4

在 R 中,我需要一个有效的解决方案来打乱列表中包含的元素,保留元素的总数和局部元素的大小(在这种情况下,列表的每个元素都是一个向量)

a<-LETTERS[1:6]
b<-LETTERS[6:10]
c<-LETTERS[c(9:15)]

l=list(a,b,c)
> l
[[1]]
[1] "A" "B" "C" "D" "E" "F"

[[2]]
[1] "F" "G" "H" "I" "J"

[[3]]
[1] "I" "J" "K" "L" "M" "N" "O"

改组应该随机选择列表中的字母(无需替换)并将它们放在列表中任何向量的随机位置。

我希望我已经清楚了!谢谢 :-)

4

2 回答 2

10

您可以尝试使用第一个列表的骨架重新创建第二个列表,并用第一个列表的所有元素填充它,如下所示:

u<-unlist(l)
l2<-relist(u[sample(length(u))],skeleton=l)
> l2
[[1]]
[1] "F" "A" "O" "I" "S" "Q"

[[2]]
[1] "R" "P" "K" "F" "G"

[[3]]
 [1] "A" "N" "M" "J" "H" "G" "E" "B" "T" "C" "D" "L"

希望这可以帮助!

于 2013-08-23T18:20:18.620 回答
3

像这样...?

> set.seed(1)
> lapply(l, sample)
[[1]]
[1] "B" "F" "C" "D" "A" "E"

[[2]]
[1] "J" "H" "G" "F" "I"

[[3]]
[1] "J" "M" "O" "L" "N" "K" "I"
于 2013-08-23T18:22:30.257 回答