind(vals=xx, cnts=tabulate(match(x, xx)))
Check this out! User wants to count outcomes by categories and make a sincle table list. Thomas Lumley says (2005-01-04), "table() will count frequencies, and interaction() will create a variable with all combinations of a set of variables, so something like
>table(interaction(f1,f2))works." [top]
Index the vector by group:
t.test(x[sex == 1], x[sex == 2])
should do the trick. You could try:
summary(aov(sex~group))
which is equivalent but assumes equal variances. (from Mark Myatt)
The power of the test can be calculated:
power.t.test()
see help(power.t.test)
[top]Use shapiro.test(x) for testing whether x comes from a normal distribution. X must be a vector or variable.
[top]Question: I want summary information by subgroup.
Answer: This can be done with tapply, or the by() function, which is a "wrapper" (that means "simplifying enhancement") of tapply.
Use like so. If you have groups defined by a factor A, to get summaries of variables x through z in dataframe "fred",
>attach(fred) >by(fred[,x:z], A, summary)
You can make a list of factors for a multidimensional set of groups, and many different functions can be put in place of "summary". See the by documentation.
Before by, we did it like this, and you may need it for some reason.
Get the average of a variable "rt" for each subgroup created by combinations of values for 3 factors.
tapply(rt,list(zf, xf, yf), mean, na.rm=T)
alternatively,
ave(rt, zf, xf, yf, FUN=function(x)mean(x,na.rm=T))
(from Peter Dalgaard)
[top]1. Rather than
lm(dataframe$varY~dataframe$varX)
, do
lm(varY~varX,data=dataframe)
2. To fit a model with a "lagged" variable, note thereis no built in lag function for ordinary vectors. lag in the ts packagedoes a different thing. So consider:
mylag<-function(x,d=1) {
n<-length(x)
c(rep(NA,d),x)[1:n]
}
lm(y~mylag(x))
[top]
Brian Ripley: By convention summary methods create a new object of class print.summary.foo which is then auto-printed.
Basically, you find out what's in a result, then grab it. Suppose a fitted model is "fm3DNase1":
> names(summary(fm3DNase1))(from Rashid Nassar)
[1] "formula" "residuals" "sigma" "df" "cov.unscaled"
[6] "correlation" "parameters"> summary(fm3DNase1)$sigma[1] 0.01919449
> summary(fm3DNase1)$parameters
Estimate Std. Error t value Pr(>|t|)
Asym 2.345179 0.07815378 30.00724 2.164935e-13
xmid 1.483089 0.08135307 18.23027 1.218523e-10
scal 1.041454 0.03227078 32.27237 8.504308e-14# for the standard errors:
> summary(fm3DNase1)$parameters[,2]
Asym xmid scal
0.07815378 0.08135307 0.03227078
The t-values are
summary(fm3DNase1)$coefficients[, "t value"]
and the standard errors are
summary(fm3DNase1)$coefficients[, "Std. Error"]
You can also use
vcov(fm3DNase1)
to get the estimated covaraince matrix of the estimates.
[top]grp is a factor. THen:
R> lm(y ~ grp:x, data=foo)
Call: lm(formula = y ~ grp:x, data = foo)
Coefficients: (Intercept) grpa.x grpb.x grpc.x -0.80977 0.01847 0.13124 0.14259 Martyn
[top]summary(lm.sml <- lm(y ~ x1)) summary(lm.big <- lm(y ~ x1 + x2 + x3 + x4)) anova(lm.sml, lm.big)now will do a test of b2 = b3 = b4 = 0
Look at
demo(lm.glm)
where models "l1" and "l0" are compared that way. (from Martin Maechler)
Here's another one along the same lines. Bill Venables (Feb.2,00) said the philosophy is "You fit a larger model that contains the given model as a special case and test one within the other." "Suppose you wonder if a variable x Suppose you have only one predictor with repeated values, say x, and you are testing a simple linear regression model. Then you can do the test using
inner.mod <- lm(y ~ x, dat) outer.mod <- lm(y ~ factor(x), dat) anova(inner.mod, outer.mod)Test for lack of fit done. If you have several predictors defining the repeated combinations all you need do is paste them together, for example, and make a factor from that. (from Bill Venables)
Suppose your data frame x has some column, say, ID, which identifies the various cases, and you fitted
fit1 <- lm(y ~ rhs, data=df)Now do
fit2 <- lm(y ~ factor(ID), data=df) anova(fit1, fit2, test="F")e.g.
set.seed(123) df <- data.frame(x = rnorm(10), ID=1:10)[rep(1:10, 1+rpois(10, 3)), ] df$y <- 3*df$x+rnorm(nrow(df)) fit1 <- lm(y ~ x, data=df) fit2 <- lm(y ~ factor(ID), data=df) anova(fit1, fit2, test="F")there is an R package lmtest on CRAN, which is full of tests for linear models. (from Brian Ripley)
The models with factor() as the indep variable are the most general because they individually fit each category, whereas using the variable X assumes linearity (from me!)
Here's another example. "I want to know how much variance of pre/post-changes in relationship satisfaction can be explained by changes in iv1 and iv2 (dv ~ iv1 + iv2), by changes in iv3 and iv4 (dv ~ iv3 + iv4) and I want to know wether a model including all of the iv's explains a significant amount of variance over these two models (dv ~ iv1 + iv2 + iv3 + iv4)."
model.0 <- lm(dv ~ 1) model.A <- lm(dv ~ iv1 + iv2) model.B <- lm(dv ~ iv3 + iv4) model.AB <- lm(dv ~ iv1 + iv2 + iv3 + iv4)(from Ragnar Beer) [top]anova(model.AB, model.A, model.0) anova(model.AB, model.B, model.0)
Suppose you fit a model "lm1". That model is supposed to include methods fitted() and predict().
