list 如何从R中的列表中进行子集
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/9624169/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How to subset from a list in R
提问by user1257894
I have a rather simple task but haven't find a good solution.
我有一个相当简单的任务，但还没有找到一个好的解决方案。
> mylist
[[1]]
[1] 1 2 3 4 5 6 7 8 9 10
[[2]]
[1] "a" "b" "c" "d" "e" "f" "g" "h" "i" "j" "k" "l" "m" "n" "o" "p" "q" "r" "s" "t" "u" "v" "w" "x" "y" "z"
[[3]]
[1] 25 26 27 28 29 30 31 32
y <- c(3,5,9)
I would like to extract from mylistthe sub-elements 3,5, and 9 of each component in the list.
I have tried, sapply[mylist,"[[",y]but not luck!, and others like vapply, lapply, etc.. Thanks in advance for your help
我想从mylist 中提取列表中每个组件的子元素 3,5 和 9。我已经尝试过了，sapply[mylist,"[[",y]但运气不佳！还有其他人喜欢 vapply、lapply 等。在此先感谢您的帮助
Mauricio Ortiz
毛里西奥·奥尔蒂斯
回答by sgibb
You could use sapply(mylist, "[", y):
你可以使用sapply(mylist, "[", y)：
mylist <- list(1:5, 6:10, 11:15)
sapply(mylist, "[", c(2,3))
回答by joran
Try using [instead of [[(and depending on what you're after you light actually want lapply).
尝试使用[而不是[[（并取决于您点亮后实际想要的东西lapply）。
From ?'[[':
来自?'[['：
The most important distinction between [, [[ and $ is that the [ can select more than one element whereas the other two select a single element.
[、[[ 和 $ 之间最重要的区别是 [ 可以选择多个元素，而其他两个选择单个元素。
回答by USER_1
Using lapply:
使用lapply：
# create mylist
list1<-1:10
list2<-letters[1:26]
list3<-25:32
mylist<-list(list1,list2,list3)
# select 3,5,9th element from each list
list.2 <- lapply(mylist, function(x) {x[c(3,5,9)]})
回答by DanO
purrrprovides another solution for solving these kinds of list manipulations within the tidyverse
purrr提供了另一种解决方案来解决tidyverse 中的这些类型的列表操作
library(purrr)
library(dplyr)
desired_values <- c(1,3)
mylist <- list(1:5, letters[1:6], 11:15) %>%
purrr::map(`[`,desired_values)
mylist
回答by toby544
I don't think sgibb's answer gives what you would want. I suggest making a new function:
我不认为 sgibb 的回答给出了你想要的。我建议创建一个新功能：
subsetList <- function(myList, elementNames) {
lapply(elementNames, FUN=function(x) myList[[x]])
}
Then you can use it like this:
然后你可以像这样使用它：
x <- list(a=3, b="hello", c=4.5, d="world")
subsetList(x, c("d", "a"))
subsetList(x, c(4, 1))
These both give
这些都给
[[1]]
[1] "world"
[[2]]
[1] 3
which is what you would want, I think.
这就是你想要的，我想。
回答by Umaomamaomao
An easy way to subset repeated named elements of a list, similar to other answers here.
一种对列表的重复命名元素进行子集化的简单方法，类似于此处的其他答案。
(so I can find it next time I look this question up)
（所以我下次查这个问题时可以找到它）
E.g., subset the "b"elements from a repeating list where each element includes an "a"and "b"sub-element:
例如，"b"从重复列表中对元素进行子集化，其中每个元素都包含一个"a"和"b"子元素：
mylist <- list(
list(
"a" = runif(3),
"b" = runif(1)
),
list(
"a" = runif(3),
"b" = runif(1)
)
)
mylist
#> [[1]]
#> [[1]]$a
#> [1] 0.7547490 0.6528348 0.2339767
#>
#> [[1]]$b
#> [1] 0.8815888
#>
#>
#> [[2]]
#> [[2]]$a
#> [1] 0.51352909 0.09637425 0.99291650
#>
#> [[2]]$b
#> [1] 0.8407162
blist <- lapply(
X = mylist,
FUN = function(x){x[["b"]]}
)
blist
#> [[1]]
#> [1] 0.8815888
#>
#> [[2]]
#> [1] 0.8407162
Created on 2019-11-06 by the reprex package(v0.3.0)
由reprex 包(v0.3.0)于 2019 年 11 月 6 日创建
回答by user1234357
There are better ways of doing this, but here's a quick solution.
有更好的方法可以做到这一点，但这里有一个快速的解决方案。
# your values
list1<-1:10
list2<-letters[1:26]
list3<-25:32
# put 'em together in a list
mylist<-list(list1,list2,list3)
# function
foo<-function(x){x[c(3,5,9)]}
# apply function to each of the element in the list
foo(mylist[[1]])
foo(mylist[[2]])
foo(mylist[[3]])
# check the output
> foo(mylist[[1]])
[1] 3 5 9
> foo(mylist[[2]])
[1] "c" "e" "i"
> foo(mylist[[3]])
[1] 27 29 NA
