one solution is to set NAs to 0, e.g.,
m <- matrix(1:3, 3, 3)
x <- list(m, m+3, m+6)
x[[1]][1] <- NA
x. <- lapply(x, function (x) {x[is.na(x)] <- 0; x} )
Reduce("+", x.)
I hope it helps.
Best,
Dimitris
On 5/4/2012 11:19 AM, Evgenia wrote:
I have a list ( in my real problem a double list y[[1:24]][[1:15]] but I
think the solution would be the same)
m<- matrix(1:3, 3, 3)
x<- list(m, m+3, m+6)
and as I want to have the sum of elements I use Reduce(`+`, x)
having as result
Reduce(`+`, x)
[,1] [,2] [,3]
[1,] 12 12 12
[2,] 15 15 15
[3,] 18 18 18
How can I take the sum of the list elements ignoring NA element
For example if I have
x[[1]][1]<-NA
Then I want to have
[,1] [,2] [,3]
[1,] 11 12 12
[2,] 15 15 15
[3,] 18 18 18
instead of
Reduce(`+`, x)
[,1] [,2] [,3]
[1,] NA 12 12
[2,] 15 15 15
[3,] 18 18 18
Thanks in advance
Evgenia
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--
Dimitris Rizopoulos
Assistant Professor
Department of Biostatistics
Erasmus University Medical Center
Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands
Tel: +31/(0)10/7043478
Fax: +31/(0)10/7043014
Web: http://www.erasmusmc.nl/biostatistiek/
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