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 = s
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
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 wrote:
> Hi,
>
> I have three matrices which could be, for example:
> A = 0,
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