代替解析，一个更简单的选择是使用 map2 遍历列表与 map 之后。使用 map2 ，我们传递感兴趣的列和基于命名的 list 'faclist'

Instead of the parsing, an easier option is to use map2 after looping over the list with map. With map2, we pass the columns of interest and the labels to be changed based on the named list 'faclist'

library(dplyr)
library(purrr)
ldat1 <- map(ldat, ~  {
     .x[names(faclist)] <- map2(.x %>% 
                             dplyr::select(names(faclist)), 
                         faclist, ~ factor(.x, labels= .y))
       .x} )

-输出

str(ldat1[[1]])
#'data.frame':  39 obs. of  7 variables:
# $ col1: Factor w/ 1 level "Sales": 1 1 NA NA NA NA NA NA 1 NA ...
# $ col2: Factor w/ 1 level "OPS": 1 1 1 1 1 NA NA NA NA 1 ...
# $ col3: Factor w/ 1 level "Management": 1 1 NA NA NA NA NA 1 NA NA ...
# $ col4: Factor w/ 1 level "HR": 1 NA NA NA NA NA NA NA NA NA ...
# $ col5: Factor w/ 2 levels "Local","Overseas": 1 2 1 1 1 2 1 2 2 1 ...
# $ cc1 : num  1 1 1 1 1 1 1 1 1 1 ...
# $ cc2 : num  1 NA 1 1 1 NA 1 NA NA 1 ...
str(ldat1[[2]])
#'data.frame':  18 obs. of  6 variables:
# $ col1: Factor w/ 1 level "Sales": 1 NA NA NA NA NA NA NA 1 NA ...
#$ col2: Factor w/ 1 level "OPS": 1 1 1 1 NA 1 1 1 1 1 ...
# $ col3: Factor w/ 1 level "Management": 1 NA NA NA NA NA NA NA NA NA ...
# $ col4: Factor w/ 1 level "HR": 1 NA NA NA NA NA NA NA NA 1 ...
# $ col5: Factor w/ 2 levels "Local","Overseas": 1 1 1 1 1 1 1 1 1 1 ...
# $ cc1 : num  1 1 1 1 1 1 1 1 1 1 ...

如果它不是列表，而是两个向量，则只需更改 names（faclist）和带有 labels $的 list 'faclist'向量

If it is not a list, but two vectors, then just change the names(faclist) with the 'col_names' vector and the list 'faclist' with labels vector

ldat1 <- map(ldat, ~  {
     .x[col_names] <- map2(.x %>% 
                             dplyr::select(col_names), 
                         labels, ~ factor(.x, labels= .y))
       .x} )