If you want the predicted value for each nonmissing case, do
> predict(lm1)
predict has a lot more power because it can tell you the predicted values for a certain set of inputs into the model. Give it a dataframe as the second argument. That new data frame must have (at least) all of the variables in the original model. For example,
x<-c(1,3,2,1,4,1,4,NA) y<-c(4,3,4,1,4,1,4,2) mod1 <-m(y~x) testdataframe<-data.frame(x=c(1))Or:
predict(mod1,newdata=testdataframe)
> lot <- c(30,20,60,80,40,50,60,30,70,60)
> hours <-c(73,50,128,170,87,108,135,69,148,132)
> z1 <- lm(hours~lot)
> newdf <- data.frame(lot=80)
> predict(z1,newdata=newdf,interval="confidence",level=0.90)
fit lwr upr[1,] 170 166.9245 173.0755Please note, there is a difference between "confidence" and "prediction" in this command. One is the confidence interval of the point estimate, the other is the confidence interval for a prediction about an individual case.
Read carefully the documentation on your model. If you use glm to fit a model, you can read the docs for it with
?predict.glm. In particular, the predict method for glm can calculate various types of predictions, the value of the "linear predictor" X(b-hat) or it can give the transformed value, such as the probability in a logistic regression. The option there is type="link" for the X(b-hat) or type="response" for the probability that corresponds to X(b-hat). Know what I mean, either X(b-hat) or 1/(1-exp(-X(b-hat))).
Lately, I've been teaching regression and it is really really hard for newcomers to get predicted values for "interesting" cases. We've gone through all kinds of tedious routines to calculate the "new data frame" so that the output from predict is interesting and useful for graphs.
Suppose your old data frame is "olddf" and you want to create a new data frame for a problem with predictors X1,X2,X3,X4. One idea is to set X2, X3, X4 at their means and then allow X1 to vary from 1 to 10. This works:
mymod1 <- glm(Y~X1+X2+X3+X4, data=olddf, family=binomial(link=logit)) newdf <- data.frame(X1=1:10,X2=mean(olddf$X2),X3=mean(olddf$X3),X4=mean(olddf$X4)) p1 <- predict(mymodl1,newdata=data.frame,type="response") plot(1:10,p1, main="Logistic demonstration graph with X2,X3,X4 at mean values")
That provides a plot that people like.
It is impossibly tedious to keep re-generating data frames to explore the variables. There are some shortcuts. In R's base, there is a function called expand.grid (more on that in a moment). In the Design library from Prof. Harrell, the "datadist" procedure is used to create information about the variables that is used in automatically creating that kind of plot. However, sometimes, the plots are not exactly what one might like. But usually they are pretty good, at least as a starting point. To make a plot demonstrating X1, it will set other variables to their means if they are numeric and to their modes if they are not numeric.
In the end, however, a person may want to calculate predicted values for a variety of different outputs. Here is where the function "expand.grid" comes in very handy. If one specifies a list of values to be considered for each variable, then expand.grid will create a "mix and match" data frame to represent all possibilities.
> x <- c( 1, 2, 3, 4 ) > y <- c( 22.1, 33.2, 44.4 ) > expand.grid(x,y) Var1 Var2 1 1 22.1 2 2 22.1 3 3 22.1 4 4 22.1 5 1 33.2 6 2 33.2 7 3 33.2 8 4 33.2 9 1 44.4 10 2 44.4 11 3 44.4 12 4 44.4So, if you create the new data frame with
newdf <- expand.grid(X1,X2,X3,X4), then put that as newdata into predict and one easily obtains all of the desired information.
Note many models have a "fitted" attribute and there is the accessor function fitted. If you use fitted() on a model, it may not return what you expect. For lm() it gives the same as predict, but not for glm(). The default for predict() gives the "linear predictor", while fitted gives the value of "mu" (inverse link applied to linear predictor). fitted(mod) equals predict(mod,ndf,type="response").
x <- rnorm(100) y <- rnorm(100) mylm<-glm((y>0)~x,family=binomial) > attributes(mylm) $names [1] "coefficients" "residuals" "fitted.values" [4] "effects" "R" "rank" [7] "qr" "family" "linear.predictors" [10] "deviance" "aic" "null.deviance" [13] "iter" "weights" "prior.weights" [16] "df.residual" "df.null" "y" [19] "converged" "boundary" "model" [22] "call" "formula" "terms" [25] "data" "offset" "control" [28] "method" "contrasts" "xlevels"> predict(mylm) 1 2 3 4 5 6 7 8 9 10
0.7341672 0.3246603 0.2955893 0.4058782 0.5045060 0.5282107 0.3429243 0.1159356 0.3394576 0.4889709
> fitted(mylm)
1 2 3 4 5 6 7 8 9 10
0.6757191 0.5804596 0.5733639 0.6000991 0.6235177 0.6290657 0.5849007 0.5289515 0.5840588 0.6198640
> predict(mylm,type="response") 1 2 3 4 5 6 7 8 9 10 0.6757191 0.5804596 0.5733639 0.6000991 0.6235177 0.6290657 0.5849007 0.5289515 0.5840588 0.6198640
To make a table showing how factors affect the fitted or predicted values, Brian Ripley says (2003-11-29) ":
allergy$fitted <- fitted(allergy.fit.main.2int)and you permute the dimensions by reordering the factors.... xtabs generates a call to table() which could be used directly.
xtabs(fitted ~ t + f + c, data=allergy)
In this case there were no repeated (and no missing) combinations of factors: if there has been just predict the model at all combinations and apply xtabs/table to the predictions."
