[R] rowSums()

[Chuck Cleland]

On 9/24/2008 10:38 AM, Marc Schwartz wrote:
> 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.

  Based on the difference in behavior for Sums and Means, this might be
another possibility:

rowMeans(testDat, na.rm=TRUE) * rowSums(!is.na(testDat))

[1]  1 NA  6

> 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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-- 
Chuck Cleland, Ph.D.
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