On 2012-07-23 21:48, R. Michael Weylandt wrote:
Perhaps something like:
Reduce(function(x,y){x[is.na(x)] <- 0; y[is.na(y)] <- 0; x + y}, list(A,B,C))
Not the most elegant, but it will get the job done.
Michael
I like Reduce(), but here are a couple more solutions:
1.
tmp <- mapply(FUN = sum, A, B, C,
MoreArgs = list(na.rm = TRUE))
matrix(tmp, nrow(A))
2.
Using the abind package to create a 3D array and
then using apply() to sum over the appropriate
columns in the array:
require(abind)
tmp <- abind(A, B, C, along = ncol(A) + 1)
apply(tmp, 1:2, sum, na.rm = TRUE)
2a.
Both of the above solutions generate zeros where
one might prefer NA. Here's a way to get the NAs:
sum2 <- function(x){
ifelse(all(is.na(x)), NA, sum(x, na.rm = TRUE))
}
apply(tmp, 1:2, sum2)
Peter Ehlers
On Mon, Jul 23, 2012 at 3:47 PM, Thiago Couto <couto.thiag...@gmail.com> wrote:
Hi,
I have three matrices which could be, for example:
A = 0, NA
NA, 3
B = 1, NA
0, NA
C = 1, NA
1, 1
(The point is that they all may have NA's in some cells)
QUESTION: How do I perform a element-by-element sum of the elements of
these three matrices (A + B + C), ignoring NA's, to obtain:
D = 2, NA
1, 4
In reality I am handling much larger matrices (not just 2x2).
Thank you for any help!
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