[R] rowSums()

[Adaikalavan Ramasamy]

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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