[top]Use the I() function to protect the ^2 and ^3 from being evaluated as part of the formula, ie.
formula = Response ~ Var1 + I(Var1^2) + I(Var1^3)
(from Douglas Bates)
I was doing this in a practical class yesterday. There is another way,
Response ~ poly(Var1, 3)
which fits the same cubic model by using orthogonal polynomials. The fitted values will be the same, but the coefficients are those of the orthogonal polynomials, not the terms in the polynomial. You might like to compare the two approaches. (from Brian D. Ripley)
[top]a$fstatistic returns the actual f-test and df, but not the p-value. Get a linear model, "grab" the f part of the output, use the f distribution (pf).
foo <- lm(y~x) a <- summary(foo) f.stat <- a$fstatistic p.value <- 1-pf(f.stat["value"],f.stat["numdf"],f.stat["dendf"])(from Guido Masarotto) [top]
Begin with:
data(stackloss) fm <- lm(stack.loss ~ ., data=stackloss) anova(fm)
Add more models:
fm2 <- update(fm1, . ~ . - Water.Temp) fm3 <- update(fm2, . ~ . - Air.Flow) anova(fm2, fm1) anova(fm3, fm2)Note that the SSqs are all the same, but the sequential table compares them to the residual MSq, not to the next-larger model (from Brian D. Ripley) [top]
Mark M. Span asked this: I have a four-parameter, biexponential model. What I want to evaluate is whether the parameter values differ between (groups of) tasks.
myfunc <-formula(x ~ a*exp(-b*age) + (c*exp(d*age)) )both x (representing response time) and age are variables in the current scope. At the moment I fit the model for each task seperately. But now I cannot test the parameters for equality.
Answer: Use a model parametrizing a,b,c by groups and one without, and use anova to compare the models. A worked example is in V&R3, page 249. (from Brian Ripley)
[top]Use the nls library
library(nls)
... get data
model.fit <- nls(y ~ exp(x))
plot(model.fit)
[top]
(from Halvorsen) I have the following simple function:
testcarfunction(pow){ ob <-glm(Pound~CG+Age+Vage,
data=car,weights=No,subset=No>0,family=quasi(link=power(pow),
var=mu^2)) deviance(ob) }But trying to run it gives:> > testcar(1/2)> > Error in power(pow) : Object "pow" not found
A more reliable version (replacement) is
eval(substitute(glm(Pound~CG+Age+Vage,data=car,(from Thomas Lumley) [top]
weights=No subset=No>0, family=quasi(link=power(pow),var=mu^2)),
list(pow=pow)))
var(X)
or, if you want the covariance matrix of a fitted model, don't forget you can take the results from summary() as described above.:
Also note package MASS has a generic function vcov to pull out variance-covariance estimates of coefficient estimates, so you can do
> vcov(fm3DNase1)
Asym xmid scal
Asym 0.006108 0.0062740 0.0022720
xmid 0.006274 0.0066183 0.0023794
scal 0.002272 0.0023794 0.0010414
too. As it's generic, it works for most fitted model objects.
[top]
In R, I can use
library(ctest) binom.test(15, 20)to get a confidence interval, but under the assumption of an infinite population. (from RINNER Heinrich)
To get an estimate corrected for a finite population, the stubs must be in phyper:
uniroot(function(x)phyper(15,x,100-x,20)-0.025,,0,100)$root uniroot(function(x)phyper(14,x,100-x,20)-0.975,,0,100)$root(from Peter Dalgaard) [top]
More generally, the vcov function in the MASS package extracts If you need variance-covariance matrices, try the MASS library's vcov function. Or, to just save the diagnoal "standard errors" of coefficients, do
library(MASS) stdErrVector<-sqrt(diag(vcov(lm.D9)))
Or, the hard way,
> example(lm) > summary(lm.D9) > names(.Last.value) > summary(lm.D9)$coeff > summary(lm.D9)$coeff[, "Std. Error"][top]
> summary(c(123456,1,2,3), digits=7) Min. 1st Qu. Median Mean 3rd Qu. Max. 1.00 1.75 2.50 30865.50 30866.25 123456.00
and you get what you want. In most cases the number of "significant digits" printed out by summary methods in R is truncated to 3 or 4. (from Robert Gentleman)
[top]Question: In a split-plot design if there is a missing subunit summary gives me a table with two rows for the same factor, one in the error within section and one in the section using error between units. With no data missing the table is "normal". How does one interpret the table when data is missing?, or is it that aov cannot cope with missing values in this case?
Answer (from Brian Ripley): aov() does cope, but the conventional analysis is indeed as you describe. Factors do have effects in more than one stratum. For example, consider the oats example in Venables & Ripley (1999). If I omit observation 70, I get
> summary(oats.aov)
Error: B
Df Sum of Sq Mean Sq F Value Pr(F)
Nf 1 841.89 841.890 0.2242563 0.6605042
Residuals 4 15016.57 3754.142
Error: V %in% B
Df Sum of Sq Mean Sq F Value Pr(F)
Nf 1 1100.458 1100.458 1.747539 0.2187982
V 2 1156.821 578.410 0.918521 0.4335036
Residuals 9 5667.471 629.719
Error: Within
Df Sum of Sq Mean Sq F Value Pr(F)
Nf 3 19921.29 6640.431 37.23252 0.0000000
Nf:V 6 408.96 68.160 0.38217 0.8864622
Residuals 44 7847.41 178.350
so Nf appears in all three strata, not just the last one. The `recovery of intra-block information is needed'.
