rowSums with all NA

Question

df <- data.frame(a = c(1, 1, NA, 0, 1, 0),
                 b = c(0, 1, NA, NA, 0, 1),
                 c = c(NA, 0, NA, 0, 1, NA),
                 d = c(1, NA, NA, 1, 1, 0))

rowSums(df)
#[1] NA NA NA NA  3 NA
rowSums(df, na.rm=T)
#[1] 2 2 0 1 3 1

The first I get, but my assumption and hope was that the third observation would return NA. Is there a way to get it to return NA for the third observation?

See here : http://stackoverflow.com/questions/18189753/rowsums-but-keeping-na-values (multiply by an `ifelse`) — etienne, Nov 12 '15 at 13:11

Roland · Accepted Answer · 2015-11-12T13:19:56.827

9

Here is one option:

rowSums(df, na.rm = TRUE) * NA ^ (rowSums(!is.na(df)) == 0)
# [1]  2  2 NA  1  3  1

This uses that anything ^ 0 equals 1 in R.

edited Nov 12 '15 at 13:19

answered Nov 12 '15 at 13:15

Roland

127,288
10
191
288

score 0 · Answer 2 · answered Nov 12 '15 at 13:14

0

not the best solution, but

a<-rowSums(df,na.rm=T)
a[a==0 & is.na(rowSums(df))]<-NA

should work

answered Nov 12 '15 at 13:14

zugabe

54
5

and what about columns with zeroes? you could not distinguish them from NA columns! – sam Nov 12 '15 at 13:18
You could do `a[rowSums(!is.na(df)) == 0] <- NA` instead – David Arenburg Nov 12 '15 at 13:24
a column with zeroes shouldn't be transferred as is.na(rowSums(df)) equals F. Or have I overlooked a problem? – zugabe Nov 12 '15 at 13:26

rowSums with all NA

2 Answers2

Linked