rowSums()

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