我想使用tidyverse重命名一些列在与Tibbles关于元素I的名称的列表中在映射函数调用中使用。示例：
c("star","wars") %>%
set_names() %>%
map(~starwars %>%
mutate(col_1 = .x) %>% #this as an example for the use of other functions that I have to use in the real data
select(col_1 ,everything()) %>%
rename_at(vars(name:eye_color), list(~str_c(.,"_",.x)))) <- the issue
结果：
$star
# A tibble: 87 x 15
col_1 name_name height_height mass_mass hair_color_hair~
<chr> <chr> <int> <dbl> <chr>
1 star Luke Sky~ 172 77 blond
2 star C-3PO 167 75 NA
3 star R2-D2 96 32 NA
4 star Darth Va~ 202 136 none
5 star Leia Org~ 150 49 brown
6 star Owen Lars 178 120 brown, grey
7 star Beru Whi~ 165 75 brown
8 star R5-D4 97 32 NA
9 star Biggs Da~ 183 84 black
10 star Obi-Wan ~ 182 77 auburn, white
# ... with 77 more rows, and 10 more variables:
# skin_color_skin_color <chr>, eye_color_eye_color <chr>,
# birth_year <dbl>, sex <chr>, gender <chr>, homeworld <chr>,
# species <chr>, films <list>, vehicles <list>,
# starships <list>
$wars
# A tibble: 87 x 15
col_1 name_name height_height mass_mass hair_color_hair~
<chr> <chr> <int> <dbl> <chr>
1 wars Luke Sky~ 172 77 blond
2 wars C-3PO 167 75 NA
3 wars R2-D2 96 32 NA
4 wars Darth Va~ 202 136 none
5 wars Leia Org~ 150 49 brown
6 wars Owen Lars 178 120 brown, grey
7 wars Beru Whi~ 165 75 brown
8 wars R5-D4 97 32 NA
9 wars Biggs Da~ 183 84 black
10 wars Obi-Wan ~ 182 77 auburn, white
# ... with 77 more rows, and 10 more variables:
# skin_color_skin_color <chr>, eye_color_eye_color <chr>,
# birth_year <dbl>, sex <chr>, gender <chr>, homeworld <chr>,
# species <chr>, films <list>, vehicles <list>,
# starships <list>
我想要创建的内容：
$star
# A tibble: 87 x 15
col_1 name_star height_star mass_star hair_color_star~
<chr> <chr> <int> <dbl> <chr>
1 star Luke Sky~ 172 77 blond
2 star C-3PO 167 75 NA
3 star R2-D2 96 32 NA
4 star Darth Va~ 202 136 none
5 star Leia Org~ 150 49 brown
6 star Owen Lars 178 120 brown, grey
7 star Beru Whi~ 165 75 brown
8 star R5-D4 97 32 NA
9 star Biggs Da~ 183 84 black
10 star Obi-Wan ~ 182 77 auburn, white
# ... with 77 more rows, and 10 more variables:
# skin_color_star <chr>, eye_color_star <chr>,
# birth_year <dbl>, sex <chr>, gender <chr>, homeworld <chr>,
# species <chr>, films <list>, vehicles <list>,
# starships <list>
$wars
# A tibble: 87 x 15
col_1 name_wars height_wars mass_wars hair_color_wars~
<chr> <chr> <int> <dbl> <chr>
1 wars Luke Sky~ 172 77 blond
2 wars C-3PO 167 75 NA
3 wars R2-D2 96 32 NA
4 wars Darth Va~ 202 136 none
5 wars Leia Org~ 150 49 brown
6 wars Owen Lars 178 120 brown, grey
7 wars Beru Whi~ 165 75 brown
8 wars R5-D4 97 32 NA
9 wars Biggs Da~ 183 84 black
10 wars Obi-Wan ~ 182 77 auburn, white
# ... with 77 more rows, and 10 more variables:
# skin_color_wars <chr>, eye_color_wars <chr>,
# birth_year <dbl>, sex <chr>, gender <chr>, homeworld <chr>,
# species <chr>, films <list>, vehicles <list>,
# starships <list>
出于某种原因，rename_at会查看"；。x〃；作为"来自列的名称而不是从地图上看。怎么可能利用map函数的.x呢？
一种解决方法是在rename_at()中使用匿名函数而不是公式，以避免.x和.引用同一对象的问题。您也可以为map()函数和rename_at()中的波浪号使用匿名函数。
c("star","wars") %>%
set_names() %>%
map(~starwars %>%
mutate(col_1 = .x) %>%
select(col_1 ,everything()) %>%
rename_at(vars(name:eye_color), function(var) str_c(var, "_", .x)))
$star
# A tibble: 87 x 15
col_1 name_star height_star mass_star hair_color_star skin_color_star
<chr> <chr> <int> <dbl> <chr> <chr>
1 star Luke Sky~ 172 77 blond fair
2 star C-3PO 167 75 NA gold
3 star R2-D2 96 32 NA white, blue
4 star Darth Va~ 202 136 none white
5 star Leia Org~ 150 49 brown light
6 star Owen Lars 178 120 brown, grey light
7 star Beru Whi~ 165 75 brown light
8 star R5-D4 97 32 NA white, red
9 star Biggs Da~ 183 84 black light
10 star Obi-Wan ~ 182 77 auburn, white fair
# ... with 77 more rows, and 9 more variables: eye_color_star <chr>,
# birth_year <dbl>, sex <chr>, gender <chr>, homeworld <chr>, species <chr>,
# films <list>, vehicles <list>, starships <list>
$wars
# A tibble: 87 x 15
col_1 name_wars height_wars mass_wars hair_color_wars skin_color_wars
<chr> <chr> <int> <dbl> <chr> <chr>
1 wars Luke Sky~ 172 77 blond fair
2 wars C-3PO 167 75 NA gold
3 wars R2-D2 96 32 NA white, blue
4 wars Darth Va~ 202 136 none white
5 wars Leia Org~ 150 49 brown light
6 wars Owen Lars 178 120 brown, grey light
7 wars Beru Whi~ 165 75 brown light
8 wars R5-D4 97 32 NA white, red
9 wars Biggs Da~ 183 84 black light
10 wars Obi-Wan ~ 182 77 auburn, white fair
# ... with 77 more rows, and 9 more variables: eye_color_wars <chr>,
# birth_year <dbl>, sex <chr>, gender <chr>, homeworld <chr>, species <chr>,
# films <list>, vehicles <list>, starships <list>
在dplyr1.0.0版本中，rename_at()已被rename_with()取代(但不会消失(。rename_with()代码看起来像：
... %>%
rename_with(.fn = function(var) str_c(var, "_", .x), .cols = name:eye_color))
使用@aosmith的答案的另一种选择是强制表达一个化解表达式：
library(rlang)
c("star","wars") %>%
set_names() %>%
map(~starwars %>%
mutate(col_1 = .x) %>%
select(col_1 ,everything()) %>%
rename_at(vars(name:eye_color), list(~str_c(.,"_",!!quo(.x)))))
我认为使用不同符号的匿名函数肯定是更安全、更清晰的方法，但这是为了表明这是可能的。