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）中添加一个列，以便第四列指示组件的名称。这是我要寻找的例子。

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)))

我尝试过

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

但无济于事。我了解为什么以上是不起作用的，但无法找到不同的解决方案。感谢您的帮助！

Answer 1

我们可以使用imap

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

或基本r与map

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

的

$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

以lapply 序列或名称

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