How to use purrr functions inside dplyr's select_if - r

Question

I'm trying to find the shortest possible dplyr-purr combination.
Can I reduce the following statement which combines select_if() and map_df() ?
training.set.imputed %>%
select_if(~sum(is.na(.))>0) %>% map_df(~sum(is.na(.)))
I tried this:
training.set.imputed %>%
select_if(~sum(is.na(.))>0, .funs = ~sum(is.na(.)))
which throws this error:
Error: nm must be NULL or a character vector the same length as x
What does this mean?
Any ideas how to form the .funs term?

Answer 1

The .funs argument in select_if requires a renaming function, and not a mutating function, so you can do something like this with it, but you can't mutate the variable values:
tibble(blah = 1:2, bleh = 3:4, bluh = c(NA, NA)) %>%
select_if(~ sum(is.na(.x)) > 0, .funs = toupper)
#### OUTPUT ####
# A tibble: 2 x 1
BLUH
<lgl>
1 NA
2 NA
If you insist on using a combination of purrr and dplyr, then this is probably your best bet (Edit: I just noticed that G. Grothendieck gave this answer, but I'll include it anyway for the sake of completeness.):
df %>%
map_df(~ sum(is.na(.))) %>%
select_if(~ . > 0)
#### OUTPUT ####
# A tibble: 1 x 2
b d
<int> <int>
1 4 1
However, you can simplify it a little by just using dplyr's summarize_if:
df %>%
summarise_if(anyNA, ~ sum(is.na(.)))
#### OUTPUT ####
# A tibble: 1 x 2
b d
<int> <int>
1 4 1
Since you're really just after column sums, base R might offer the most concise option:
colSums(is.na(df)) %>%
.[. > 0]
#### OUTPUT ####
b d
4 1
Data
structure(list(a = c(2L, 2L, 5L, 10L, 10L, 18L, 18L, 19L, 11L,
14L, 12L, 10L, 4L, 16L, 5L, 5L, 11L, 2L, 14L, 7L), b = c(10L,
20L, 16L, NA, 6L, 1L, 11L, 12L, 12L, 12L, 8L, NA, NA, 8L, 11L,
19L, 8L, 9L, NA, 19L), c = c(11L, 11L, 20L, 8L, 15L, 4L, 17L,
4L, 4L, 11L, 20L, 11L, 6L, 12L, 17L, 7L, 14L, 18L, 15L, 19L),
d = c(19L, 16L, 17L, 14L, 8L, 19L, 7L, 6L, 6L, 13L, 7L, 19L,
11L, 17L, NA, 10L, 3L, 3L, 3L, 2L), e = c(12L, 17L, 14L,
5L, 8L, 19L, 8L, 3L, 17L, 1L, 2L, 6L, 5L, 17L, 14L, 5L, 8L,
2L, 8L, 2L)), row.names = c(NA, -20L), class = c("tbl_df",
"tbl", "data.frame"))

Answer 2

I assume you want the number of NAs in each column keeping only columns that have at least 1 NA.
!) This avoids the code duplication and does not give an error. First calculate the number of NAs in each column and then pick out the columns that are greater than 0.
# test input - BOD comes with R
BOD[1,2] <- NA
BOD %>%
map_df(~ sum(is.na(.))) %>%
select_if(~ . > 0)
giving:
# A tibble: 1 x 1
demand
<int>
1 1
2) This first selects out those columns with at least one NA and then finds the number of NAs in those columns giving the same result:
BOD %>%
select_if(anyNA) %>%
map_df(~ sum(is.na(.)))