我想将命名列表的列表转换为数据框,其中某些列表缺少列。我可以使用已弃用的rbind_all
成功地做到这一点,但不能使用替换的bind_rows
示例
缺少列的列表(el3
缺少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
没有丢失的列
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
如何使用不推荐使用的功能复制rbind_all
行为?
答案 0 :(得分:4)
请在?bind_rows
中阅读以下示例:
# 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,您可以尝试以下方法:
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
答案 1 :(得分:0)
这是使用map_dfr
软件包中的purrr
的一种解决方法。
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