How can I remove an element from a list?

Question

I have a list and I want to remove a single element from it. How can I do this?

I've tried looking up what I think the obvious names for this function would be in the reference manual and I haven't found anything appropriate.

Answer 1

tidyverse

There are several options from the purrr package:

Easily keep or discard by name or index

discard_at and keep_at allows you to keep elements by name OR position:

library(purrr)

l <- list("a" = 1:2, "b" = 3:4, "d" = list("e" = 5:6, "f" = 7:8))

discard_at(l, "d")
# $a
# [1] 1 2
# 
# $b
# [1] 3 4

keep_at(l, c("a", "b"))
# $a
# [1] 1 2
# 
# $b
# [1] 3 4

Keep or discard with combination of name and index (works for nested lists)

pluck and assign_in work well with nested values and you can access it using a combination of names and/or indices:

# select values (by name and/or index)
all.equal(pluck(l, "d", "e"), pluck(l, 3, "e"), pluck(l, 3, 1))
[1] TRUE

# or if element location stored in a vector use !!!
pluck(l, !!! as.list(c("d", "e")))
[1] 5 6

# remove values (modifies in place)
pluck(l, "d", "e") <- NULL

# assign_in to remove values with name and/or index (does not modify in place)
assign_in(l, list("d", 1), NULL)
$a
[1] 1 2

$b
[1] 3 4

$d
$d$f
[1] 7 8

Or you can remove values using modify_list by assigning zap() or NULL:

all.equal(list_modify(l, a = zap()), list_modify(l, a = NULL))
[1] TRUE

Keep or discard using predicate function

You can remove or keep elements using a predicate function with discard and keep:

# remove numeric elements ("d" is a list)
discard(l, is.numeric)
$d
$d$e
[1] 5 6

$d$f
[1] 7 8

# keep numeric elements
keep(l, is.numeric)
$a
[1] 1 2

$b
[1] 3 4

Answer 2

Assume that 'remove' means making it 'NULL' without changing the overall order.

mylist <- list("a", "b", "c", "d", "e", "f", "g")

# remove rule (or)
# 1. remove letter "e" & "f"
# 2. remove the 2nd value

mapply(
  \(x, y) if (x %in% c("e", "f") | y == 2) { NULL } else { x },
  mylist,
  1:length(mylist)
)

# output:
# [[1]]
# [1] "a"
# 
# [[2]]
# NULL
# 
# [[3]]
# [1] "c"
# 
# [[4]]
# [1] "d"
# 
# [[5]]
# NULL
# 
# [[6]]
# NULL
# 
# [[7]]
# [1] "g"

Answer 3

If you don't want to modify the list in-place (e.g. for passing the list with an element removed to a function), you can use indexing: negative indices mean "don't include this element".

x <- list("a", "b", "c", "d", "e")  # example list

x[-2]        # without 2nd element

x[-c(2, 3)]  # without 2nd and 3rd

Also, logical index vectors are useful:

x[x != "b"]  # without elements that are "b"

This works with dataframes, too:

df <- data.frame(number = 1:5, name = letters[1:5])

df[df$name != "b", ]      # rows without "b"

df[df$number %% 2 == 1, ] # rows with odd numbers only

Answer 4

If you only want to remove the first occurrence of the element "b" and leave the rest

x <- c("a", "b", "b", "c", "d", "e")

which(x == "b")
# [1] 2 3

which(x == "b")[1]
# [1] 2

x[-which(x == "b")[1]]
# [1] "a" "b" "c" "d" "e"

Answer 5

Using lapply and grep:

lst <- list(a = 1:4, b = 4:8, c = 8:10)
# say you want to remove a and c
toremove<-c("a","c")
lstnew<-lst[-unlist(lapply(toremove, function(x) grep(x, names(lst)) ) ) ]
# or
pattern<-"a|c"
lstnew<-lst[-grep(pattern, names(lst))]
# or
lst %>% purrr::discard(names(.) == "a") # use %in% for a set

Answer 6

I don't know R at all, but a bit of creative googling led me here: http://tolstoy.newcastle.edu.au/R/help/05/04/1919.html

The key quote from there:

I do not find explicit documentation for R on how to remove elements from lists, but trial and error tells me

myList[[5]] <- NULL

will remove the 5th element and then "close up" the hole caused by deletion of that element. That suffles the index values, So I have to be careful in dropping elements. I must work from the back of the list to the front.

A response to that post later in the thread states:

For deleting an element of a list, see R FAQ 7.1

And the relevant section of the R FAQ says:

... Do not set x[i] or x[[i]] to NULL, because this will remove the corresponding component from the list.

Which seems to tell you (in a somewhat backwards way) how to remove an element.

Answer 7

Here is a simple solution that can be done using base R. It removes the number 5 from the original list of numbers. You can use the same method to remove whatever element you want from a list.