If you have an unbalanced layout (e.g. with a missing value) use lme to fit a model. V&R do this example in lme too.
[top]> aov(cbind(y1,y2,y3)~x)
[top]Bartlett's test, I like (personally). SPSS now uses Levine's test, some say incorrectly and nonrobustly. R's offers the Fligner-Killeen (median) test, fligner.test() in ctest.
The test "does an anova on a modified response variable that is the absolute value of the difference between an observation and the median of its group (more robust than Levene's original choice, the mean)". (from Brian Ripley)
Incidentally, if you insist on having Levine's calculation, Brian Ripley showed how: "
>Levene <- function(y, group)
{
group <- as.factor(group) # precautionary
meds <- tapply(y, group, median)
resp <- abs(y - meds[group])
anova(lm(resp ~ group))[1, 4:5]
}
> data(warpbreaks)
> attach(warpbreaks)
> Levene(breaks, tension)
F value Pr(>F)
group 2.818 0.06905
I could (and probably would) dress it up with a formula interface,
but that would obscure the simplicity of the calculation."
[top]
x <- runif(100) w <- runif(100) y <- x^2.4*w^3.2+rnorm(100,0,0.01) plot(x,y) plot(w,y) library(nls) fit <- nls(y~x^a*w^b,start=list(a=2,b=3))l.est <- lm(log(y) ~ log(x)+log(w)-1) fit <- nls(y~x^a*w^b,start=list(a=coef(l.est)[1],b=coef(l.est)[2]))
Error is a fairly common result of starting with parameter values that are far away from the true values, or have very different scales, so that the numeric-derivative part of nls fails. Various solutions are to do an approximate least-squares regression to start things off (also the "self-starting" models in the nls package), or to provide an analytic derivative. (all from Ben Bolker)
[top]If you use the nls function in R to fit the model in the parameterization that Guido described, you can graphically examine the profile likelihood with
pr <- profile(nlsObject) plot(pr)For an example, try
library(nls) example(plot.profile.nls)(from Douglas Bates) [top]
This is about the polr package, but the general point is that to compare likelihood models, you have to use your head and specify the comparison. Brian Ripley said:
As in the example on the help page
options(contrasts=c("contr.treatment", "contr.poly"))
data(housing)
house.plr <- polr(Sat ~ Infl + Type + Cont, weights = Freq, data = housing)
house.plr0 <- polr(Sat ~ 1, weights = Freq, data = housing)
Note that -2 LOG L is not actually well-defined as it depends on the
grouping of the data assumed, and so is not a statistic. I assume
that you want differences of that quantity, as in
house.plr0$deviance - house.plr$deviance 169.7283[top]
Hiroto Miyoshi described an experiment with 2 groups and pre/post measurement of a dichotomous dependent variable. Frank Harrell had this excellent response:
"There are several ways to go. GEE is one, random effects models another. One other approach is to install the Hmisc and Design packages (http://hesweb1.med.virginia.edu/biostat/s) and do (assume id is the unique subject identifier):
f <- lrm(y ~ x1 + x2*x3 + ..., x=T, y=T) # working independence model g <- robcov(f, id) # cluster sandwich variance adjustment h <- bootcov(f, id, B=100) # cluster bootstrap adjustment summary(g) # etc.[top]
Compute the Odds Ratio and its confidence intervall, from a logistic model (glm(.......,family=binomial....).
I show a simple function to do this in my introductory notes available from: http://www.myatt.demon.co.uk
Basically it is:
lreg.or <- function(model)
{
lreg.coeffs <- coef(summary(salex.lreg))
lci <- exp(lreg.coeffs[ ,1] - 1.96 * lreg.coeffs[ ,2])
or <- exp(lreg.coeffs[ ,1])
uci <- exp(lreg.coeffs[ ,1] + 1.96 * lreg.coeffs[ ,2])
lreg.or <- cbind(lci, or, uci)
lreg.or
}
(from Mark Myatt)
[top]
Since the lme/nlme support in R is cutting-edge and Doug Bates is an author and frequent R contributor, this is one of the packages that is worth looking at. When I see useful tidbits, I'm putting them here:
Test whether one should use a model with mixed effects or just a plain old linear model:
Bates wrote (2005-01-05) "I would recommend the likelihood ratio test against a linear model fit by lm. The p-value returned from this test will be conservative because you are testing on the boundary of the parameter space."
[top]You have to load the ts library! try
library(ts) example(acf)
There are other packages with time series components. Try
help.search("time")
to see whatever you have installed. Under my packages section of this document, I keep a running tally of packages that have time series stuff in them
Many time series procedures want to operate on a "time series" object, not raw numbers or a data frame. You can create such an object for "somedataframe" by:
myTS <- ts(as.matrix(somedataframe), start=c(1992,1), frequency=12)
Note start and frequence designate time attributes.
[top]This uses the ts package's filter(): A simple exponential weighted moving average procedure would be
ewma <- function (x, lambda = 1, init = 0)
{
y <- filter(lambda*x, filter=1-lambda, method="recursive", init=init)
return (y)
}
Using ewma as a forecast function is known as exponential smoothing.
