Repeating a user-defined function using replicate() or sapply()

Question

I have defined a custom function, like this:

my.fun = function() {

      for (i in 1:1000) {
      ...
        for (j in 1:20) {
          ...
        }
      }

 return(output)

}

which returns an output matrix, output, composed by 1000 rows and 20 columns.

What I need to do is to repeat the function say 5 times and to store the five output results into a brand new matrix, say final, but without using another for-loop (this for making the code clearer, and also because in a second moment I would like to try to parallelize these additional 5 repetitions).

Hence final should be a matrix with 5000 rows and 20 columns (the rationale behind these 5 repetitions is that within the two for-loops I use, among other functions, sample).

I tried to use final <- replicate(5, my.fun()), which correctly computes the five replications, but then I have to "manually" put the elements into a brand new 5000 x 20 matrix.. is there a more elgant way to do so? (maybe using sapply()?). Many thanks

IRTFM · Accepted Answer · 2013-01-23T23:27:11.713

As is stands you probably have an array with three dimensions. If you wanted to have a list you would have added simplify=FALSE. Try this:

do.call( rbind, replicate(5, my.fun(), simplify=FALSE ) )

Or you can use aperm in the case where "final" is still an array:

fun <- function() matrix(1:10, 2,5)
final <- replicate( 2, fun() )
> final
, , 1

     [,1] [,2] [,3] [,4] [,5]
[1,]    1    3    5    7    9
[2,]    2    4    6    8   10

, , 2

     [,1] [,2] [,3] [,4] [,5]
[1,]    1    3    5    7    9
[2,]    2    4    6    8   10

> t( matrix(aperm(final, c(2,1,3)), 5,4) )
     [,1] [,2] [,3] [,4] [,5]
[1,]    1    3    5    7    9
[2,]    2    4    6    8   10
[3,]    1    3    5    7    9
[4,]    2    4    6    8   10

There may be more economical matrix operations. I just haven't discovered one yet.

Many thanks for the help. By the way, about the three dimensions array, you were right :) — Stefano Lombardi, Jan 23 '13 at 22:36

score 7 · Answer 2 · answered Jan 23 '13 at 22:03

7

If you replace replicate with rlply from the plyr package, you can use do.call with rbind:

library(plyr)
do.call(rbind, rlply(5, my.fun()))

If you'd rather not rely on the plyr package, you can always do:

do.call(rbind, lapply(1:5, function(i) my.fun()))

answered Jan 23 '13 at 22:03

David Robinson

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score 7 · Answer 3 · answered Jan 23 '13 at 22:52

7

Depends on which package you use for parallel computing, but here's how I would do it (hide it in a loop using sapply, just like replicate).

library(snowfall)
sfInit(parallel = TRUE, cpus = 4, type = "SOCK")
# sfExport() #export appropriate objects that will be needed inside a function, if applicable
# sfLibrary() #call to any special library
out <- sfSapply(1:5, fun = my.fun, simplify = FALSE)
sfStop()

answered Jan 23 '13 at 22:52

Roman Luštrik

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Thanks a lot, this is very interesting, and by the way I was planning to use `snowfall`. – Stefano Lombardi Jan 23 '13 at 22:57

score 0 · Answer 4 · edited May 10 '15 at 07:30

0

Try this:

final <- replicate(5, my.fun(), simplify = "matrix")

You will get the result of 'final' in the form of matrix.

edited May 10 '15 at 07:30

Sandy Muspratt

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answered May 10 '15 at 05:42

LifeLongLearner

21
4

I thought that `simplify` can only accept `FALSE`, `TRUE` or `"array"`. The default `TRUE` returns a vector or matrix if appropriate. – Valentin_Ștefan Mar 26 '18 at 15:28

Repeating a user-defined function using replicate() or sapply()

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