help with element-by-element sum with NA

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.thiagoba at 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!

         [[alternative HTML version deleted]]

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help with element-by-element sum with NA

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