[R] rowSums()
Marc Schwartz
marc_schwartz at comcast.net
Wed Sep 24 16:38:23 CEST 2008
on 09/24/2008 09:06 AM 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
The behavior you observe is documented in ?rowSums in the Value section:
If there are no values in a range to be summed over (after removing
missing values with na.rm = TRUE), that component of the output is set
to 0 (*Sums) or NA (*Means), consistent with sum and mean.
So:
> sum(c(NA, NA), na.rm = TRUE)
[1] 0
As per the definition of the sum of an empty set being 0, which I got
burned on myself a while back.
You could feasibly use:
Res <- rowSums(testDat, na.rm = TRUE)
is.na(Res) <- rowSums(is.na(testDat)) == ncol(testDat)
HTH,
Marc Schwartz
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