将变量名传递到dplyr 1.1.0中mutate（）函数的左侧[重复]

Question

library(dplyr)
library(forcats)

# toy data
df <- tibble(d_1_a = factor(c(letters[1:5],NA,NA)),
             d_1_b = factor(c(NA,NA,letters[1:5])))

muteFunct <- function(d, i) {
  d %>% mutate(newFact = forcats::fct_explicit_na(.data[[i]]))
}

muteFunct(df, "d_1_a")

# output
#   d_1_a d_1_b newFact  
#   <fct> <fct> <fct>    
# 1 a     NA    a        
# 2 b     NA    b        
# 3 c     a     c        
# 4 d     b     d        
# 5 e     c     e        
# 6 NA    d     (Missing)
# 7 NA    e     (Missing)

muteFunct2 <- function(d, i) {
  d %>% mutate(!!.data[[i]] := forcats::fct_explicit_na(.data[[i]]))
}

muteFunct2(df, "d_1_a")

Error: The LHS of `:=` must be a string or a symbol
Run `rlang::last_error()` to see where the error occurred. 

Answer 1

在LHS中，我们可以传递字符串并进行赋值(:=)。由于i传递的是列名，所以只需执行！！i:=

muteFunct2 <- function(d, i) {
   d %>%
     mutate(!! i:= forcats::fct_explicit_na(.data[[i]]))
 }

-测试

muteFunct2(df, "d_1_a")
# A tibble: 7 x 2
  d_1_a     d_1_b
  <fct>     <fct>
1 a         <NA> 
2 b         <NA> 
3 c         a    
4 d         b    
5 e         c    
6 (Missing) d    
7 (Missing) e    

为了更新现有的列，我们还可以使用跨。其优点是它也可以包含多个列

muteFunct2 <- function(d, i) {
   d %>%
     mutate(across(all_of(i),  ~ forcats::fct_explicit_na(.)))
 }

muteFunct2(df, "d_1_a")
muteFunct2(df, names(df))
# A tibble: 7 x 2
  d_1_a     d_1_b    
  <fct>     <fct>    
1 a         (Missing)
2 b         (Missing)
3 c         a        
4 d         b        
5 e         c        
6 (Missing) d        
7 (Missing) e