数据集包含许多包含 NA 或 1 值的列,有点像这样:
> data_frame(a = c(NA, 1, NA, 1, 1), b=c(1, NA, 1, 1, NA))
# A tibble: 5 x 2
a b
<dbl> <dbl>
1 NA 1.00
2 1.00 NA
3 NA 1.00
4 1.00 1.00
5 1.00 NA
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所需的输出:用列名作为字符串替换所有 1 个值,
> data_frame(a = c(NA, 'a', NA, 'a', 'a'), b=c('b', NA, 'b', 'b', NA))
# A tibble: 5 x 2
a b
<chr> <chr>
1 <NA> b
2 a <NA>
3 <NA> b
4 a b
5 a <NA>
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这是我在 transmute_all 中使用匿名函数的尝试:
> data_frame(a = c(NA, 1, NA, 1, 1), b=c(1, NA, 1, 1, NA)) %>%
+ transmute_all(
+ funs(function(x){if (x == 1) deparse(substitute(x)) else NA})
+ )
Error in mutate_impl(.data, dots) :
Column `a` is of unsupported type function
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编辑:尝试#2:
> data_frame(a = c(NA, 1, NA, 1, 1), b=c(1, NA, 1, 1, NA)) %>%
+ transmute_all(
+ funs(
+ ((function(x){if (!is.na(x)) deparse(substitute(x)) else NA})(.))
+ )
+ )
# A tibble: 5 x 2
a b
<lgl> <chr>
1 NA b
2 NA b
3 NA b
4 NA b
5 NA b
Warning messages:
1: In if (!is.na(x)) deparse(substitute(x)) else NA :
the condition has length > 1 and only the first element will be used
2: In if (!is.na(x)) deparse(substitute(x)) else NA :
the condition has length > 1 and only the first element will be used
>
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一种选择是 map2
library(purrr)
map2_df(df1, names(df1), ~ replace(.x, .x==1, .y))
# A tibble: 5 x 2
# a b
# <chr> <chr>
#1 NA b
#2 a NA
#3 NA b
#4 a b
#5 a NA
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或者正如@Moody_Mudskipper 评论的那样
imap_dfr(df1, ~replace(.x, .x==1, .y))
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在base R,我们可以做
df1[] <- names(df1)[col(df1) *(df1 == 1)]
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df1 <- data_frame(a = c(NA, 1, NA, 1, 1), b=c(1, NA, 1, 1, NA))
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小智 5
如果你想坚持使用dplyr你几乎已经有了的解决方案
library(dplyr)
df <- data_frame(a = c(NA, 1, NA, 1, 1), b = c(1, NA, 1, 1, NA))
df %>%
transmute_all(funs(ifelse(. == 1, deparse(substitute(.)), NA)))
#> # A tibble: 5 x 2
#> a b
#> <chr> <chr>
#> 1 <NA> b
#> 2 a <NA>
#> 3 <NA> b
#> 4 a b
#> 5 a <NA>
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