CODEWITHSUNDEEP
rpurrr
ℕʘʘḆḽḘ
ℕʘʘḆḽḘ

Reputation: 19395

filter out the longest string in purr

Consider this example

list('test', 'one')

I would like to subset this list and only keep the longest string in the list. Using purrr::keep does not seem to work.

> list('test', 'one') %>% keep(~ nchar(.x) == max(nchar(.)))
[[1]]
[1] "test"

[[2]]
[1] "one"

Any ideas? Thanks!

Upvotes: 2

Views: 95

Answers (2)

Julius Vainora
Julius Vainora

Reputation: 48241

If l is stored, then indeed base R seems best:

l <- list('test', 'one')

# If you want only the first one or there is a unique element
l[which.max(nchar(l))]
# [[1]]
# [1] "test"

# General
l[nchar(l) == max(nchar(l))]
# [[1]]
# [1] "test"

Now with keep we may do

list('test', 'one') %>% keep(function(x) nchar(x) == max(nchar(.)))
# [[1]]
# [1] "test"

The issue appears to be that both . and .x are just individual elements of the list in ~ nchar(.x) == max(nchar(.)).

Upvotes: 1

YOLO
YOLO

Reputation: 21739

You can do simple:

k <- list('test', 'one')
k[which.max(lapply(k, nchar))]

[[1]]
[1] "test"

Upvotes: 1

Related Questions