下面是如何在R中删除列表的最后一个元素:

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

如果x可能是一个向量，那么你需要创建一个新对象:

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

处理列表和向量

这个怎么样?还是使用下标

> 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

我想补充的是，如果它是一个命名列表，你可以简单地使用。

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

所以你可以覆盖原来的列表

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

从列表l中删除名为a的元素。

还可以使用magrittr包的提取函数从列表中进行负索引以删除列表项。

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

purrr包中有几个选项没有被提到:

Pluck和assign_in可以很好地使用嵌套值，你可以使用名称和/或索引的组合来访问它:

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

或者你可以通过分配zap()或NULL使用modify_list删除值:

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

你可以使用带有discard和keep的谓词函数来删除或保留元素:

# 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

在命名列表的情况下，我发现这些helper函数很有用

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