[R] Using apply with more than one matrix
Waichler, Scott R
Scott.Waichler at pnnl.gov
Thu May 1 17:30:11 CEST 2014
Thank you, A.K. I learned from both of your solutions. I find the one that uses alply easier to follow and understand intuitively. I guess I'll want to learn more about what plyr can do. I've been using R for years but hadn't pushed vectorization far enough in my code. Now my computing will be more efficient.
Thanks also to David and the others who responded to my question. It all helped. --Scott Waichler
> -----Original Message-----
> From: arun [mailto:smartpink111 at yahoo.com]
> Sent: Thursday, May 01, 2014 6:24 AM
> To: R. Help
> Cc: Waichler, Scott R
> Subject: Re: [R] Using apply with more than one matrix
>
> Hi,
> Sorry, a typo in the previous function:
> ---------------------------------
> if (condition1[i] && !is.na(indx3)) {
> arr[x1][indx3] + m2[i] ###### should be mat2[i]
> } else NA
>
> -----------
>
> Also, you can try:
> library(plyr)
> evaluateNew <- function(arr, mat1, mat2){ if (!all(dim(mat1) ==
> dim(mat2))) {
> stop("both matrices should have equal dimensions")
> }
> indx1 <- unlist(alply(arr, c(1,2), function(x) {x[!is.na(x)][1]}))
> condition1 <- !is.na(mat1) & mat1 > mat2 ifelse(condition1, indx1+mat2,
> NA) } evaluateNew(a1, m1, m2)
>
> # [,1] [,2]
> #[1,] NA 1.66
> #[2,] 3.14 NA
> A.K.
>
>
>
>
> On Thursday, May 1, 2014 5:49 AM, arun <smartpink111 at yahoo.com> wrote:
>
>
> Hi,
>
> You may try:
> evaluate <- function(arr, mat1, mat2) {
> if (!all(dim(mat1) == dim(mat2))) {
> stop("both matrices should have equal dimensions")
> }
> indx1 <- as.matrix(do.call(expand.grid, lapply(dim(arr), sequence)))
> indx2 <- paste0(indx1[, 1], indx1[, 2])
> condition1 <- !is.na(mat1) & mat1 > mat2
> ans <- sapply(seq_along(unique(indx2)), function(i) {
> x1 <- indx1[indx2 %in% unique(indx2)[i], ]
> indx3 <- which(!is.na(arr[x1]))[1]
> if (condition1[i] && !is.na(indx3)) {
> arr[x1][indx3] + m2[i]
> } else NA
> })
> dim(ans) <- dim(mat1)
> ans
> }
> evaluate(a1,m1,m2)
> # [,1] [,2]
> #[1,] NA 1.66
> #[2,] 3.14 NA
>
> A.K.
>
>
>
> On Thursday, May 1, 2014 2:53 AM, "Waichler, Scott R"
> <Scott.Waichler at pnnl.gov> wrote:
> > I would ask you to look at this loop-free approach and ask if this is
> > not equally valid?
> >
> > ans <- matrix(NA, ncol=2, nrow=2)
> > ind.not.na <- which(!is.na(a1))
> > ans[] <- condition1*a1[,,ind.not.na[1]]+ m2 # two matrices of equal
> >dimensions, one logical.
> > ans
> > [,1] [,2]
> > [1,] NA 1.66
> > [2,] 2.74 NA
>
> Thanks, I am learning something. I didn't know you could multiply a
> logical object by a numerical one. But notice the answer is not the same
> as mine, because I am doing an operation on the vector of values a1[i,j,]
> first.
>
> I tried a modification on sapply below, but it doesn't work because I
> haven't referenced the 3d array a1 properly. So I guess I must try to get
> a 2d result from a1 first, then use that in matrix arithmetic.
>
> Sapply or mapply may work, I haven't used these much and will try to learn
> better how to use them. Your use of sapply looks good; but I'm trying to
> understand if and how I can bring in the operation on a1. This doesn't
> work:
>
> evaluate <- function(idx) {
> ind.not.na <- which(!is.na(a1[idx,])) ])) # doesn't work; improper
> indexing for a1
> if(length(ind.not.na) > 0) {
> return(condition1*(a1[idx,ind.not.na[1]] + m2[idx])) # doesn't work;
> improper indexing for a1
> }
> }
> vec <- sapply(seq(length(m2)), evaluate)
>
> Scott Waichler
>
>
> > -----Original Message-----
> > From: David Winsemius [mailto:dwinsemius at comcast.net]
> > Sent: Wednesday, April 30, 2014 8:46 PM
> > To: Waichler, Scott R
> > Cc: Bert Gunter; r-help at r-project.org
> > Subject: Re: [R] Using apply with more than one matrix
> >
> >
> > On Apr 30, 2014, at 6:03 PM, Waichler, Scott R wrote:
> >
> > > Here is a working example with no random parts. Thanks for your
> > patience and if I'm still off the mark with my presentation I'll drop
> > the matter.
> > >
> > > v <- c(NA, 1.5, NA, NA,
> > > NA, 1.1, 0.5, NA,
> > > NA, 1.3, 0.4, 0.9)
> > > a1 <- array(v, dim=c(2,2,3))
> > > m1 <- matrix(c(NA, 1.5, 2.1, NA), ncol=2, byrow=T)
> > > m2 <- matrix(c(1.56, 1.64, 1.16, 2.92), ncol=2)
> > > condition1 <- !is.na(m1)& m1 > m2
> > >
> > > ans <- matrix(NA, ncol=2, nrow=2) # initialize for(i in 1:2) {
> > >for(j in 1:2) {
> > > ind.not.na <- which(!is.na(a1[i,j,]))
> > > if(condition1[i,j] && length(ind.not.na) > 0) ans[i,j] <-
> > >a1[i,j,ind.not.na[1]] + m2[i,j] } } ans
> > > [,1] [,2]
> > > [1,] NA 1.66
> > > [2,] 3.14 NA
> >
> > I would ask you to look at this loop-free approach and ask if this is
> > not equally valid?
> >
> > ans <- matrix(NA, ncol=2, nrow=2)
> > ind.not.na <- which(!is.na(a1))
> > ans[] <- condition1*a1[,,ind.not.na[1]]+ m2 # two matrices of equal
> >dimensions, one logical.
> > ans
> > [,1] [,2]
> > [1,] NA 1.66
> > [2,] 2.74 NA
> > >
> > > Let me try asking again in words. If I have multiple matrices or
> > > slices
> > of 3d arrays that are the same dimension, is there a way to pass them
> > all to apply, and have apply take care of looping through i,j?
> >
> > I don't think `apply` is the correct function for this. Either
> > `mapply` or basic matrix operation seem more likely to deliver correct
> results:
> >
> >
> > > I understand that apply has just one input object x. I want to work
> > > on
> > more than one array object at once using a custom function that has
> > this
> > characteristic: in order to compute the answer at i,j I need a result
> > from higher order array at the same i,j.
> >
> > If you want to iterate over matrix indices you can either use the
> > vector version e.g. m2[3] or the matrix version, m2[2,1[.
> >
> > vec <- sapply(seq(length(m2) , function(idx) m2[idx]*condition1[idx] )
> >
> >
> >
> > > This is what I tried to demonstrate in my example above.
> > >
> > > Thanks,
> > > Scott
> >
> > David Winsemius
> > Alameda, CA, USA
>
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