Remove an element of a list by name

Question

I'm working with a long named list and I'm trying to keep/remove elements that match a certain name, within a tidyverse context, similar to

dplyr::select(contains("pattern"))

However, I'm having issues figuring it out.

library(tidyverse)

a_list <- 
  list(a = "asdfg",
       b = "qwerty",
       c = "zxcvb")

a_list %>% pluck("a") # works
a_list %>% pluck(contains("a")) #does not work

a_list[2:3] # this is what I want
a_list %>% pluck(-"a") # but this does not work

Based on [this question](https://stackoverflow.com/questions/46983716/does-a-multi-value-purrrpluck-exist), it sounds like `pluck` is a replacement for `[[` not `[` and so there isn't a built-in way to accomplish `a_list[2:3]`. `[[` only selects [one element](https://stackoverflow.com/a/1169495/8099834) — Hallie Swan, Jan 26 '18 at 19:12

score 22 · Accepted Answer · answered Sep 25 '18 at 13:32

To remove by name you could use:

a_list %>% purrr::list_modify("a" = NULL)
$`b`
[1] "qwerty"

$c
[1] "zxcvb"

I'm not sure the other answers are using the name of the element, rather than the element itself for selection. The example you gave is slightly confusing since the element 'a' both contains 'a' in it's value AND is called 'a'. So it's easy to get mixed up. To show the difference I'll modify the example slightly.

b_list <- 
  list(a = "bsdfg",
       b = "awerty",
       c = "zxcvb")

b_list %>% purrr::list_modify("a" = NULL)

returns

$`b`
[1] "awerty"

$c
[1] "zxcvb"

but

purrr::discard(b_list,.p = ~stringr::str_detect(.x,"a"))

returns

$`a`
[1] "bsdfg"

$c
[1] "zxcvb"

score 15 · Answer 2 · answered Jan 26 '18 at 19:26

15

Keeping it full tidyverse, you could do,

purrr::discard(a_list,.p = ~stringr::str_detect(.x,"a"))

answered Jan 26 '18 at 19:26

joran

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If your code/package doesn't have stringr as a dependency: `purrr::discard(a_list,.p = ~grepl("a", .x))` – wibeasley Nov 22 '22 at 19:12

score 13 · Answer 3 · edited Sep 07 '21 at 15:51

13

I know this is old, but one simple way to do this :

a_list[['a']] <- NULL

edited Sep 07 '21 at 15:51

Anoushiravan R

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answered Sep 07 '21 at 15:49

Parisa

448
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11

Base code is often the most efficient! (in terms of typing) – JebLuvsStats May 25 '23 at 14:28

score 5 · Answer 4 · answered Jan 26 '18 at 20:08

5

using base R:

a_list[!grepl("a",unlist(a_list))]
$b
[1] "qwerty"

$c
[1] "zxcvb"

answered Jan 26 '18 at 20:08

Onyambu

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Ferroao · Answer 5 · 2020-10-08T19:24:13.657

5

Similar to previous answer but searching names as in OP

within(a_list, rm(a))

a_list[!grepl("^a$",names(a_list))]
a_list[grepl("^a$",names(a_list))]<-NULL

a_list[-which(names(a_list)=="a")]
a_list[-which(names(a_list)!="a")]<-NULL
a_list[ which(names(a_list)=="a")]<-NULL

edited Oct 08 '20 at 19:24

answered May 20 '19 at 14:04

Ferroao

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28
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score 0 · Answer 6 · answered Feb 09 '22 at 18:09

Same as Parisa's answer, just fits better into tidyverse pipe chains:

library(magrittr)

a_list <-
    list(a = "asdfg",
         b = "qwerty",
         c = "zxcvb")

a_list %>%
magrittr::inset(c('a', 'c'), NULL)

Another alternative notation:

a_list %>%
`[<-`(c('a', 'b'), NULL)

Remove an element of a list by name

6 Answers6