Treat character as column name

Question

I am trying to treat a character string as a column name in case_when:

Data:

> df <- data.frame(col1 = c(NA, 1, 1, 0, 0),
                   col2 = c(NA, 1, 1, 0, 0),
                   col1_add=1:5, 
                   col2_add=c(6:9,NA))
> df
  col1 col2 col1_add col2_add
1   NA   NA        1        6
2    1    1        2        7
3    1    1        3        8
4    0    0        4        9
5    0    0        5       NA

Attempt:

df %>% mutate_at(c("col1_add", "col2_add"), ~case_when(as.name(str_remove(., "_add"))==1 & .>3 ~ 0, TRUE ~ .))

How can str_remove(., "_add") be treated as a column name ( col1, col2 )?

Answer 1

We can use across with cur_column() ( cur_column() - returns the column name, where the str_remove is used to remove the substring "_add", use get to get the value of the corresponding column, apply the logic in case_when

library(dplyr)
library(stringr)
df %>% 
  mutate(across(ends_with("_add"), 
   ~ case_when(get(str_remove(cur_column(), "_add")) == 1 & 
        .x >3 ~ 0L, TRUE ~ .x)))

-output

  col1 col2 col1_add col2_add
1   NA   NA        1        6
2    1    1        2        0
3    1    1        3        0
4    0    0        4        9
5    0    0        5       NA

NOTE: It was previously possible to get the column name with deparse/substitute from mutate_at , but now it is not possible in the newer dplyr versions. In addition, the _at/_all are deprecated in favor of across