如何根据每个组件的名称在列表的所有组件中创建一个新列？

Question

这是一个示例列表：

list1 <- list(a = structure(list(a = c(1, 2, 4, 5), b = c(1, 1, 1, 1), 
    d = c(2, 2, 2, 2)), class = "data.frame", row.names = c(NA, 
-4L)), b = structure(list(a = c(1, 2, 4, 5), b = c(1, 1, 1, 1
), d = c(2, 2, 2, 2)), class = "data.frame", row.names = c(NA, 
-4L)), d = structure(list(a = c(1, 2, 4, 5), b = c(1, 1, 1, 1
), d = c(2, 2, 2, 2)), class = "data.frame", row.names = c(NA, 
-4L)))

我想在列表的每个组件（即 a、b、d）中添加一列，以便第四列指示组件的名称。这是我正在寻找的示例。

list2 <- list(a = structure(list(a = c(1, 2, 4, 5), b = c(1, 1, 1, 1), 
    d = c(2, 2, 2, 2), name = structure(c(1L, 1L, 1L, 1L), class = "factor", .Label = "a")), class = "data.frame", row.names = c(NA, 
-4L)), b = structure(list(a = c(1, 2, 4, 5), b = c(1, 1, 1, 1
), d = c(2, 2, 2, 2), name = structure(c(1L, 1L, 1L, 1L), class = "factor", .Label = "b")), class = "data.frame", row.names = c(NA, 
-4L)), d = structure(list(a = c(1, 2, 4, 5), b = c(1, 1, 1, 1
), d = c(2, 2, 2, 2), name = structure(c(1L, 1L, 1L, 1L), class = "factor", .Label = "c")), class = "data.frame", row.names = c(NA, 
-4L)))

我试过

的变体

list2 <- lapply(list1, function(x) cbind(list1, name = names(x[1])))

但无济于事。我明白为什么上面的方法不起作用，但无法找到不同的解决方案。感谢您的帮助！

Answer 1

我们可以用imap

library(purrr)
library(dplyr)
imap(list1, ~ .x %>%
           mutate(name = .y))

或在 base R 中与 Map

Map(cbind, list1, name = names(list1))

-输出

$a
  a b d name
1 1 1 2    a
2 2 1 2    a
3 4 1 2    a
4 5 1 2    a

$b
  a b d name
1 1 1 2    b
2 2 1 2    b
3 4 1 2    b
4 5 1 2    b

$d
  a b d name
1 1 1 2    d
2 2 1 2    d
3 4 1 2    d
4 5 1 2    d

使用lapply，可以通过遍历序列或名称来完成

lapply(names(list1), function(x) cbind(list1[[x]], name = x))

如何根据每个组件的名称在列表的所有组件中创建一个新列？

How to create a new column in all components of a list based on the name of each component?

r

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