suppress NAs in paste()
Regarding the bounty
Ben Bolker's paste2
-solution produces a ""
when the strings that are pasted contains NA
's in the same position. Like this,
> paste2(c("a","b", "c", NA), c("A","B", NA, NA))
[1] "a, A" "b, B" "c" ""
The fourth element is an ""
instead of an NA
Like this,
[1] "a, A" "b, B" "c" NA
I'm offering up this small bounty for anyone who can fix this.
Original question
I've read the help page ?paste
, but I don't understand how to have R ignore NA
s. I do the following,
foo <- LETTERS[1:4]
foo[4] <- NA
foo
[1] "A" "B" "C" NA
paste(1:4, foo, sep = ", ")
and get
[1] "1, A" "2, B" "3, C" "4, NA"
What I would like to get,
[1] "1, A" "2, B" "3, C" "4"
I could do like this,
sub(', NA$', '', paste(1:4, foo, sep = ", "))
[1] "1, A" "2, B" "3, C" "4"
but that seems like a detour.
For the purpose of a "true-NA": Seems the most direct route is just to modify the value returned by paste2
to be NA
when the value is ""
paste3 <- function(...,sep=", ") {
L <- list(...)
L <- lapply(L,function(x) {x[is.na(x)] <- ""; x})
ret <-gsub(paste0("(^",sep,"|",sep,"$)"),"",
gsub(paste0(sep,sep),sep,
do.call(paste,c(L,list(sep=sep)))))
is.na(ret) <- ret==""
ret
}
val<- paste3(c("a","b", "c", NA), c("A","B", NA, NA))
val
#[1] "a, A" "b, B" "c" NA
I know this question is many years old, but it's still the top google result for r paste na
. I was looking for a quick solution to what I assumed was a simple problem, and was somewhat taken aback by the complexity of the answers. I opted for a different solution, and am posting it here in case anyone else is interested.
bar <- apply(cbind(1:4, foo), 1,
function(x) paste(x[!is.na(x)], collapse = ", "))
bar
[1] "1, A" "2, B" "3, C" "4"
In case it isn't obvious, this will work on any number of vectors with NA
s in any positions.
IMHO, the advantage of this over the existing answers is legibility. It's a one-liner, which is always nice, and it doesn't rely on a bunch of regexes and if/else statements which may trip up your colleagues or future self. Erik Shitts' answer mostly shares these advantages, but assumes there are only two vectors and that only the last of them contains NA
s.
My solution doesn't satisfy the requirement in your edit, because my project has the opposite requirement. However, you can easily solve this by adding a second line borrowed from 42-'s answer:
is.na(bar) <- bar == ""