Unexpected behaviour of apply()

Le vendredi 08 mars 2013 ? 09:29 +0100, Pierrick Bruneau a ?crit :

Hello everyone,

Considering the following code sample :

----
indexes <- function(vec) {
    vec <- which(vec==TRUE)
    return(vec)
}

This is essentially which(), what did you write such a convoluted
function to get the same result?

mat <- matrix(FALSE, nrow=10, ncol=10)
mat[1,3] <- mat[3,1] <- TRUE
----

Issuing apply(mat, 1, indexes) returns a 10-cell list, as expected.
Now if I do:

----
mat[1,3] <- mat[3,1] <- FALSE
apply(mat, 1, indexes)
----

I would expect a 10-cell list with integer(0) in each cell - instead I get
integer(0), which wrecks my further process.
From ?apply:

If each call to ?FUN? returns a vector of length ?n?, then ?apply?
     returns an array of dimension ?c(n, dim(X)[MARGIN])? if ?n > 1?.
     If ?n? equals ?1?, ?apply? returns a vector if ?MARGIN? has length
     1 and an array of dimension ?dim(X)[MARGIN]? otherwise.  If ?n? is
     ?0?, the result has length 0 but not necessarily the ?correct?
     dimension.

Note especially the last sentence.

Is there a simple way to get the result I expect (and the only consistent
one, IMHO) ?

One of the interests of apply() is that it combines the return values
from all function calls into a convenient form, but this can indeed be a
problem if you cannot know in advance what this form will be. If you
need a list in all cases, then just call lapply():
lapply(seq(nrow(mat)), function(i) which(mat[i,]))


Regards

Thanks by advance for your help,

Pierrick Bruneau
http://www.bruneau44.com

	[[alternative HTML version deleted]]

______________________________________________
R-help at 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.

This is nice fodder for 'The R Inferno' -- thanks.

As Milan said, 'which' will suffice as the function.

Here is a specialized function that only returns a
list and is only implemented to work with matrices.
It should solve your current dilemma.

applyL <-
function (X, MARGIN, FUN, ...)
{
         stopifnot(length(dim(X)) == 2, length(MARGIN) == 1)

         FUN <- match.fun(FUN)
         ans <- vector("list", dim(X)[MARGIN])
         if(MARGIN == 1) {
                 for(i in seq_along(ans)) {
                         ans[[i]] <- FUN(X[i,], ...)
                 }
         } else {
                 for(i in seq_along(ans)) {
                         ans[[i]] <- FUN(X[,i], ...)
                 }
         }
         names(ans) <- dimnames(X)[[MARGIN]]
         ans
}


Pat

Hello everyone,

Considering the following code sample :

----
indexes <- function(vec) {
     vec <- which(vec==TRUE)
     return(vec)
}
mat <- matrix(FALSE, nrow=10, ncol=10)
mat[1,3] <- mat[3,1] <- TRUE
----

Issuing apply(mat, 1, indexes) returns a 10-cell list, as expected.
Now if I do:

----
mat[1,3] <- mat[3,1] <- FALSE
apply(mat, 1, indexes)
----

I would expect a 10-cell list with integer(0) in each cell - instead I get
integer(0), which wrecks my further process.
Is there a simple way to get the result I expect (and the only consistent
one, IMHO) ?

Thanks by advance for your help,

Pierrick Bruneau
http://www.bruneau44.com

	[[alternative HTML version deleted]]

______________________________________________
R-help at 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.

Patrick Burns
pburns at pburns.seanet.com
twitter: @burnsstat @portfolioprobe
http://www.portfolioprobe.com/blog
http://www.burns-stat.com
(home of:
  'Impatient R'
  'The R Inferno'
  'Tao Te Programming')

An embedded and charset-unspecified text was scrubbed...
Name: not available
URL: <https://stat.ethz.ch/pipermail/r-help/attachments/20130308/74dca850/attachment.pl>