Try, e.g.:
x <- ts(diffinv(rnorm(100))) y <- ewma(x,lambda=0.2,init=x[1]) plot(x) lines(y,col="red")(from Adrian Tripletti) [top]
I need a better list here, but:
1. glm in R is not PROC GLM in SAS. In R the "G" is for "generalized", cf. GLIM or GENMOD in SAS. The SAS GLM (General Linear Model) functionality comes via lm() and friends in R. (from Peter Dalgaard)
[top]x1 <- seq(from=-1,to=2, length=20) x2 <- (rep(1:5,4))/4 x <- cbind(1,x1,x2) y <- c(0,1,1,1,1,1,0,0,1,1,1,0,1,0,1,1,1,1,1,1) glm.fit(x,y,fam=binomial(link=logit))$coeff(from Vipul Devas) Leave off the $coeff to see all output. [top]
For a dichotomous dependent variable, use the glm function in R:
aLogitModel <- glm(y~x, family=binomial(link=logit))
or
aProbitModel <- glm(y~x, family=binomial(link=probit))
See Jim Lindsey's package "ordinal": http://alpha.luc.ac.be/~plindsey/publications.html
And also Proportional Odds Logistic Regression in the MASS package
The VGAM package has a variety of these models: http://www.stat.auckland.ac.nz/~yee/VGAM
[top]library(modreg)
Users report confusion about the difference between lowess() in S and loess in R. Martin Maechler advised us, "loess() by default is not resistant/robust where as lowess() is. ... I would recommend using loess(....., family = "sym")"
[top]R-base has "lowess"
The package "modreg" has "loess". The latter is a linear modeling-alike tool,and it can be used with other R-base tools for dealing with linear models, like the residuals function:
> l <- loess(y1~x1) > plot(l) > r <- residuals(l)r is now a variable holding residuals. [top]
User asked to estimate what we call a "contextual regression," a model of individuals, many of whom are in the same subsets (organizations, groups, etc). This is a natural case for a mixed regression model
José C. Pinheiro and Douglas M. Bates, Mixed Effect Models in S and S-Plus (Springer,2000).
The nlme package for R has a procedure lme that estimates these models.
In MASS, one can find the glmmPQL procedure, which can fit generalized linear models with random parameters. It repeatedly calls the lme function from the nlme package to estimate the models.
See also Jim Lindsey's repeated measurements package (available at http://alpha.luc.ac.be/~lucp0753/rcode.html).
[top]Some timeseries tools are distributed with R itself, in the ts library. type
library(ts)?ts?arima0
DSE:Paul Gilbert said "My dse package handles multiple endogenousvariables (y's) and multiple input or conditioning variables (which are often calledexogenous variables, but sometimes are not truly exogenous). Integrating models arehandled but you might need to consider what estimation technique to use. I have notimplemented any integration tests (volunteers?). DSE also handles state space models.It does not implement time varying models or non-linear models, although it wouldprovide a valuable framework for anyone who would like to do that." Look on CRAN!Or check at Paul's site, http://www.bank-banque-canada.ca/pgilbert.
For unevenly spaced time series data, take a look atLindsey's growth and repeated libraries: www.luc.ac.be/~jlindsey/rcode.html (from Jim Lindsey)
In the "nlme" package, there is a continuous AR(1) class, corCAR1, whichshould be used in the case when the times are unequally spaced andmeasured on a continuous scale. (from Jose Pinheiro). The usage of both lme and nlme is described inJose C. Pinheiro and Douglas M. Bates, Mixed-Effects Models in S and S-PLUS,Springer,2000.
[top]The car and lmtest packages both have functions for Durbin-Watson tests. (John Fox)
[top]You can do a censored regression with the survival5 library
library(survival5) survreg(Surv(y,c), dist="normal")where y is the censored dep-var and c is 1 if observed and 0 otherwise. (from Dan Powers) [top]
-you can write your models using your favourite ode solver library (limited C or Fortran) and then dyn.load() the model, so you can use it inside R. It is not too difficult, and if you are used to a certain library where you have your models, is probably the easiest way.
-there is a preliminary porting to R of the nls2 library, which is a library that makes nonlinear regression from differential equation models. It is a bit difficult to install, and is also not easygoing, but the nonlinear regression works well. In that library there is a way to actually compile a differential equation model and integrate it using the lsode routine from ODEPACK. So if you want to see how to plug a ode solver with R in a more general way you can start having a look at it in http://www-bia.inra.fr/J/AB/nls2/welcome.html (from Jesus Maria Frias Celayeta)
[top]Consider
http://rpgsql.sourceforge.net/
Brian Ripley said "RODBC works well with MySQL, in my experience better than RMySQL."
There is a package RMySQL.
and other packages in CRAN's contrib/Devel.
[top]The VR bundle of packages has linear and quadratic DA (lda, qda) in the MASS package, neural nets in nnet, and several other methods including k nearest neighbour in the class package. You find it in the contrib section of CRAN.
Classification trees can be fitted using both the tree and rpart packages. (Friedrich Leisch)
[top]loglin() in the base package and loglm() in package MASS (from Thomas Lumley)
[top]The gnlr and gnlr3 functions in my gnlm library at www.luc.ac.be/~jlindsey/rcode.html will do this for a wide selection of distributions.... The gnlr function in my gnlm library will fit quite a variety of different overdispersed Poisson- and binomial-based distributions (i.e. phi different from one) using their exact likelihoods. (from Jim Lindsey)
Maximize the likelihood directly. There are worked examples in the Venables & Ripley on-line exercises and answers. Now for the gamma there are some shortcuts, as the MLE of the mean is the sample mean, and gamma.shape.glm in package MASS will fit the shape parameter by MLE (from Brian D. Ripley)
The log-likelihood function for the parameters alpha and beta given the data x is sum(dgamma(x, shape = alpha, scale = beta, log = TRUE)) To determine the maximum likelihood estimates, simply put the negative of the log-likelihood into the optim function.
