What does it mean to unroll the loop?
Thanks!
Ranjan
On Fri, 10 Jul 2015 07:07:15 -0700 Jeff Newmiller <jdnew...@dcn.davis.ca.us>
wrote:
> Strictly speaking, the answer is yes because you can unroll the loop, but
> that probably is not what you really want or need to do.
>
> Your example seems about right, but it is not clear how you plan to make 44
> conform with 5500.
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> Sent from my phone. Please excuse my brevity.
>
> On July 10, 2015 6:44:19 AM PDT, Ranjan Maitra
> <maitra.mbox.igno...@inbox.com> wrote:
> >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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> >>
>
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