R: removing NULL elements from a list

R

R Problem Overview


mylist <- list(NULL, NULL, NULL, NULL, NULL, NULL, NULL, NULL, NULL, NULL, 
    123, NULL, 456)

> mylist
[[1]]
NULL

[[2]]
NULL

[[3]]
NULL

[[4]]
NULL

[[5]]
NULL

[[6]]
NULL

[[7]]
NULL

[[8]]
NULL

[[9]]
NULL

[[10]]
NULL

[[11]]
[1] 123

[[12]]
NULL

[[13]]
[1] 456

My list has 13 elements, 11 of which are NULL. I would like to remove them, but preserve the indices of the elements that are nonempty.

mylist2 = mylist[-which(sapply(mylist, is.null))]
> mylist2
[[1]]
[1] 123

[[2]]
[1] 456

This removes the NULL elements just fine, but I don't want the nonempty elements to be reindexed, i.e, I want mylist2 to look something like this, where the indices of the nonempty entries are preserved.

> mylist2
[[11]]
[1] 123

[[13]]
[1] 456

R Solutions


Solution 1 - R

The closest you'll be able to get is to first name the list elements and then remove the NULLs.

names(x) <- seq_along(x)

## Using some higher-order convenience functions
Filter(Negate(is.null), x)
# $`11`
# [1] 123
# 
# $`13`
# [1] 456

# Or, using a slightly more standard R idiom
x[sapply(x, is.null)] <- NULL
x
# $`11`
# [1] 123
# 
# $`13`
# [1] 456

Solution 2 - R

Simply use mylist[lengths(mylist) != 0].

Function lengths() was introduced in R 3.2.0 (April 2015).

Solution 3 - R

The purrr package, included in Tidyverse, has elegant and fast functions for working with lists:

require(tidyverse)

# this works
compact(mylist)

# or this
mylist %>% discard(is.null)
 
# or this
# pipe "my_list" data object into function "keep()", make lambda function inside "keep()" to return TRUE FALSE.
mylist %>% keep( ~ !is.null(.) )
 

All above options are from Purrr. Output is:

[[1]] 
[1] 123

[[2]] 
[1] 456

Note: compact() was in plyr, but dplyr superseded plyr, and compact() stayed around but moved to purrr. Anyway, all the functions are within the parent package tidyverse.



Here's a link to the Purrr cheat sheet download:

https://rstudio.com/resources/cheatsheets/

Or to view the Purrr cheatsheet directly in a browser:

https://evoldyn.gitlab.io/evomics-2018/ref-sheets/R_purrr.pdf

Solution 4 - R

There's a function that automatically removes all the null entries of a list, and if the list is named, it maintains the names of the non-null entries.

This function is called compact from the package plyr.

l <- list( NULL, NULL, foo, bar)
names(l) <- c( "one", "two", "three", "four" )

plyr::compact(l)

If you want to preserve the indexes of the non-null entries, you can name the list as it is done in the post before and then compact your list:

names(l) <- seq_along(l)
plyr::compact(l)

Solution 5 - R

If you want to keep the names you can do

a <- list(NULL, NULL, NULL, NULL, NULL, NULL, NULL, NULL, NULL, NULL, 
          123, NULL, 456)
non_null_names <- which(!sapply(a, is.null))
a <- a[non_null_names]
names(a) <- non_null_names
a

You can then access the elements like so

a[['11']]
num <- 11
a[[as.character(num)]]
a[[as.character(11)]]
a$`11`

You can't get them in the neat [[11]], [[13]] notation, though, because those represent numerical indices.

Solution 6 - R

This solution works with nested list as well

rlist::list.clean(myNestedlist ,recursive = T)

Solution 7 - R

Here it is with convenient chaining notation

library(magrittr)

mylist %>%
  setNames(seq_along(.)) %>%
  Filter(. %>% is.null %>% `!`, .)

Solution 8 - R

here's a very simple way to do it using only base R functions:

names(mylist) <- 1:length(mylist)
mylist2 <- mylist[which(!sapply(mylist, is.null))]

Attributions

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionAdrianView Question on Stackoverflow
Solution 1 - RJosh O'BrienView Answer on Stackoverflow
Solution 2 - RF. PrivéView Answer on Stackoverflow
Solution 3 - RHayward ObladView Answer on Stackoverflow
Solution 4 - RDavid Rodriguez LoresView Answer on Stackoverflow
Solution 5 - RFelipe GerardView Answer on Stackoverflow
Solution 6 - RAdam Lee PerelmanView Answer on Stackoverflow
Solution 7 - RbramtaylView Answer on Stackoverflow
Solution 8 - RitsmisterbrownView Answer on Stackoverflow