A worked-out example is given as xmp06.12 in the Devore5 library (available in the contrib section at CRAN). Use
library(Devore5) help(xmp06.12) # shows the description with comments example(xmp06.12) # run the example to see the result(from Douglas Bates) [top]
CRAN should have bootstrap and boot, two packages worth looking at. Most people say that "boot" is better, more actively maintained.
[top]Robust Regression is available in S, its name is rreg.
There's better robust regression in the VR bundle of packages.
>library(MASS) >help(rlm) >library(lqs) >help(lqs)(from Thomas Lumley
You could try function rlm in package MASS (in the VR bundle). That has an MM option, and was originally written to work around problems with the (then) version of lmRobMM in S-PLUS. It uses lqs in package lqs, and that is substantailly faster than the ROBETH-based code (as I started with that route). (from Brian D. Ripley)
[top]hmctest() in package lmtest performs the Harrison-McCabe-Test against heteroskedasticity. (from Torsten Hothom)
[top]The contrib package tseries (by Adrian Trapletti) contains Garch tools and also:
NelPlo Nelson-Plosser Macroeconomic Time Series garch Fit GARCH Models to Time Series get.hist.quote Download Historical Finance Data jarque.bera.test Jarque-Bera Test na.remove NA Handling Routines for Time Series
It can also do White's test, not the general heteroskedasticity test, but a more specific White statistic which tests for neglected nonlinearity in the conditional mean, essentially for a regression equation or linear time series model.
Adrian Trapletti, author of tseries package, also added "One very simple way to test conditional heteroskedasticity (time series setup) is to use the Box.test (package ts) on the squared values of a time series as
Box.test(x^2, lag = 1, type="Ljung")
[top]http://cm.bell-labs.com/cm/ms/departments/sia/project/locfit/index.html
[top]The int function in my rmutil library will handle this. www.luc.ac.be/~jlindsey/rcode.html (from Jim Lindsey)
[top]boxcox() in MASS
[top]Try the GIS package GRASS:
http://www.baylor.edu/~grass/statsgrasslist.html
from which the package described in Computers & Geosciences (2000), 26, pp. 1043-1052, may be accessed. There is a mailing-list too, and more progress is described in work with Markus Neteler, who is coordinating GRASS 5.0 development. The current package is GRASS_0.1-4, and is at:
http://reclus.nhh.no/gc00/GRASS/GRASS_0.1-4.tar.gz
Please contact me if this may be helpful - the package is under development, and needs user feedback! (from Roger Bivand Roger.Bivand@nhh.no)
See also a recent paper in Computers & Geosciences:
TI: Using the R statistical data analysis language on GRASS 5.0 GIS data base files AU: R.S. Bivand SO: Computers & Geosciences v.26, no.9/10, pp. 1043-1052. PY: 2000
Paulo Justantiano pointed out there is also:
- the package "spatial" is included with MASS
- the package "sgeostats" is available at the CRAN web site (for geostatistical analysis only)
- "splancs" is available at ftp://reclus.nhh.no/pub/R/splancs-2.01-1.tar.gz (for point process only)
- "geoR" is available at: http://www.maths.lancs.ac.uk/~ribeiro/geoR.html (for geostatistical analysis only)
Rainer Hurling added, have a look at
http://reclus.nhh.no/gc00/gc009.htm
The GIS software 'GRASS' (http://www.geog.uni-hannover.de/grass/index2.html) works fine with database system 'PostgreSQL' (http://www.postgresql.org) and 'R' (http://www.R-project.org). For linking between GRASS and PostgreSQL see http://www.geog.uni-hannover.de/grass/projects/postgresqlgrass and between GRASS and R see ftp://reclus.nhh.no/pub/R .
[top]Brian Ripley said:
Depends on what you mean (multinomial?). There is what I understand by multinomial logistic in function multinom in package nnet in the VR bundle, and nnet there can do several other variants on this theme.
Most such models are not hard to program and fit by a general-purpose optimizer like optim().
One alternative I've not tried:
David Clayton said:
You can also fit these models using the gee() package. The method is not ML, but there are certain other advantages to this method.
See
http://www.mrc-bsu.cam.ac.uk/pub/publications/submitted/david_clayton/ordcat2.ps
for details.
[top]See "nlme", available at CRAN.
[top]If you want parametric models, the gnlr and gnlr3 functions in my gnlm library will do all this: www.luc.ac.e/~jlindsey/rcode.html
[top]See cmdscale in the package "mva" and sammon and isoMDS in "MASS". There are some more in multiv, or at least more of the same. (from Brian D. Ripley)
[top]Notes on the use of R for psychology experiments and questionnaires by Jonathan Baron and Yuelin Li at
http://www.psych.upenn.edu/~baron/rpsych.htm and http://www.psych.upenn.edu/~baron/rpsych.pdf and http://www.psych.upenn.edu/~baron/rpsych.tex
It is intended for students and others who are doing research in psychology. What makes it "for psychology" is that it tries to cover the major statistical methods that psychologists use (particularly those who study humans).
