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 <smartpink...@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.waich...@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:dwinsem...@comcast.net]
> Sent: Wednesday, April 30, 2014 8:46 PM
> To: Waichler, Scott R
> Cc: Bert Gunter; r-help@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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