Applying is.na() function in a vector with pipe function

Question

I was experimenting with the pipe function from dplyr, and running below code without success -

library(dplyr)
12 %>% ifelse(is.na(.), FALSE, TRUE)
### Error in ifelse(., is.na(.), FALSE, TRUE) : unused argument (TRUE)

Any pointer why I am getting this error? What is the correct approach if I want to use pipe?

A similar situation was discussed recently https://stackoverflow.com/questions/60106626/piping-a-vector-into-all-to-test-equality/60106676#60106676 — tmfmnk, Feb 07 '20 at 12:55
Please think about the value of logical conditions, such as `is.na`, before coding constructs like `ifelse(logical_condition, FALSE, TRUE)`. The result of `logical_condition` **already is a logical value**. — Rui Barradas, Feb 07 '20 at 13:03
@RuiBarradas In my code I was trying convert a NA to FALSE to eliminate the chance of any break in subsequent codes — Bogaso, Feb 07 '20 at 13:12
Does this answer your question? [Using the %>% pipe, and dot (.) notation](https://stackoverflow.com/questions/42385010/using-the-pipe-and-dot-notation) — Brian, Feb 07 '20 at 14:19
OK, but something like `%>% mutate(new_value = !is.na(.))` is better. — Rui Barradas, Feb 07 '20 at 15:54

score 2 · Answer 1 · answered Feb 07 '20 at 13:13

2

Another option is to unnest the functions further.

12 %>% is.na %>% ifelse(FALSE, TRUE)
# TRUE

answered Feb 07 '20 at 13:13

AndS.

G. Grothendieck · Answer 2 · 2020-02-07T14:08:59.583

2

We can use is.na directly in the pipeline rather than nesting it:

12 %>% is.na %>% ifelse(FALSE, TRUE)

or

12 %>% is.na %>% `!`

or

library(magrittr)
12 %>% is.na %>% not

or

12 %>% (is.na %>% Negate)

edited Feb 07 '20 at 14:08

answered Feb 07 '20 at 13:14

G. Grothendieck

Carles · Accepted Answer · 2020-02-07T13:18:44.180

0

There are several possibilities:

Using brackets: As @tmfmnk pointed out, one should use {} to get that behavior in pipe:

 library(dplyr)
 vec <- c(12,13, NA)
 vec %>% {ifelse(is.na(.), FALSE,TRUE)}
 # TRUE  TRUE FALSE

Unnesting: Alternatively, as @Ands pointed out, one can unnest the functions:

12 %>% is.na %>% ifelse(FALSE, TRUE)

edited Feb 07 '20 at 13:18

answered Feb 07 '20 at 12:59

Carles

score 0 · Answer 4 · answered Feb 07 '20 at 13:00

0

library(dplyr)
12 %>% ifelse(is.na(.), FALSE)
#FALSE

this works for me

answered Feb 07 '20 at 13:00

Carbo

But `c(12, NA) %>% ifelse(is.na(.), FALSE)` doesn't remove the `NA`, which was the question's point. – Rui Barradas Feb 07 '20 at 15:58

4 Answers4