将 data.frames 列表中的列表取消列出到单个 data.frame答案

【问题标题】：Unlist a list within a list of data.frames into a single data.frame将 data.frames 列表中的列表取消列出到单个 data.frame
【发布时间】：2018-05-18 19:09:37
【问题描述】：

dat <- list(list(structure(list(ID = 1, Gender = structure(1L, .Label = "Male", class = "factor"), 
    Phrase = structure(1L, .Label = "Hello", class = "factor"), 
    Phrase2 = structure(1L, .Label = "Goodbye", class = "factor")), .Names = c("ID", 
"Gender", "Phrase", "Phrase2"), row.names = c(NA, -1L), class = "data.frame"), 
    structure(list(ID = 1, Gender = structure(1L, .Label = "Female", class = "factor"), 
        Phrase = structure(1L, .Label = "Good afternoon", class = "factor"), 
        Phrase2 = structure(1L, .Label = "Goodbye", class = "factor")), .Names = c("ID", 
    "Gender", "Phrase", "Phrase2"), row.names = c(NA, -1L), class = "data.frame")), 
    list(structure(list(ID = 2, Gender = structure(1L, .Label = "Male", class = "factor"), 
        Phrase = structure(1L, .Label = "Hello", class = "factor"), 
        Phrase2 = structure(1L, .Label = "Good afternoon", class = "factor")), .Names = c("ID", 
    "Gender", "Phrase", "Phrase2"), row.names = c(NA, -1L), class = "data.frame"), 
        structure(list(ID = 2, Gender = structure(1L, .Label = "Female", class = "factor"), 
            Phrase = structure(1L, .Label = "Goodbye", class = "factor"), 
            Phrase2 = structure(1L, .Label = "Goodbye", class = "factor")), .Names = c("ID", 
        "Gender", "Phrase", "Phrase2"), row.names = c(NA, -1L
        ), class = "data.frame")))

> dat
[[1]]
[[1]][[1]]
  ID Gender Phrase Phrase2
1  1   Male  Hello Goodbye

[[1]][[2]]
  ID Gender         Phrase Phrase2
1  1 Female Good afternoon Goodbye


[[2]]
[[2]][[1]]
  ID Gender Phrase        Phrase2
1  2   Male  Hello Good afternoon

[[2]][[2]]
  ID Gender  Phrase Phrase2
1  2 Female Goodbye Goodbye

我有一个名为 dat 的 data.frames 列表，其中每个列表元素中都有多个条目（例如按性别）。如何将这个 data.frames 列表组合成一个看起来像这样的 data.frame？

  ID Gender         Phrase        Phrase2
1  1   Male          Hello        Goodbye
2  1 Female Good afternoon        Goodbye
3  2   Male          Hello Good afternoon
4  2 Female        Goodbye        Goodbye

我尝试过ldply(dat, data.frame)、do.call("rbind", dat) 和rbind.fill(dat) 均无济于事。

【问题讨论】：

标签： r list dataframe

【解决方案1】：

我们可以试试

library(tidyverse)
map_df(dat, bind_rows)
# ID Gender         Phrase        Phrase2
#1 1   Male          Hello        Goodbye
#2  1 Female Good afternoon        Goodbye
#3  2   Male          Hello Good afternoon
#4  2 Female        Goodbye        Goodbye

【讨论】：

【解决方案2】：

使用tidyverse 包的另一种方式。

library(tidyverse)

dat %>% flatten() %>% bind_rows()
#   ID Gender         Phrase        Phrase2
# 1  1   Male          Hello        Goodbye
# 2  1 Female Good afternoon        Goodbye
# 3  2   Male          Hello Good afternoon
# 4  2 Female        Goodbye        Goodbye

【讨论】：