I guess this would be the fastest way would be:
rs <- rowSums( testDat, na.rm=T)
rs[ which( rowMeans(is.na(testDat)) == 1 ) ] <- NA
since both rowSums and rowMeans are internally coded in C.
Regards, Adai
Doran, Harold wrote:
Say I have the following data:
testDat <- data.frame(A = c(1,NA,3), B = c(NA, NA, 3))
testDat
A B
1 1 NA
2 NA NA
3 3 3
rowsums() with na.rm=TRUE generates the following, which is not desired:
rowSums(testDat[, c('A', 'B')], na.rm=T)
[1] 1 0 6
rowsums() with na.rm=F generates the following, which is also not
desired:
rowSums(testDat[, c('A', 'B')], na.rm=F)
[1] NA NA 6
I see why this occurs, but what I hope to have returned would be:
[1] 1 NA 6
To get what I want I could do the following, but normally my ideas are
bad ideas and there are codified and proper ways to do things.
rr <- numeric(nrow(testDat))
for(i in 1:nrow(testDat)) rr[i] <- if(all(is.na(testDat[i,]))) NA else
sum(testDat[i,], na.rm=T)
rr
[1] 1 NA 6
Is there a "proper" way to do this? In my real data, nrow is over
100,000
Thanks,
Harold
sessionInfo()
R version 2.7.2 (2008-08-25)
i386-pc-mingw32
locale:
LC_COLLATE=English_United States.1252;LC_CTYPE=English_United
States.1252;LC_MONETARY=English_United
States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] MiscPsycho_1.2 lattice_0.17-13 statmod_1.3.6
loaded via a namespace (and not attached):
[1] grid_2.7.2
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and provide commented, minimal, self-contained, reproducible code.