## stackListItems.R
## Paul Johnson
## 2010-09-07
## I asked about this in r-help last week and promised
## a summary of answers.
## We face this problem all the time. A procedure
## generates a list of data frames. How to stack them
## together?
## The short answer is that the plyr package's rbind.fill
## method is probably the fastest method that is not
## prone to trouble and does not require much user caution.
## result <- rbind.fill(mylist)
## A slower alternative that also works is
## result <- do.call("rbind", mylist)
## That is always available in R and it works well enough, even
## though it is not quite as fast. Both of these are much faster than
## a loop that repeatedly applies "rbind".
## Truly blazing speed can be found if we convert this into
## matrices, but that is not possible if the list actually
## contains data frames.
## Here is a test case
df1 <- data.frame(x=rnorm(100),y=rnorm(100))
df2 <- data.frame(x=rnorm(100),y=rnorm(100))
df3 <- data.frame(x=rnorm(100),y=rnorm(100))
df4 <- data.frame(x=rnorm(100),y=rnorm(100))
mylist <- list(df1, df2, df3, df4)
## Here's the way we have done it. We understand this,
## we believe the result, it is easy to remember. It is
## also horribly slow for a long list.
resultDF <- mylist[[1]]
for (i in 2:4) resultDF <- rbind(resultDF, mylist[[i]])
## It works better to just call rbind once, as in:
resultDF2 <- rbind( mylist[[1]],mylist[[2]],mylist[[3]],mylist[[4]])
## That is faster because it calls rbind only once.
## But who wants to do all of that typing? How tiresome.
## Thanks to Erik Iverson in r-help, I understand that
resultDF3 <- do.call("rbind", mylist)
## is doing the EXACT same thing.
## Erik explained that "do.call( "rbind", mylist)"
## is *constructing* a function call from the list of arguments.
## It is shorthand for "rbind(mylist[[1]], mylist[[2]], mylist[[3]])"
## assuming mylist has 3 elements.
## Check the result:
all.equal( resultDF2, resultDF3)
## You often see people claim it is fast to allocate all
## of the required space in one shot and then fill it in.
## I got this algorithm from code in the
## "complete" function in the "mice" package.
## It allocates a big matrix of 0's and
## then it places the individual data frames into that matrix.
m <- 4
nr <- nrow(df1)
nc <- ncol(df1)
resultDF4 <- as.data.frame(matrix(0, nrow = nr*m, ncol = nc))
for (j in 1:m) resultDF4[(((j-1)*nr) + 1):(j*nr), ] <- mylist[[j]]
## This is a bit error prone for my taste. If the data frames have
## different numbers of rows, some major code surgery will be needed.
##
## Dennis Murphy pointed out the plyr package, by Hadley Wickham.
## Dennis said " ldply() in the plyr package. The following is the same
## idea as do.call(rbind, l), only faster."
library("plyr")
resultDF5 <- ldply(mylist, rbind)
all.equal(resultDF, resultDF5)
## Plyr author Hadley Wickham followed up with "I think all you want here is rbind.fill:"
resultDF6 <- rbind.fill(mylist)
all.equal(resultDF, resultDF6)
## Gabor Grothendieck noted that if the elements in mylist were matrices, this would all work faster.
mylist2 <- lapply(mylist, as.matrix)
matrixDoCall <- do.call("rbind", mylist2)
all.equal(as.data.frame(matrixDoCall), resultDF)
## Gabor also showed a better way than 'system.time' to find out how
## long this takes on average using the rbenchmark package. Awesome!
#> library(rbenchmark)
#> benchmark(
#+ df = do.call("rbind", mylist),
#+ mat = do.call("rbind", L),
#+ order = "relative", replications = 250
#+ )
## To see the potentially HUGE impact of these changes, we need to
## make a bigger test case. I just used system.time to evaluate, but
## if this involved a close call, I'd use rbenchmark.
phony <- function(i){
data.frame(w=rnorm(1000), x=rnorm(1000),y=rnorm(1000),z=rnorm(1000))
}
mylist <- lapply(1:1000, phony)
### First, try my usual way
resultDF <- mylist[[1]]
system.time(
for (i in 2:1000) resultDF <- rbind(resultDF, mylist[[i]])
)
## wow, that's slow:
## user system elapsed
## 168.040 4.770 173.028
### Now do.call method:
system.time( resultDF3 <- do.call("rbind", mylist) )
all.equal(resultDF, resultDF3)
## Faster! Takes one-twelfth as long
## user system elapsed
## 14.64 0.85 15.49
### Third, my adaptation of the complete function in the mice
### package:
m <- length(mylist)
nr <- nrow(mylist[[1]])
nc <- ncol(mylist[[1]])
system.time(
resultDF4 <- as.data.frame(matrix(0, nrow = nr*m, ncol = nc))
)
colnames(resultDF4) <- colnames(mylist[[1]])
system.time(
for (j in 1:m) resultDF4[(((j-1)*nr) + 1):(j*nr), ] <- mylist[[j]]
)
all.equal(resultDF, resultDF4)
##Disappointingly slow on the big case:
# user system elapsed
# 80.400 3.970 84.573
### That took much longer than I expected, Gabor's
### hint about the difference between matrix and data.frame
### turns out to be important. Do it again, but don't
### make the intermediate storage thing a data.frame:
mylist2 <- lapply(mylist, as.matrix)
m <- length(mylist2)
nr <- nrow(mylist2[[1]])
nc <- ncol(mylist2[[1]])
system.time(
resultDF4B <- matrix(0, nrow = nr*m, ncol = nc)
)
colnames(resultDF4B) <- colnames(mylist[[1]])
system.time(
for (j in 1:m) resultDF4B[(((j-1)*nr) + 1):(j*nr), ] <- mylist2[[j]]
)
### That's FAST!
### user system elapsed
### 0.07 0.00 0.07
all.equal(resultDF, as.data.frame(resultDF4B))
### Now the two moethods from plyr.
system.time( resultDF5 <- ldply(mylist, rbind))
## Just about as fast, much less error prone
## user system elapsed
## 1.290 0.000 1.306
all.equal(resultDF, resultDF5)
system.time(resultDF6 <- rbind.fill(mylist))
## user system elapsed
## 0.450 0.000 0.459
all.equal(resultDF, resultDF6)
## Gabor was right. If we have matrices, do.call is
## just about as good as anything.
system.time(matrixDoCall <- do.call("rbind", mylist2) )
## user system elapsed
## 0.030 0.000 0.032
all.equal(as.data.frame(matrixDoCall), resultDF)