Summing list with NA elements
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 -- View this message in context: http://r.789695.n4.nabble.com/Summing-list-with-NA-elements-tp4608167.html Sent from the R help mailing list archive at Nabble.com.
______________________________________________ R-help at r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
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/