Prof. Baron made a nice "reference card" as well: http://www.psych.upenn.edu/~baron/refcard.pdf
[top]http://www.stat.Berkeley.EDU/users/terry/zarray/Html/Rintro.html
[top]"S Poetry" is available online at http://www.seanet.com/%7Epburns/Spoetry/
http://www.insightful.com/resources/doc/default.html Splus guides - quite useful, I think, for general approaches especially (free downloads)
http://www.insightful.com/resources/biblio.html list of Splus books - too many!
http://hesweb1.med.virginia.edu/biostat/s/doc/splus.pdf guide to Splus for SAS users (or anyone, really)
and the other documents in CRAN contributed documents
Also, I'd check here, the R site for contributed docs, periodically: http://cran.r-project.org/other-docs.html
[top]http://www.statslab.cam.ac.uk/~pat
[top]These are useful!
http://euclid.math.mcgill.ca/keith/
[top]Mark Myatt offers some notes and example datasets that I found useful: http://www.myatt.demon.co.uk/ He offers everything in one big bundle called "Rex.zip"
[top]Hans Ehrbar's online course http://www.econ.utah.edu/ehrbar/ecmet.pdf
Jim Lindsey's R code for Introductory Statistics: A Modelling Approach. Oxford University Press (1995) and Models for Repeated Measurements. Oxford University Press (1999, Second Edition) http://alpha.luc.ac.be/~jlindsey/books.html See also the instructor's manual: www.luc.ac.be/~jlindsey/publications.html
[top]OK, you probably find this in CRAN, but I kept forgetting:
http://maths.newcastle.edu.au/~rking/R/
Jonathan Baron's site has a search tool as well, http://search.R-project.org/
[top]The easiest thing is to use some editor, Emacs or Winedit, write your R code, and then run it into R, and if it fails, change it in the editor, run it it again. Emacs with ESS installed will make this work fine. I've heard of plugins for other editors, did not try them, though.
If you don't like that, you can type commands directly into the R window and they have some handy things.
Up and down arrows cycle through the history of commands.
If you type in a function and want to revise it:
fix(f) or g <- edit(f) should bring up an editor.
I personally don't see any point in this, though, since you should just use an editor.
[top]When you quit R, since version 1 or so, it asks if you want to save the workspace. If you say yes, it creates a file .RData. The next time you start R in this working directory, it opens all those objects in .RData again. Type "objects()" to see.
If you want that, OK. If you don't, say NO at quit time, or erase the .RData file before starting R.
[top]If you try
> a <- c(1,2,3)
> save(a, file="test.RData")
> rm(a)
> load("test.RData")
> a
[1] 1 2 3
(from Peter Dalgaard, 12/29/2001)
[top]
objects() lists all objects
attributes(anObject)
str(anObject) -- diagnostic info on anObject
page(anObject) -- pages through anObject
rm(anObject) -- removes an object
rm(list=ls()) -- removes all objects
[top]Suppose you want to remove objects beginning with lm (lm1, lm2, lm3 etc).
rm(list=objects(pattern="^lm.*")) (from Kjetil Kjernsmo)
Don't do this to remove all objects with names like g01, g02, etc: rm(list=ls(pat="g*")). That removes everything!
To remove objects whose names have "g" anywhere in the string, use rm(list=ls(pat="g")), or names starting with "g" one can use rm(list=ls(pat="^g")).
The "^" and "$" can be used to delimit the beginning and ending of a string, respectively. (from Rashid Nassar)
[top]consider sink("filename")
The file .Rhistory contains commands typed on Unix systems.
Emacs users can run R in a shell and save everything in that shell to a file.
Ripley added: But if you are on Windows you can save the whole console record to a file, or cut-and-paste sections. (And that's what I do on Unix, too.)
[top]In your HOME directory, the file ~/.Rprofile. For example, if you think the stars on significance tests are stupid, follow Brian Ripley's example:
options(show.signif.stars=FALSE) ps.options(horizontal=FALSE)There is a system-wide Rprofile you could edit as well in the distribution [top]
Commands to the operating system: system("insert command here between quotes").
For the particular case of changing directories use getwd/setwd [I think system() spawns its own shell, so changing directories with system() isn't persistent] . For example:
> getwd() ## start in my home directory
[1] "/home/ben"
> system("cd ..") ## this executes OK but ...
> getwd()
[1] "/home/ben" ## I'm still in the same working directory
> setwd("..") ## this is the way to change working directory
NULL
> getwd() ## now I've moved up one
[1] "/home"
(from Ben Boker)
[top]
> date() [1] "Tue Jan 30 10:22:39 2001" > Sys.time() [1] "2001-01-30 10:21:29" > unclass(Sys.time()) [1] 980871696This is the number of seconds since the epoch (Jan 1, 1970). I use it as a "time stamp" on files sometimes. [top]
The function file.exists() will do this
[top]Do ?list.files().
list.files() gives back a vector of character strings, one for each file matching the pattern. Want file names ending in .dat? Try:
myDat <- list.files(pattern="\\.dat$")Then iterate over this myDat object, using the elements myDat[[i]] if you do something to each file.
Please note patterns in R are supposed to be regular expressions, not ordinary shell wildcards. Regular expressions are fairly standard in unix, but each version has its wrinkles. To see about the grep command, or about apropos, where regexp is used, do:
?grep or ?apropos
Martin Maechler spelled out a few of the basics for me:
"." means ``any arbitrary character''
"*" means ``the previous [thing] arbitrary many (0,1,...) times''
where [thing] is a single character (or something more general
for advanced usage)
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install.packages()
will download a package and install it!
It will get it from the network where you say, as in
install.packages("VR",lib="c:/r/library",CRAN="http://cran.r-project.org"). I have stared at the help pages for install.packages for many hours and still can't figure how to install a package that I've already downloaded and saved on disk. So maybe its just better to do it this way. Note that the package name, such as "VR", does not include the "zip" extension that is the end of the file name on CRAN.
