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R:如何重新排列列表对象

转载 作者:行者123 更新时间:2023-12-04 00:18:30 24 4
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mylist <- list(matrix(c(1, 2, 3, "red", "blue", "green", "apple", "banana", "orange"), ncol = 3, nrow = 3), 
matrix(c(2, 3, 9, "green", "blue", "green", "apple", "guava", "orange"), ncol = 3, nrow = 3),
matrix(c(43, 90, 1, "black", "red", "green", "orange", "watermelon", "orange"), ncol = 3, nrow = 3))
> mylist
[[1]]
[,1] [,2] [,3]
[1,] "1" "red" "apple"
[2,] "2" "blue" "banana"
[3,] "3" "green" "orange"

[[2]]
[,1] [,2] [,3]
[1,] "2" "green" "apple"
[2,] "3" "blue" "guava"
[3,] "9" "green" "orange"

[[3]]
[,1] [,2] [,3]
[1,] "43" "black" "orange"
[2,] "90" "red" "watermelon"
[3,] "1" "green" "orange"

mylist 中,我有 3 个子列表。每个子列表包含一个 3x3 矩阵,其中第一列包含数字,第二列包含颜色,第三列包含水果。

我想重新排列这些子列表,使第一个子列表中的矩阵包含数字,第二个包含颜色,第三个包含水果。也就是说,我希望它看起来像这样:

[[1]]
[,1] [,2] [,3]
[1,] 1 2 43
[2,] 2 3 90
[3,] 3 9 1

[[2]]
[,1] [,2] [,3]
[1,] "red" "green" "black"
[2,] "blue" "blue" "red"
[3,] "green" "green" "green"

[[3]]
[,1] [,2] [,3]
[1,] "apple" "apple" "orange"
[2,] "banana" "guava" "watermelon"
[3,] "orange" "orange" "orange"

我的直觉是使用 lapply 循环遍历矩阵的列并将它们存储在一个新列表中,但我不确定如何有效地执行此操作。

最佳答案

您可以绑定(bind)、按列拆分并转换回矩阵:

lapply(asplit(do.call(rbind, mylist), 2), matrix, 3)

[[1]]
[,1] [,2] [,3]
[1,] "1" "2" "43"
[2,] "2" "3" "90"
[3,] "3" "9" "1"

[[2]]
[,1] [,2] [,3]
[1,] "red" "green" "black"
[2,] "blue" "blue" "red"
[3,] "green" "green" "green"

[[3]]
[,1] [,2] [,3]
[1,] "apple" "apple" "orange"
[2,] "banana" "guava" "watermelon"
[3,] "orange" "orange" "orange"

关于R:如何重新排列列表对象,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/62378817/

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