当所有元素均为NA时作为操作结果的NA

Question

This might be slightly silly but I would appreciate a better way to deal with this problem. I have a dataframe as the following

a <- matrix(1,5,3)
a[1:2,2] <- NA
a[1,c(1,3)] <- NA
a[3:5,2] <- 2
a[2:5,3] <- 3 
a <- data.frame(a)
colnames(a) = c("First", "Second", "Third")

I want to sum only some of, say, the columns but I would like to keep the NAs when all elements in the summed columns are NA. In short, if I sum First and Second columns I want to get something like

mySum <- c(NA, 1, 3, 3, 3)

Neither of the two options below provides what I want

rowSums(a[, c("First", "Second")])
rowSums(a[, c("First", "Second")], na.rm=TRUE)

but on the positive side I have resolved this by using a combination of is.na and all

mySum <- rowSums(a[, c("First", "Second")], na.rm=TRUE)
iNA = apply(a[, c("First", "Second")], 2, is.na)
iAllNA = apply(iNA, 1, all)
mySum[iAllNA] = NA

This feels slightly awkward though so I was wondering if there is a smarter way to handle this.

Answer 1

Using apply with margin = 1 for every row if all the row elements are NA we return NA or else we return the sum of them.

apply(a[c("First", "Second")], 1, function(x) 
                          ifelse(all(is.na(x)), NA, sum(x, na.rm = TRUE)))

#[1] NA  1  3  3  3

Answer 2

mycols = c("First", "Second")
replace(x = rowSums(a[mycols], na.rm = TRUE),
        list = rowSums(is.na(a[mycols])) == length(mycols),
        values = NA)
#[1] NA  1  3  3  3