代码之家 › 专栏 › 技术社区 › ahmohamed

dplyr:包含rbind_all和bind_行的列表到数据帧的列表

dplyr r

2

ahmohamed · 技术社区 · 7 年前

我想将一个命名列表列表转换为一个数据框,其中有些缺少列。我可以成功地做到这一点 rbind_all 但不是用替代品 bind_rows

例子缺少列的列表( el3 丢失的 b )

ex = list(el1=c(a=1, b=2, c=3), el2=c(a=2, b=3, c=4), el3=c(a=3, c=5))

rbind_all(ex)
# A tibble: 3 x 3
      a     b     c
  <dbl> <dbl> <dbl>
1     1     2     3
2     2     3     4
3     3    NA     5


> bind_rows(ex)
Error in bind_rows_(x, .id) : Argument 3 must be length 3, not 2

不缺少列

ex2 = list(el1=c(a=1, b=2, c=3), el2=c(a=2, b=3, c=4), el3=c(a=3, b=4, c=5))

rbind_all(ex2)
# A tibble: 3 x 3
      a     b     c
  <dbl> <dbl> <dbl>
1     1     2     3
2     2     3     4
3     3     4     5

bind_rows(ex2) # Output is transposed for some reason
# A tibble: 3 x 3
    el1   el2   el3
  <dbl> <dbl> <dbl>
1     1     2     3
2     2     3     4
3     3     4     5

如何复制 RB 使用非弃用函数的行为?

2 回复 | 直到 7 年前

1

3

mt1022 7 年前

请在中阅读此示例 ?bind_rows :

# Note that for historical reasons, lists containg vectors are
# always treated as data frames. Thus their vectors are treated as
# columns rather than rows, and their inner names are ignored:
ll <- list(
  a = c(A = 1, B = 2),
  b = c(A = 3, B = 4)
)
bind_rows(ll)

# You can circumvent that behaviour with explicit splicing:
bind_rows(!!!ll)

因此,在您的情况下,您可以尝试:

ex = list(el1=c(a=1, b=2, c=3), el2=c(a=2, b=3, c=4), el3=c(a=3, c=5))
bind_rows(!!!ex)

# # A tibble: 3 x 3
#       a     b     c
#   <dbl> <dbl> <dbl>
# 1     1     2     3
# 2     2     3     4
# 3     3    NA     5

ex2 = list(el1=c(a=1, b=2, c=3), el2=c(a=2, b=3, c=4), el3=c(a=3, b=4, c=5))
bind_rows(!!!ex2)

# # A tibble: 3 x 3
#       a     b     c
#   <dbl> <dbl> <dbl>
# 1     1     2     3
# 2     2     3     4
# 3     3     4     5

2

0

www 7 年前

以下是一个使用 map_dfr 从 purrr 包裹。

library(dplyr)
library(purrr)

map_dfr(ex, ~as_tibble(t(.)))
# # A tibble: 3 x 3
#       a     b     c
#   <dbl> <dbl> <dbl>
# 1     1     2     3
# 2     2     3     4
# 3     3    NA     5