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
On 08/03/2013 08:29, Pierrick Bruneau wrote:
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
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--
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and provide commented, minimal, self-contained, reproducible code.
