suppress NAs in paste()

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 NAs 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 NAs.

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 == ""

I found a dplyr/tidyverse solution to that question, which is rather elegant in my opinion.

library(tidyr)
foo <- LETTERS[1:4] 
foo[4] <- NA 
df <- data.frame(foo, num = 1:4)
df %>% unite(., col = "New.Col",  num, foo, na.rm=TRUE, sep = ",")
>    New.Col
  1:     1,A
  2:     2,B
  3:     3,C
  4:       4