Question

我正在寻找一种tidyverse / purrr方法来向列表中添加元素。例如：

library(tidyverse)

l <- list(c("a", "a", "b"), c("c", "d"), c("e", "f", "e", "g"))
l
[[1]]
[1] "a" "a" "b"

[[2]]
[1] "c" "d"

[[3]]
[1] "e" "f" "e" "g"

如何构建管道%>%以返回如下内容：

desired <- list(
  list(vec = c("a", "a", "b"), length = 3, unique = 2),
  list(vec = c("c", "d"), length = 2, unique = 2),
  list(vec = c("e", "f", "e", "g"), length = 4, unique = 3)
)

desired
[[1]]
[[1]]$vec
[1] "a" "a" "b"

[[1]]$length
[1] 3

[[1]]$unique
[1] 2


[[2]]
[[2]]$vec
[1] "c" "d"

[[2]]$length
[1] 2

[[2]]$unique
[1] 2


[[3]]
[[3]]$vec
[1] "e" "f" "e" "g"

[[3]]$length
[1] 4

[[3]]$unique
[1] 3

我知道我可以使用l %>% map(length)或l %>% map(unique)映射单个函数，但是我想在列表中添加新元素并在一个管道中完成。

Answer 1

l <- list(c("a", "a", "b"), c("c", "d"), c("e", "f", "e", "g"))

Purr方法：

l %>% 
  map(~ list(vec = .,
             length = length(.),
             unique = length(unique(.))))

输出

[[1]]
[[1]]$`vec`
[1] "a" "a" "b"

[[1]]$length
[1] 3

[[1]]$unique
[1] 2


[[2]]
[[2]]$`vec`
[1] "c" "d"

[[2]]$length
[1] 2

[[2]]$unique
[1] 2


[[3]]
[[3]]$`vec`
[1] "e" "f" "e" "g"

[[3]]$length
[1] 4

[[3]]$unique
[1] 3

但是我会尽可能地坚持使用数据帧（小标题）：

l %>%
  tibble(vec = .) %>%
  mutate(length = map_dbl(vec, length),
         unique = map_dbl(vec, ~length(unique(.))))

输出：

# A tibble: 3 x 3
  vec       length unique
  <list>     <dbl>  <dbl>
1 <chr [3]>      3      2
2 <chr [2]>      2      2
3 <chr [4]>      4      3

tidyverse purrr方法向列表中添加元素

1 个答案: