How can I remove an element from a list?

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

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.

Hope that helps, or at least leads you in the right direction.

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.

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

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

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

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

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

There are a few options in the purrr package that haven't been mentioned:

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

library(purrr)

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

# 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

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

# remove numeric elements
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

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