[R] Indexing multi-dimensional table
Henrik Bengtsson
hb at biostat.ucsf.edu
Thu Dec 22 11:33:55 CET 2011
On Thu, Dec 22, 2011 at 4:34 AM, David A Vavra <davavra at verizon.net> wrote:
> I want to take slices of a multi-dimensional table (or array) without
> knowing the number of dimensions in advance.
>
>
>
> As a test I tried using (in this example a 3d table):
>
>
>
> do.call(`[`, list(tbl, x,NULL,NULL)]
>
>
>
> where I built the list on the fly. It works great as long as I only want the
> first dimension however when I try a different dimension, say with
> list(tbl,NULL,x,NULL), I get "<0 x 0> matrix" as the result. I thought this
> was because I wasn't calling the right function but there is no `[.table` or
> `[.matrix` or even `[.array`.
>
>
>
> Am I going about this the wrong way?
>From help("[", package="base"): "An index value of NULL is treated as
if it were integer(0).".
> y <- array(1:24, dim=c(2,3,4))
> dimnames(y)[[1]] <- sprintf("a%d", 1:dim(y)[1])
> dimnames(y)[[2]] <- sprintf("b%d", 1:dim(y)[2])
> dimnames(y)[[3]] <- sprintf("c%d", 1:dim(y)[3])
> y[NULL,2:3,1,drop=FALSE]
, , c1
b2 b3
> y[integer(0),2:3,1,drop=FALSE]
, , c1
b2 b3
I don't think there is an easy way to achieve:
> y[,2:3,1,drop=FALSE]
, , c1
b2 b3
a1 3 5
a2 4 6
using do.call("[") without explicitly specify the indices for that
"missing" dimension, i.e.
> y[seq(length=dim(y)[1]),2:3,1,drop=FALSE]
, , c1
b2 b3
a1 3 5
a2 4 6
If you're willing to use R.utils you can do:
library("R.utils");
> extract(y, indices=list(2:3,1), dims=c(2,3), drop=TRUE);
b2 b3
a1 3 5
a2 4 6
My $.02
/Henrik
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> DAV
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> [[alternative HTML version deleted]]
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