我有一个很大的列表，其中包含四个不同结构的数据框元素。每个数据框都有一个ID变量，可以多次列出。
下面是我的列表的简单版本。
set.seed(123)
fruit <- data.frame(
id = rep(1:10, 3),
crop = rep(c("apple", "pear", "banana"), each = 10),
consumed = sample(0:20, 30, replace=TRUE),
sold = sample(1:10, 30, replace = TRUE))
veg <- data.frame(
id = rep(1:10, 2),
crop = rep(c("potatoe", "onion"), each = 10),
consumed = sample(0:20, 20, replace=TRUE),
gifted = sample(0:10, 20, replace=TRUE)
)
milk <- data.frame(
id = 1:10,
item = "milk",
consumed = sample(1:6, 10, replace = T),
produced = sample(1:20, 10, replace = T),
processed = sample(0.5:3.5, 10, replace = T),
sold = sample(0.5:3.5, 10, replace = T))
food <- list(fruit, veg, milk)
food
我想按 ID 拆分列表，其中 ID 是第一级，水果、蔬菜和牛奶成为第二级，或者找到另一种方法按 ID 跨列表元素进行分组。基本上，我希望一个特定 ID 的所有内容都集中在一个地方，以应用我编写的函数来汇总该特定 ID 的所有数据。
我尝试了一些地图选项。
也许你正在追求这个？
lst <- unlist(lapply(food, split, ~id), recursive = FALSE)
split(lst, names(lst))
这给出了
$`1`
$`1`$`1`
id crop consumed sold
1 1 apple 14 9
11 1 pear 4 5
21 1 banana 13 4
$`1`$`1`
id crop consumed gifted
1 1 potatoe 1 2
11 1 onion 11 2
$`1`$`1`
id item consumed produced processed sold
1 1 milk 6 14 2.5 3.5
$`10`
$`10`$`10`
id crop consumed sold
10 10 apple 13 2
20 10 pear 3 10
30 10 banana 6 1
$`10`$`10`
id crop consumed gifted
10 10 potatoe 15 5
20 10 onion 7 9
$`10`$`10`
id item consumed produced processed sold
10 10 milk 4 20 3.5 3.5
$`2`
$`2`$`2`
id crop consumed sold
2 2 apple 18 9
12 2 pear 18 8
22 2 banana 16 6
$`2`$`2`
id crop consumed gifted
2 2 potatoe 3 9
12 2 onion 13 7
$`2`$`2`
id item consumed produced processed sold
2 2 milk 6 19 2.5 1.5
$`3`
$`3`$`3`
id crop consumed sold
3 3 apple 13 10
13 3 pear 8 2
23 3 banana 10 8
$`3`$`3`
id crop consumed gifted
3 3 potatoe 12 1
13 3 onion 2 2
$`3`$`3`
id item consumed produced processed sold
3 3 milk 3 7 3.5 2.5
$`4`
$`4`$`4`
id crop consumed sold
4 4 apple 2 7
14 4 pear 2 1
24 4 banana 6 6
$`4`$`4`
id crop consumed gifted
4 4 potatoe 4 9
14 4 onion 13 7
$`4`$`4`
id item consumed produced processed sold
4 4 milk 2 9 0.5 3.5
$`5`
$`5`$`5`
id crop consumed sold
5 5 apple 9 5
15 5 pear 7 9
25 5 banana 20 6
$`5`$`5`
id crop consumed gifted
5 5 potatoe 18 1
15 5 onion 6 0
$`5`$`5`
id item consumed produced processed sold
5 5 milk 5 7 3.5 3.5
$`6`
$`6`$`6`
id crop consumed sold
6 6 apple 17 7
16 6 pear 6 9
26 6 banana 11 7
$`6`$`6`
id crop consumed gifted
6 6 potatoe 19 9
16 6 onion 2 6
$`6`$`6`
id item consumed produced processed sold
6 6 milk 6 2 1.5 3.5
$`7`
$`7`$`7`
id crop consumed sold
7 7 apple 10 5
17 7 pear 9 6
27 7 banana 14 1
$`7`$`7`
id crop consumed gifted
7 7 potatoe 13 5
17 7 onion 14 10
$`7`$`7`
id item consumed produced processed sold
7 7 milk 5 16 1.5 2.5
$`8`
$`8`$`8`
id crop consumed sold
8 8 apple 4 6
18 8 pear 8 5
28 8 banana 9 6
$`8`$`8`
id crop consumed gifted
8 8 potatoe 2 3
18 8 onion 20 6
$`8`$`8`
id item consumed produced processed sold
8 8 milk 3 13 3.5 0.5
$`9`
$`9`$`9`
id crop consumed sold
9 9 apple 19 9
19 9 pear 18 9
29 9 banana 12 2
$`9`$`9`
id crop consumed gifted
9 9 potatoe 7 0
19 9 onion 4 6
$`9`$`9`
id item consumed produced processed sold
9 9 milk 4 19 0.5 3.5
这里有一个选项，先使用
tidyr::nest()purrr::reduce()idmilkfruitvegnames(food) <- c("fruit", "veg", "milk")
lapply(names(food), \(f) nest(food[[f]], .by=id,.key =f )) %>%
reduce(full_join, by="id")
输出：
id fruit veg milk
<int> <list> <list> <list>
1 1 <tibble [3 × 3]> <tibble [2 × 3]> <tibble [1 × 5]>
2 2 <tibble [3 × 3]> <tibble [2 × 3]> <tibble [1 × 5]>
3 3 <tibble [3 × 3]> <tibble [2 × 3]> <tibble [1 × 5]>
4 4 <tibble [3 × 3]> <tibble [2 × 3]> <tibble [1 × 5]>
5 5 <tibble [3 × 3]> <tibble [2 × 3]> <tibble [1 × 5]>
6 6 <tibble [3 × 3]> <tibble [2 × 3]> <tibble [1 × 5]>
7 7 <tibble [3 × 3]> <tibble [2 × 3]> <tibble [1 × 5]>
8 8 <tibble [3 × 3]> <tibble [2 × 3]> <tibble [1 × 5]>
9 9 <tibble [3 × 3]> <tibble [2 × 3]> <tibble [1 × 5]>
10 10 <tibble [3 × 3]> <tibble [2 × 3]> <tibble [1 × 5]>