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。

使用lapply和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))]

只是想快速添加(因为我在任何答案中都没有看到它)，对于一个命名列表，你也可以做l["name"] <- NULL。例如:

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

rlist包(http://cran.r-project.org/web/packages/rlist/index.html)用于处理各种列表操作。

示例(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"))

结果:

# $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

单行从列表中删除Null元素:

x = x(((酸式焦磷酸钠(x, is.null) arr.ind = TRUE)))

干杯

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