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使用tidyverse将列表子元素修改,提取并连接到R中的data.frame

[英]Modify, extract, and concatenate list sub-elements into a data.frame in R with tidyverse

I'm trying to find an elegant way to work with list structures in R. In particular, in this case, I'd like to extract sub-elements from a list, modify them based on their associated data in that list, and concatenate them into a data frame. 我试图找到一种在R中使用列表结构的优雅方法。特别是在这种情况下,我想从列表中提取子元素,然后根据该列表中的关联数据对其进行修改,然后进行连接他们变成一个数据帧。 Perhaps easier with an example: 举个例子也许更容易:

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"))

I can concatenate data1 elements like this: 我可以像这样串联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

But I would like to add a "dataset" column, which is populated by the name of the list element from whence the data was taken: 但是我想添加一个“数据集”列,该列由获取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

Is there a way to do this nicely with the tidyverse? 有没有一种方法可以很好地使用tidyverse? I'd also be open to data.table solutions. 我也愿意接受data.table解决方案。

Thanks, Allie 谢谢,艾莉

Provide an .id parameter to map_df , which will create a column giving the name of the list: map_df提供一个.id参数,这将创建一个给出列表名称的列:

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

Or map_dfr should work as well: 或者map_dfr应该也可以工作:

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

map_dfr has an .id argument: map_dfr具有.id参数:

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

giving: 给予:

        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

Restructure as a "tidy" table with list columns... 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

Then "unnest" data1: 然后“嵌套”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

Or unnest data2: 或不必要的数据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

Here is an option to do this on multiple datasets in the list 这是对list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     

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