Question

我有以下数据框：

output <- list(structure(list(member = structure(list(name = c("Mick.HMSC-ad", 
"John.HMSC-ad", "Paul.HMSC-ad"), band = c("Stones", "Beatles", 
"Beatles")), .Names = c("name", "band"), class = c("tbl_df", 
"tbl", "data.frame"), row.names = c(NA, -3L)), instrument = structure(list(
    name = c("John", "Paul", "Keith"), plays = c("guitar.HMSC-ad", 
    "bass.HMSC-ad", "guitar.HMSC-ad")), .Names = c("name", "plays"
), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA, 
-3L))), .Names = c("member", "instrument")), structure(list(member = structure(list(
    name = c("Mick.HMSC-bm", "John.HMSC-bm", "Paul.HMSC-bm"), 
    band = c("Stones", "Beatles", "Beatles")), .Names = c("name", 
"band"), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA, 
-3L)), instrument = structure(list(name = c("John", "Paul", "Keith"
), plays = c("guitar.HMSC-bm", "bass.HMSC-bm", "guitar.HMSC-bm"
)), .Names = c("name", "plays"), class = c("tbl_df", "tbl", "data.frame"
), row.names = c(NA, -3L))), .Names = c("member", "instrument"
)))

在这种情况下，我有2个列表，但实际上还有更多。

看起来像这样：

> output
[[1]]
[[1]]$member
# A tibble: 3 x 2
          name    band
         <chr>   <chr>
1 Mick.HMSC-ad  Stones
2 John.HMSC-ad Beatles
3 Paul.HMSC-ad Beatles

[[1]]$instrument
# A tibble: 3 x 2
   name          plays
  <chr>          <chr>
1  John guitar.HMSC-ad
2  Paul   bass.HMSC-ad
3 Keith guitar.HMSC-ad


[[2]]
[[2]]$member
# A tibble: 3 x 2
          name    band
         <chr>   <chr>
1 Mick.HMSC-bm  Stones
2 John.HMSC-bm Beatles
3 Paul.HMSC-bm Beatles

[[2]]$instrument
# A tibble: 3 x 2
   name          plays
  <chr>          <chr>
1  John guitar.HMSC-bm
2  Paul   bass.HMSC-bm
3 Keith guitar.HMSC-bm

我想要做的是提取$member元素并将它们绑定到一个数据框中：

          name    band
         <chr>   <chr>
1 Mick.HMSC-ad  Stones
2 John.HMSC-ad Beatles
3 Paul.HMSC-ad Beatles
4 Mick.HMSC-bm  Stones
5 John.HMSC-bm Beatles
6 Paul.HMSC-bm Beatles

我该怎么做？

Answer 1

您可以使用purrr::map_df从每个子列表中提取member元素，将各个元素行绑定到数据框（？map_df ）自动：

purrr::map_df(output, ~ .x$member)

# A tibble: 6 x 2
#          name    band
#         <chr>   <chr>
#1 Mick.HMSC-ad  Stones
#2 John.HMSC-ad Beatles
#3 Paul.HMSC-ad Beatles
#4 Mick.HMSC-bm  Stones
#5 John.HMSC-bm Beatles
#6 Paul.HMSC-bm Beatles

使用dplyr的{{1}}解决方案：

bind_rows

Answer 2

包含lapply，do.call和rbind的传统基础R方法。仅提取每个列表的member部分，并rbind将它们组合在一起。

do.call("rbind", lapply(output, function(x) x[['member']]))


#          name    band
#1 Mick.HMSC-ad  Stones
#2 John.HMSC-ad Beatles
#3 Paul.HMSC-ad Beatles
#4 Mick.HMSC-bm  Stones
#5 John.HMSC-bm Beatles
#6 Paul.HMSC-bm Beatles

Answer 3

我们可以使用rbindlist

中的data.table

library(data.table)
rbindlist(lapply(output, function(x) x$member))
#           name    band
#1: Mick.HMSC-ad  Stones
#2: John.HMSC-ad Beatles
#3: Paul.HMSC-ad Beatles
#4: Mick.HMSC-bm  Stones
#5: John.HMSC-bm Beatles
#6: Paul.HMSC-bm Beatles

如何提取数据帧列表的子集并将它们绑定为一个

3 个答案: