如何将 `mutate_at` 用于 tibble 中的列表列？

Question

我有一个结构如下的小标题：

df <- 
  tibble(
    x = 1:3, 
    light_93 = list(1:3, 5:7, 18:20),
    light_94 = list(3:5, 9:11, 18:20),
    light_95 = list(5:7, 44:46, 30:32))

我想创建多个新列，给出每个 light_ 列列表的平均值。所以我想要这个结果：

out <- 
  df %>% 
  mutate(light_93_mean = map_dbl(light_93, mean),
         light_94_mean = map_dbl(light_94, mean),
         light_95_mean = map_dbl(light_95, mean))

我可以使用mutate_at 自动执行此操作吗？ （我有数百个列表列。）我不知道如何让它在一个小标题中工作。

Answer 1

在mutate_at 的vars 参数中指定要应用的列，然后在每个列中使用map 循环list 并获取mean

library(dplyr)
library(purrr)
df %>%
     mutate_at(vars(starts_with('light')), 
        list(mean = ~ map_dbl(., mean)))
# A tibble: 3 x 7
#      x light_93  light_94  light_95  light_93_mean light_94_mean light_95_mean
#  <int> <list>    <list>    <list>            <dbl>         <dbl>         <dbl>
#1     1 <int [3]> <int [3]> <int [3]>             2             4             6
#2     2 <int [3]> <int [3]> <int [3]>             6            10            45
#3     3 <int [3]> <int [3]> <int [3]>            19            19            31

---

或者使用带有across和mutate的devel版本

df %>% 
     mutate(across(starts_with('light'), ~ map_dbl(., mean), names = "{col}_mean"))
# A tibble: 3 x 7
#      x light_93  light_94  light_95  light_93_mean light_94_mean light_95_mean
#  <int> <list>    <list>    <list>            <dbl>         <dbl>         <dbl>
#1     1 <int [3]> <int [3]> <int [3]>             2             4             6
#2     2 <int [3]> <int [3]> <int [3]>             6            10            45
#3     3 <int [3]> <int [3]> <int [3]>            19            19            31

也可以应用不同功能的不同列集

df %>% 
    mutate(across(starts_with('light'), ~ map_dbl(., mean), names = "{col}_mean"),
           across(matches('(94|95)$'), ~ map_dbl(., sum), names = "{col}_sum"))
# A tibble: 3 x 9
#      x light_93  light_94  light_95  light_93_mean light_94_mean light_95_mean light_94_sum light_95_sum
#  <int> <list>    <list>    <list>            <dbl>         <dbl>         <dbl>        <dbl>        <dbl>
#1     1 <int [3]> <int [3]> <int [3]>             2             4             6           12           18
#2     2 <int [3]> <int [3]> <int [3]>             6            10            45           30          135
#3     3 <int [3]> <int [3]> <int [3]>            19            19            31           57           93

Answer 2

在base R中，我们可以使用grep选择以"light"开头的列，并计算每个列表的mean并添加为新列。

cols <- grep('^light', names(df), value = TRUE)
df[paste0(cols, "_mean")] <- lapply(df[cols], function(x) sapply(x, mean))
df

# A tibble: 3 x 7
#      x light_93  light_94  light_95  light_93_mean light_94_mean light_95_mean
#  <int> <list>    <list>    <list>            <dbl>         <dbl>         <dbl>
#1     1 <int [3]> <int [3]> <int [3]>             2             4             6
#2     2 <int [3]> <int [3]> <int [3]>             6            10            45
 #3    3 <int [3]> <int [3]> <int [3]>            19            19            31