将嵌套（3级）列表转换为长/高格式数据框

Question

我有一个嵌套列表，有3个级别：

m = list(try1 = list(list(court = c("jack", "queen", "king"),
                          suit = list(diamonds = 2, clubs = 5)), 
                     list(court = c("jack", "queen", "king"),
                          suit = list(diamonds = 45, clubs = 67))), 
         try2 = list(list(court = c("jack", "queen", "king"),
                          suit = list(diamonds = 400, clubs = 300)), 
                     list(court = c("jack", "queen", "king"),
                          suit = list(diamonds = 5000, clubs = 6000))))

> str(m)
List of 2
 $ try1:List of 2
  ..$ :List of 2
  .. ..$ court: chr [1:3] "jack" "queen" "king"
  .. ..$ suit :List of 2
  .. .. ..$ diamonds: num 2
  .. .. ..$ clubs   : num 5
  ..$ :List of 2
  .. ..$ court: chr [1:3] "jack" "queen" "king"
  .. ..$ suit :List of 2
  .. .. ..$ diamonds: num 45
  .. .. ..$ clubs   : num 67
 $ try2:List of 2
  ..$ :List of 2
  .. ..$ court: chr [1:3] "jack" "queen" "king"
  .. ..$ suit :List of 2
  .. .. ..$ diamonds: num 400
  .. .. ..$ clubs   : num 300
  ..$ :List of 2
  .. ..$ court: chr [1:3] "jack" "queen" "king"
  .. ..$ suit :List of 2
  .. .. ..$ diamonds: num 5000
  .. .. ..$ clubs   : num 6000

对于try1和try2中的每个子列表，我需要提取suit子列表并对其元素进行rbind，使得生成的数据框架为长格式，包含4列 - value（套装的值），suit（标识哪个适合价值来自，即钻石或俱乐部），iter（以确定诉讼所属的子名单，即1或2）和try（try1或try2）。

我可以使用expand.grid()和mapply()的组合实现这一目标：

grd = expand.grid(try = names(m), iter = 1:2, suit = c("diamonds", "clubs"))

grd$value = mapply(function(x, y, z) m[[x]][[y]]$suit[[z]], grd[[1]], grd[[2]], grd[[3]])

结果：

> grd
   try iter     suit value
1 try1    1 diamonds     2
2 try2    1 diamonds   400
3 try1    2 diamonds    45
4 try2    2 diamonds  5000
5 try1    1    clubs     5
6 try2    1    clubs   300
7 try1    2    clubs    67
8 try2    2    clubs  6000

但是，我想知道是否有更通用/简洁的方法来再现上述结果（最好是在基数R中）？我正在考虑从每个子列表中提取suit元素，然后在结果列表中递归使用stack()之类的东西：

rapply(m, function(x) setNames(stack(x), names(x)))

但这会引发一个错误，我不太清楚为什么，我不知道在它的位置使用什么。

Answer 1

我们可以使用map和melt的组合

library(purrr)
library(reshape2)
library(dplyr)
map_df(m, ~ .x %>%
                 map(pluck, "suit")  %>% 
                   melt, .id = 'try') 

---

或者与enframe和map

library(tibble)
map_df(m, ~ .x %>% 
              map_df(pluck, "suit") %>% 
                    map_df(~ enframe(.x, name = "iter") %>%
                       unnest, .id = "suit"), .id = 'try'  )
# A tibble: 8 x 4
#  try   suit      iter value
#  <chr> <chr>    <int> <dbl>
#1 try1  diamonds     1     2
#2 try1  diamonds     2    45
#3 try1  clubs        1     5
#4 try1  clubs        2    67
#5 try2  diamonds     1   400
#6 try2  diamonds     2  5000
#7 try2  clubs        1   300
#8 try2  clubs        2  6000