Hi all
I don't know much about R internals and what goes on behind the scenes but
to me it would seem that
lapply(dim(a), function(i) seq(length=i))
accomplishes the same as
jj <- function(i) seq_len(dim(a)[i])
lapply(seq_len(length(dim(a))), jj)
and might be easier to read. Is there a performance penalty for this? In R
2.4 you could probably use
lapply(dim(a), seq_len)
but I'm still using R 2.3.1.
I would also think that the subsetting on the left hand side of
a[index] <- apply(index, 1, f)
might cause unnecessary overhead, and since the use of expand.grid has
already ensured that all the numbers are in the correct order we could
simply use
array(apply(index, 1, f), dim=dim(a))
Thus this could all be put together in the following one liner which seems
to me to accomplish the same as do.index
do.index3 <- function(a, f) {
array( apply( as.matrix(expand.grid(lapply(dim(a), function(i)
seq(length=i)))) ,1,f) , dim=dim(a))
}
f <- function(l){
sum(unlist(l))
}
a <- array(0,c(2,3,4))
b <- array(0,c(2,2,2,2))
do.index(a,f)
do.index3(a,f)
do.index(b,f)
do.index3(b,f)
>-----Original Message-----
>From: [EMAIL PROTECTED]
>[mailto:[EMAIL PROTECTED] On Behalf Of Martin Maechler
>Sent: 16 November 2006 06:04 PM
>To: Robin Hankin
>Cc: R-help mailing list
>Subject: Re: [R] filling an array, vectorized
>
>>>>>> "Robin" == Robin Hankin <[EMAIL PROTECTED]>
>>>>>> on Thu, 16 Nov 2006 11:54:38 +0000 writes:
>
> Robin> Gabor
> Robin> Tamas
>
> Robin> yet again I find myself trumped by Gabor because I
>forget that
> Robin> TRUE is a perfectly acceptable argument to "[".
>
> Robin> Heh.
>
>
> Robin> I'll stick do.index2() in the magic package.
>
>[ calling it 'do.index' - not 'do.index2' hopefully. ]
>
>Note that there are two things I'd further change:
>
>1) expand.grid() had grown a 'KEEP.OUT.ATTRS' argument recently
> which you want to set to FALSE for efficiency
>2) jj(TRUE) maybe short and cute, but I find it ugly that it
> really relies on
> seq(<integer-vector>) |--> seq(along = <integer-vector>)
> and would prefer the more efficient and clearer
> seq_len(length(dim(a)))
>
>
>Note however that for typical cases these two changes don't
>make the result (noticeably) faster at all
>
>
>do.indexM <- function(a,f) {
> jj <- function(i) seq_len(dim(a)[i])
> index <- as.matrix(expand.grid(lapply(seq_len(length(dim(a))), jj),
> KEEP.OUT.ATTRS = FALSE) )
> a[index] <- apply(index, 1, f)
> a
>}
>
>Martin
>
> Robin> best wishes
>
>
> Robin> rksh
>
>
> Robin> On 16 Nov 2006, at 11:27, Gabor Grothendieck wrote:
>
> >> Here is minor simplification:
> >>
> >> do.index2 <- function(a,f){
> >> jj <- function(i) seq(dim(a)[i])
> >> index <- as.matrix(expand.grid(lapply(jj(TRUE), jj)))
> >> a[index] <- apply(index, 1, f)
> >> a
> >> }
> >>
> >> # test
> >> a <- array(0,c(2,3,4))
> >> identical(do.index(a, f), do.index2(a, f))
> >> b <- array(0,c(2,2,2,2))
> >> identical(do.index(b, f), do.index2(b, f))
> >>
> >>
> >> On 11/16/06, Robin Hankin <[EMAIL PROTECTED]> wrote:
> >>> Tamas
> >>>
> >>> first of all, Thank You for a really well-posed,
>interesting problem.
> >>> Answer follows.
> >>>
> >>>
> >>>
> >>> do.index <- function(a,f){
> >>> jj <- function(i){seq_len(dim(a)[i])}
> >>> index <- as.matrix(expand.grid(sapply(1:length(dim
> >>> (a)),jj,simplify=FALSE)))
> >>> a[index] <- apply(index,1,f)
> >>> return(a)
> >>> }
> >>>
> >>>
> >>>
> >>> f <- function(l){
> >>> sum(unlist(l))
> >>> }
> >>>
> >>> a <- array(0,c(2,3,4))
> >>> b <- array(0,c(2,2,2,2))
> >>>
> >>> do.index(a,f)
> >>> do.index(b,f)
> >>>
> >>>
> >>>
> >>> best wishes
> >>>
> >>> Robin
> >>>
> >>>
> >>>
> >>>
> >>>
> >>> On 15 Nov 2006, at 19:16, Tamas K Papp wrote:
> >>>
> >>> > Hi,
> >>> >
> >>> > I am sure this has come up before, but my searches
>of the archive
> >>> > didn't give any results (maybe I didn't use the
>right keywords,
> >>> but if
> >>> > I use too many, the search times out).
> >>> >
> >>> > I have a vector of dimensions n, length is not fixed, eg
> >>> >
> >>> > n <- c(4,5,7)
> >>> >
> >>> > or
> >>> >
> >>> > n <- c(19,4,5,7)
> >>> >
> >>> > and a function f that takes a vector of indices,
>same length of
> >>> n, and
> >>> > gives a scalar.
> >>> >
> >>> > I would like to fill the array
> >>> >
> >>> > a <- array(dim=n)
> >>> >
> >>> > so that (... is just notation, not R's ...)
> >>> >
> >>> > a[i,j,k,...] <- f(list(i,j,k,...))
> >>> >
> >>> > I would use loops, but since n can have different
>lengths, I don't
> >>> > know how many loops I would need beforehand. Is
>there a way to do
> >>> > this?
> >>> >
> >>> > Thanks,
> >>> >
> >>> > Tamas
> >>> >
> >>> > ______________________________________________
> >>> > [email protected] 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.
> >>>
> >>>
>
> Robin> --
> Robin> Robin Hankin
> Robin> Uncertainty Analyst
> Robin> National Oceanography Centre, Southampton
> Robin> European Way, Southampton SO14 3ZH, UK
> Robin> tel 023-8059-7743
>
> Robin> ______________________________________________
> Robin> [email protected] mailing list
> Robin> https://stat.ethz.ch/mailman/listinfo/r-help
> Robin> PLEASE do read the posting guide
>http://www.R-project.org/posting-guide.html
> Robin> and provide commented, minimal, self-contained,
>reproducible code.
>
>______________________________________________
>[email protected] 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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