Hi,
Sorry to post again, but there is a careless error in my first R code snippet:
--- begin R code ---
AA <- array(1:60, dim = c(5500,44,33))
BB <- array(1:60, dim = c(44,44,33))
arraymatprod <- function(A, B) {
EE <- array(dim=dim(A));
for (i in 1:dim(A)[3]) EE[,,i] <- A[,,i] %*% B[,,i]
EE
}
system.time(arraymatprod(AA, BB))
--- end R code ---
The second snippet is correct:
--- begin R code ---
arraymatrixproduct <- function(A, B) {
require(abind)
dA <- dim(A)[1]
AB <- abind(A, B, along = 1)
array(apply(X = AB, MARGIN = 3, FUN = (function(mat) ((mat[1:dA, ] %*%
mat[(dA+1):(dim(AB)[1]),])))), dim = dim(A))
}
system.time(arraymatrixproduct(AA, BB))
--- end R code ---
However, the second is almost twice as long as the first snippet.
Many thanks,
Ranjan
On Fri, 10 Jul 2015 08:23:49 -0500 Ranjan Maitra
<maitra.mbox.igno...@inbox.com> wrote:
> Dear friends,
>
> I have two 3-d arrays of appropriate dimensions. I want to postmultiply the
> 2-D matrix layers of the first with the 2-D matrix layers of the second. Can
> I do this easily in R avoiding loops?
>
> As an example:
>
> --- begin R code ---
>
> AA <- array(1:60, dim = c(5500,44,33))
> BB <- array(1:60, dim = c(44,44,33))
>
>
> arraymatprod <- function(A, B) {
> EE <- array(dim=dim(A));
> for (i in 1:3) EE[,,i] <- A[,,i] %*% B[,,i]
> EE
> }
>
> system.time(arraymatprod(AA, BB))
>
> --- end R code ---
>
> So, is there a way to do this without the loop?
>
> Of course, I could abind the arrays and then use apply with an appropriate
> function which would be as follows:
>
> --- begin R code ---
>
> arraymatrixproduct <- function(A, B) {
> require(abind)
> dA <- dim(A)[1]
> AB <- abind(A, B, along = 1)
> array(apply(X = AB, MARGIN = 3, FUN = (function(mat) ((mat[1:dA, ] %*%
> mat[(dA+1):(dim(AB)[1]),])))), dim = dim(A))
> }
>
> system.time(arraymatrixproduct(AA, BB))
>
> --- end R code ---
>
> However, this turns out to be slower -- perhaps because of the use of abind
> and filling the array inside the function.
>
> I just wanted suggestions to get this operation done more efficiently.
>
> Many thanks and best wishes,
> Ranjan
>
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