Question

我试图找到一种优雅的方式来处理R中的列表结构。特别是，在这种情况下，我想从列表中提取子元素，根据它们的关联来修改它们该列表中的数据，并将它们连接到数据框中。举个例子可能更容易：

mystruct <- structure(list(dataset1 = structure(list(data1 = structure(list(
    a = c(1, 2, 3), b = c(4, 5, 6)), .Names = c("a", "b"), row.names = c(NA, 
-3L), class = "data.frame"), data2 = c("a", "b", "c", "d", "e"
)), .Names = c("data1", "data2")), dataset2 = structure(list(
    data1 = structure(list(a = c(7, 8, 9), b = c(10, 11, 12)), .Names = c("a", 
    "b"), row.names = c(NA, -3L), class = "data.frame"), data2 = c("f", 
    "g", "h", "i", "j")), .Names = c("data1", "data2"))), .Names = c("dataset1", 
"dataset2"))

我可以像这样连接data1元素：

> mystruct %>% map_dfr(~.x$data1)
  a  b
1 1  4
2 2  5
3 3  6
4 7 10
5 8 11
6 9 12

但我想添加一个＆＃34;数据集＆＃34;列，由填充数据的列表元素的名称填充：

  dataset    a  b
1 dataset1   1  4
2 dataset1   2  5
3 dataset1   3  6
4 dataset2   7 10
5 dataset2   8 11
6 dataset2   9 12

有没有办法与tidyverse很好地做到这一点？我也对data.table解决方案持开放态度。

谢谢，阿利

Answer 1

为.id提供map_df参数，该参数将创建一个列，列出名称：

map_df(mystruct, 'data1', .id='dataset')

#   dataset a  b
#1 dataset1 1  4
#2 dataset1 2  5
#3 dataset1 3  6
#4 dataset2 7 10
#5 dataset2 8 11
#6 dataset2 9 12

或map_dfr也应该有效：

map_dfr(mystruct, 'data1', .id='dataset')

Answer 2

map_dfr有一个.id参数：

mystruct %>% map_dfr(~ .x$data1, .id = "id")

，并提供：

        id a  b
1 dataset1 1  4
2 dataset1 2  5
3 dataset1 3  6
4 dataset2 7 10
5 dataset2 8 11
6 dataset2 9 12

Answer 3

重组为带有列表列的“整洁”表...

library(data.table)
tabstruct = rbindlist(lapply(mystruct, lapply, list), id = TRUE)
#         .id        data1     data2
# 1: dataset1 <data.frame> a,b,c,d,e
# 2: dataset2 <data.frame> f,g,h,i,j

然后“不再”数据1：

tabstruct[, rbindlist(setNames(data1, .id), id=TRUE)]

#         .id a  b
# 1: dataset1 1  4
# 2: dataset1 2  5
# 3: dataset1 3  6
# 4: dataset2 7 10
# 5: dataset2 8 11
# 6: dataset2 9 12

或者是不必要的数据2：

tabstruct[, .(val = unlist(data2)), by=.id]
#          .id val
#  1: dataset1   a
#  2: dataset1   b
#  3: dataset1   c
#  4: dataset1   d
#  5: dataset1   e
#  6: dataset2   f
#  7: dataset2   g
#  8: dataset2   h
#  9: dataset2   i
# 10: dataset2   j

Answer 4

以下是在list

中对多个数据集执行此操作的选项

map(c('data1', 'data2'), ~  
         map2_df(mystruct, .x, ~ .x[[.y]], .id = 'id'))
#[[1]]
#        id a  b
#1 dataset1 1  4
#2 dataset1 2  5
#3 dataset1 3  6
#4 dataset2 7 10
#5 dataset2 8 11
#6 dataset2 9 12

#[[2]]
# A tibble: 5 x 3
#  id    dataset1 dataset2
#  <chr> <chr>    <chr>   
#1 1     a        f       
#2 1     b        g       
#3 1     c        h       
#4 1     d        i       
#5 1     e        j

使用tidyverse修改，提取和连接列表子元素到R中的data.frame

4 个答案: