我有一个列表，我想从其中删除一个元素。我该怎么做呢?
我试着在参考手册中查找我认为这个函数的明显名称，但我没有找到任何合适的名称。
当前回答
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
其他回答
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
使用-(负号)与元素的位置一起，例如，如果要删除第三个元素，则使用your_list[-3]
输入
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
从列表中删除单个元素
my_list[-3]
# $`a`
# [1] 3
# $b
# [1] 3
# $d
# [1] "Hello"
# $e
[1] NA
从列表中删除多个元素
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
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
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))]
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
