Hi:

Here's my test:

l <- vector('list', 1000)
for(i in seq_along(l)) l[[i]] <- data.frame(x=rnorm(100),y=rnorm(100))
system.time(u1 <- do.call(rbind, l))
   user  system elapsed
   0.49    0.06    0.60
resultDF <- data.frame()
system.time(for (i in 1:1000) resultDF <- rbind(resultDF, l[[i]]))
   user  system elapsed
  10.34    0.06   10.53
identical(u1, resultDF)
[1] TRUE

The problem with the second approach, which is really kind of an FAQ
by now, is that repeated application of rbind as a standalone function
results in 'Spaceballs: the search for more memory!' The base
object gets bigger as the iterations proceed, something new is being
added, so more memory is needed to hold both the old and new objects.
This is an inefficient time killer because as the loop proceeds,
increasingly
more time is invested in finding new memory.

Interestingly, this doesn't scale linearly: if we make a list of 10000 100 x
2
data frames, I get the following:

> l <- vector('list', 10000)
> for(i in seq_along(l)) l[[i]] <- data.frame(x=rnorm(100),y=rnorm(100))
> system.time(u1 <- do.call(rbind, l))
   user  system elapsed
  55.56   30.62   88.11
> dim(u1)
[1] 1000000       2
> str(u1)
'data.frame':   1000000 obs. of  2 variables:
 $ x: num  -0.9516 -0.6948 0.0523 2.5798 -0.0862 ...
 $ y: num  1.466 0.165 1.375 0.571 -1.099 ...
> rm(u1)
> rm(resultDF)
> resultDF <- data.frame()
# go take a shower and come back....
> system.time(for (i in 1:100000) resultDF <- rbind(resultDF, l[[i]]))
   user  system elapsed
 977.33 121.41 1130.26
> dim(resultDF)
[1] 1000000       2

This time, neither do.call nor iterative rbind did very well.

One common way around this is to pre-allocate memory and then to
populate the object using a loop, but a somewhat easier solution here
turns out to be ldply() in the plyr package. The following is the same
idea as do.call(rbind, l), only faster:

> system.time(u3 <- ldply(l, rbind))
   user  system elapsed
   6.07    0.01    6.09
> dim(u3)
[1] 1000000       2
> str(u3)
'data.frame':   1000000 obs. of  2 variables:
 $ x: num  -0.9516 -0.6948 0.0523 2.5798 -0.0862 ...
 $ y: num  1.466 0.165 1.375 0.571 -1.099 ...

HTH,
Dennis

On Sat, Sep 4, 2010 at 11:37 AM, Paul Johnson <pauljohn32@gmail.com> wrote:

> I've been doing some consulting with students who seem to come to R
> from SAS.  They are usually pre-occupied with do loops and it is tough
> to persuade them to trust R lists rather than keeping 100s of named
> matrices floating around.
>
> Often it happens that there is a list with lots of matrices or data
> frames in it and we need to "stack those together".  I thought it
> would be a simple thing, but it turns out there are several ways to
> get it done, and in this case, the most "elegant" way using do.call is
> not the fastest, but it does appear to be the least prone to
> programmer error.
>
> I have been staring at ?do.call for quite a while and I have to admit
> that I just need some more explanations in order to interpret it.  I
> can't really get why this does work
>
> do.call( "rbind", mylist)
>
> but it does not work to do
>
> sapply ( mylist, rbind).
>
> Anyway, here's the self contained working example that compares the
> speed of various approaches.  If you send yet more ways to do this, I
> will add them on and then post the result to my Working Example
> collection.
>
> ## stackMerge.R
> ## Paul Johnson <pauljohn at ku.edu>
> ## 2010-09-02
>
>
> ## rbind is neat,but how to do it to a lot of
> ## 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)
>
> ## Usually we have done a stupid
> ## loop  to get this done
>
> resultDF <- mylist[[1]]
> for (i in 2:4) resultDF <- rbind(resultDF, mylist[[i]])
>
> ## My intuition was that this should work:
> ## lapply( mylist, rbind )
> ## but no! It just makes a new list
>
> ## This obliterates the columns
> ## unlist( mylist )
>
> ## I got this idea from code in the
> ## "complete" function in the "mice" package
> ## It uses brute force to allocate 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)
> dataComplete <- as.data.frame(matrix(0, nrow = nr*m, ncol = nc))
> for (j in  1:m) dataComplete[(((j-1)*nr) + 1):(j*nr), ] <- mylist[[j]]
>
>
>
> ## I searched a long time for an answer that looked better.
> ## This website is helpful:
> ## http://stackoverflow.com/questions/tagged/r
> ## I started to type in the question and 3 plausible answers
> ## popped up before I could finish.
>
> ## The terse answer is:
> shortAnswer <- do.call("rbind",mylist)
>
> ## That's the right answer, see:
>
> shortAnswer == dataComplete
> ## But I don't understand why it works.
>
> ## More importantly, I don't know if it is fastest, or best.
> ## It is certainly less error prone than "dataComplete"
>
> ## First, make a bigger test case and use system.time to evaluate
>
> phony <- function(i){
>  data.frame(w=rnorm(1000), x=rnorm(1000),y=rnorm(1000),z=rnorm(1000))
> }
> mylist <- lapply(1:1000, phony)
>
>
> ### First, try the terse way
> system.time( shortAnswer <- do.call("rbind", mylist) )
>
>
> ### Second, try the complete way:
> m <- 1000
> nr <- nrow(df1)
> nc <- ncol(df1)
>
> system.time(
>   dataComplete <- as.data.frame(matrix(0, nrow = nr*m, ncol = nc))
>  )
>
> system.time(
>   for (j in  1:m) dataComplete[(((j-1)*nr) + 1):(j*nr), ] <- mylist[[j]]
> )
>
>
> ## On my Thinkpad T62 dual core, the "shortAnswer" approach takes about
> ## three times as long:
>
>
> ## > system.time( bestAnswer <- do.call("rbind",mylist) )
> ##    user  system elapsed
> ##  14.270   1.170  15.433
>
> ## > system.time(
> ## +    dataComplete <- as.data.frame(matrix(0, nrow = nr*m, ncol = nc))
> ## +  )
> ##    user  system elapsed
> ##   0.000   0.000   0.006
>
> ## > system.time(
> ## + for (j in  1:m) dataComplete[(((j-1)*nr) + 1):(j*nr), ] <- mylist[[j]]
> ## + )
> ##    user  system elapsed
> ##   4.940   0.050   4.989
>
>
> ## That makes the do.call way look slow, and I said "hey,
> ## our stupid for loop at the beginning may not be so bad.
> ## Wrong. It is a disaster.  Check this out:
>
>
> ## > resultDF <- phony(1)
> ## > system.time(
> ## + for (i in 2:1000) resultDF <- rbind(resultDF, mylist[[i]])
> ## +    )
> ##    user  system elapsed
> ## 159.740   4.150 163.996
>
>
> --
> Paul E. Johnson
> Professor, Political Science
> 1541 Lilac Lane, Room 504
> University of Kansas
>
> ______________________________________________
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

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