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help with element-by-element sum with NA

4 messages · Thiago Couto, arun, R. Michael Weylandt +1 more

arun
# 
HI,

Try this:
A<-matrix(c(0,NA,NA,3),ncol=2)
B<-matrix(c(1,0,NA,NA),ncol=2)
?C<-matrix(c(1,1,NA,1),ncol=2)
AB<-ifelse(is.na(A),ifelse(is.na(B),NA,B), ifelse(is.na(B), A, A+B))
?ABC<-ifelse(is.na(AB),ifelse(is.na(C),NA,C),ifelse(is.na(C),AB,AB+C))
?ABC
???? [,1] [,2]
[1,]??? 2?? NA
[2,]??? 1??? 4

A.K.




----- Original Message -----
From: Thiago Couto <couto.thiagoba at gmail.com>
To: r-help at r-project.org
Cc: 
Sent: Monday, July 23, 2012 4:47 PM
Subject: [R] help with element-by-element sum with NA

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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R. Michael Weylandt
# 
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
On Mon, Jul 23, 2012 at 3:47 PM, Thiago Couto <couto.thiagoba at gmail.com> wrote:
Peter Ehlers
#
On 2012-07-23 21:48, R. Michael Weylandt wrote:
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