If you are in the Operating System, not in an R session, you can install a package you've downloaded with
R INSTALL packageName
This works even if packageName is zipped, i.e., packageName.tar.gz will be accepted as an argument.
If you are not "root", tell R where to install the package directly from CRAN:
install.packages('tree', '~/.Rlibs')
Omitting the 2nd argument causes the routine to issue a warning andtry to install in the system package location, which is often what youwant, provided you have the required write access.
And, from the operating system shell, there is an equivalent. If you want to install the library in a private collection you can addthe -l flag to specify the location of the library. You should thenestablish an environment variable R_LIBS as a colon-separated list ofdirectories in which to look for libraries.
The call to R INSTALL would look like
R INSTALL /usr/apps/tree -l ~/.Rlibs(from Douglas Bates)
[top]R-1.2 implements a new memory approach that automatically resizes the workspace. To read more about it, type
?Memory
or
help(Memory)
You can also look at section 7.1 of the R FAQ (http://cran.r-project.org/doc/FAQ/R-FAQ.html).
I'm deleting all info about how to manually allocate memory. It is no longer recommended by the R team.
[top]Suppose you want an R program to be called in a script, and you absolutely must pass an argument to the R program to tell it what data to use or what variable name to use, etc.
Use Sys.getenv() to retrieve value you have set in your shell.
For a C shell,
$ setenv R_INFILE foobar
or BASH
$ export R_INFILE=foobar
Then in your R script grab that from the environment
> Sys.getenv("R_INFILE")
R_INFILE
"foobar"
> sink(Sys.getenv("R_INFILE"))
(from Thomas Lumley )
[top]Most unix systems have a program called script, which can keep a snapshot of text input and output. If you type
$ script filename >R ----do whatever R you want >quit() $exitThe $ are for the shell prompt. [top]
Put # at beginning of line, or
"With a good code editor none of these are any problem. There is no reason to write many lines of comments in R code: that's what the .Rd file is for. And if you want to temporarily delete a block, if(0) { ... } will work, as indeed will temporarily deleting in an editor." (from Brian D. Ripley)
Don't forget ESS has the ability to comment-out a highlighted region:
M-x comment-region
[top]See "example" at end of output in help?
Try
>?ls >example(ls)
Graphs go by too fast? do
par(ask=TRUE)
and try example(ls) again.
[top]help(function) same as ?function
help.start() fires up Netscape browser of html pages
help.search() searches in all available functions from all loaded packages
Don't forget that help.search takes regular expressions (as its help describes)
R> help.search(".*trans.*")
Help files with name or title matching `.*trans.*':
boxcox(MASS) Box-Cox Transformations for Linear Models
gilgais(MASS) Line Transect of Soil in Gilgai Territory
heart(MASS) The Stanford heart transplant data
Thomas Lumley
There is also an "apropos" function, similar in nature to emacs. If you have a vague idea of what you want, try it. Want to find how to list files? Do:
apropos("files")
Often you don't know the keyword to look for.
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Use detach(package:...), for instance:
> search() [1] ".GlobalEnv" "package:ctest" "Autoloads" "package:base" > detach(package:ctest) > search() [1] ".GlobalEnv" "Autoloads" "package:base"(from Emmanuel Paradis) [top]
With increasing frequency, I have noticed the problem that when I copy text from R, it shows up in the output with noise characters, like a quotation mark that shows up as *&\ or whatever. It turns out this is a matter of "character encoding", and in the r-help list on 2009-01-27, Duncan Murdoch made an absolutely beautiful post about it.
[top]This looks like an encoding problem: there are several different standards for encoding non-ASCII characters. All of your tools have to agree on the encoding.
To my eye it looks as though in the first case R is writing out UTF-8, and whatever you are using to look at your .tex file is assuming latin1 (some Windows programs say "ANSI", but I think that doesn't fully specify the encoding: you also need a code page, which is set somewhere in Windows control panel.)
The functions related to encodings in R are:
options(encoding="latin1") - set the default encoding
iconv(x, from="latin1", to="UTF-8") - re-encode entries, mapping each character from one encoding to the other
Encoding(x) - display the encoding of each entry (unknown means ascii or the native encoding for your platform)
Encoding(x) <- "latin1" - change the declared encoding, without changing the bytes.
> all.names(expression(sin(x+y))) [1] "sin" "+" "x" "y" > all.names(functions=FALSE,expression(sin(x+y))) [1] "x" "y" > all.vars(functions=FALSE,expression(sin(x+y))) [1] "x" "y" > all.vars(expression(sin(x+y))) [1] "x" "y"all.vars() works on a formula, terms, or expression object.
> all.vars((~j(pmin(3,g(h(x^3*y)))))) [1] "x" "y"
This function does not remove objects correctly
> testLoadSeveralHDF <- function(numFiles) { > for (i in 0:(numFiles-1)) { > filename <- paste("trial",i,".hdf",sep="") > print(filename) > hdf5load(filename) > rm(list = ls(pat="^g")) > } > }
Doug Bates said:
Add envir = .GlobalEnv to both the ls and the rm calls. That is
rm(list = ls(pat = "^g", envir = .GlobalEnv), envir = .GlobalEnv)
[top]first and second derivative of the expression wrt x
>D(expression(x^2),"x") >D(D(expression(x^2),"x"),"x")The real R FAQ 7.6 explains "How can I get eval() and D() to work?" [top]