#the original list
original_list = c(1:10)

#the list element to remove
remove = 5

#the new list (which will not contain whatever the `remove` variable equals)
new_list = c()

#go through all the elements in the list and add them to the new list if they don't equal the `remove` variable
counter = 1
for (n in original_list){
  if (n != ){
    new_list[[counter]] = n
    counter = counter + 1
  }
}

The new_list variable no longer contains 5.

new_list
# [1]  1  2  3  4  6  7  8  9 10

Answer 8

You can also negatively index from a list using the extract function of the magrittr package to remove a list item.

a <- seq(1,5)
b <- seq(2,6)
c <- seq(3,7)
l <- list(a,b,c)

library(magrittr)

extract(l,-1) #simple one-function method
[[1]]
[1] 2 3 4 5 6

[[2]]
[1] 3 4 5 6 7

Answer 9

if you'd like to avoid numeric indices, you can use

a <- setdiff(names(a),c("name1", ..., "namen"))

to delete names namea...namen from a. this works for lists

> l <- list(a=1,b=2)
> l[setdiff(names(l),"a")]
$b
[1] 2

as well as for vectors

> v <- c(a=1,b=2)
> v[setdiff(names(v),"a")]
b 
2

Answer 10

Use - (Negative sign) along with position of element, example if 3rd element is to be removed use it as your_list[-3]

Input

my_list <- list(a = 3, b = 3, c = 4, d = "Hello", e = NA)
my_list
# $`a`
# [1] 3

# $b
# [1] 3

# $c
# [1] 4

# $d
# [1] "Hello"

# $e
# [1] NA

Remove single element from list

 my_list[-3]
 # $`a`
 # [1] 3

 # $b
 # [1] 3

 # $d
 # [1] "Hello"

 # $e
 [1] NA

Remove multiple elements from list

 my_list[c(-1,-3,-2)]
 # $`d`
 # [1] "Hello"

 # $e
 # [1] NA

---

 my_list[c(-3:-5)]
 # $`a`
 # [1] 3

 # $b
 # [1] 3

---

 my_list[-seq(1:2)]
 # $`c`
 # [1] 4

 # $d
 # [1] "Hello"

 # $e
 # [1] NA

Answer 11

I would like to add that if it's a named list you can simply use within.

l <- list(a = 1, b = 2)    
> within(l, rm(a))
$b
[1] 2

So you can overwrite the original list

l <- within(l, rm(a)) 

to remove element named a from list l.

Answer 12

In the case of named lists I find those helper functions useful

member <- function(list,names){
    ## return the elements of the list with the input names
    member..names <- names(list)
    index <- which(member..names %in% names)
    list[index]    
}


exclude <- function(list,names){
     ## return the elements of the list not belonging to names
     member..names <- names(list)
     index <- which(!(member..names %in% names))
    list[index]    
}  
aa <- structure(list(a = 1:10, b = 4:5, fruits = c("apple", "orange"
)), .Names = c("a", "b", "fruits"))

> aa
## $a
##  [1]  1  2  3  4  5  6  7  8  9 10

## $b
## [1] 4 5

## $fruits
## [1] "apple"  "orange"


> member(aa,"fruits")
## $fruits
## [1] "apple"  "orange"


> exclude(aa,"fruits")
## $a
##  [1]  1  2  3  4  5  6  7  8  9 10

## $b
## [1] 4 5

Answer 13

Don't know if you still need an answer to this but I found from my limited (3 weeks worth of self-teaching R) experience with R that, using the NULL assignment is actually wrong or sub-optimal especially if you're dynamically updating a list in something like a for-loop.

To be more precise, using

myList[[5]] <- NULL

will throw the error

myList[[5]] <- NULL : replacement has length zero

or

more elements supplied than there are to replace

What I found to work more consistently is

myList <- myList[[-5]]

Answer 14

Just wanted to quickly add (because I didn't see it in any of the answers) that, for a named list, you can also do l["name"] <- NULL. For example:

l <- list(a = 1, b = 2, cc = 3)
l['b'] <- NULL

Answer 15

You can use which.

x<-c(1:5)
x
#[1] 1 2 3 4 5
x<-x[-which(x==4)]
x
#[1] 1 2 3 5

Answer 16

How about this? Again, using indices

> m <- c(1:5)
> m
[1] 1 2 3 4 5

> m[1:length(m)-1]
[1] 1 2 3 4

or

> m[-(length(m))]
[1] 1 2 3 4

Answer 17

There's the rlist package (http://cran.r-project.org/web/packages/rlist/index.html) to deal with various kinds of list operations.

Example (http://cran.r-project.org/web/packages/rlist/vignettes/Filtering.html):

library(rlist)
devs <- 
  list(
    p1=list(name="Ken",age=24,
      interest=c("reading","music","movies"),
      lang=list(r=2,csharp=4,python=3)),
    p2=list(name="James",age=25,
      interest=c("sports","music"),
      lang=list(r=3,java=2,cpp=5)),
    p3=list(name="Penny",age=24,
      interest=c("movies","reading"),
      lang=list(r=1,cpp=4,python=2)))

list.remove(devs, c("p1","p2"))

Results in:

# $p3
# $p3$name
# [1] "Penny"
# 
# $p3$age
# [1] 24
# 
# $p3$interest
# [1] "movies"  "reading"
# 
# $p3$lang
# $p3$lang$r
# [1] 1
# 
# $p3$lang$cpp
# [1] 4
# 
# $p3$lang$python
# [1] 2

Answer 18

If you have a named list and want to remove a specific element you can try:

lst <- list(a = 1:4, b = 4:8, c = 8:10)

if("b" %in% names(lst)) lst <- lst[ - which(names(lst) == "b")]

This will make a list lst with elements a, b, c. The second line removes element b after it checks that it exists (to avoid the problem @hjv mentioned).

or better:

lst$b <- NULL

This way it is not a problem to try to delete a non-existent element (e.g. lst$g <- NULL)

Answer 19

Removing Null elements from a list in single line :

x=x[-(which(sapply(x,is.null),arr.ind=TRUE))]

Cheers

Answer 20

Here is how the remove the last element of a list in R:

x <- list("a", "b", "c", "d", "e")
x[length(x)] <- NULL

If x might be a vector then you would need to create a new object:

x <- c("a", "b", "c", "d", "e")
x <- x[-length(x)]

• Work for lists and vectors