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法律研究如何在 R 中组合列表？
当我们有多个列表，但它们的的数据类型相似时，我们可能需要合并这些列表。这将很有用，因为我们可以使用一个列表名称执行计算，而不是将它们应用于多个列表中。我们可以使用 mapply 函数合并多个列表。
示例
考虑以下列表 -
> List1<-list(letters[1:5],1:5,5:1,25,c("A","B","C","D"))
> List1
[[1]]
[1] "a" "b" "c" "d" "e"
[[2]]
[1] 1 2 3 4 5
[[3]]
[1] 5 4 3 2 1
[[4]]
[1] 25
[[5]]
[1] "A" "B" "C" "D"
> List2<-list(letters[6:10],6:10,c(25,16,9,4,1),36,c("Andy","Boston","Caroline","Dinesh"))
> List2
[[1]]
[1] "f" "g" "h" "i" "j"
[[2]]
[1] 6 7 8 9 10
[[3]]
[1] 25 16 9 4 1
[[4]]
[1] 36
[[5]]
[1] "Andy" "Boston" "Caroline" "Dinesh"
> mapply(c, List1, List2, SIMPLIFY=FALSE)
[[1]]
[1] "a" "b" "c" "d" "e" "f" "g" "h" "i" "j"
[[2]]
[1] 1 2 3 4 5 6 7 8 9 10
[[3]]
[1] 5 4 3 2 1 25 16 9 4 1
[[4]]
[1] 25 36
[[5]]
[1] "A" "B" "C" "D" "Andy" "Boston" "Caroline" "Dinesh"现在假设我们还有一个 List3，那么可以按如下所示将其合并 -
> List3<-list(letters[11:15],11:15,c(6,7,8,9,10),49,c("Aaron","Betty","Corie","Donald"))
> List3
[[1]]
[1] "k" "l" "m" "n" "o"
[[2]]
[1] 11 12 13 14 15
[[3]]
[1] 6 7 8 9 10
[[4]]
[1] 49
[[5]]
[1] "Aaron" "Betty" "Corie" "Donald"
> mapply(c, List1, List2, List3, SIMPLIFY=FALSE)
[[1]]
[1] "a" "b" "c" "d" "e" "f" "g" "h" "i" "j" "k" "l" "m" "n" "o"
[[2]]
[1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
[[3]]
[1] 5 4 3 2 1 25 16 9 4 1 6 7 8 9 10
[[4]]
[1] 25 36 49
[[5]]
[1] "A" "B" "C" "D" "Andy" "Boston" "Caroline" "Dinesh" "Aaron" "Betty" "Corie" "Donald"
更新于: 2020 年 8 月 11 日
312 次浏览
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