[R] rowSums()
Adaikalavan Ramasamy
a.ramasamy at imperial.ac.uk
Wed Sep 24 16:23:50 CEST 2008
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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