How to simplify across?

Question

I have to use across() multiple times. Is there a way to simplify the following across() code?

ori_df <- data.frame(base_1 = 1:3,base_2 = 7:9,base_3 =3:1,
                     exp_1 =c(0.4,0.1,0.7),
                     exp_2 =c(0.1,0.05,0.6),
                     exp_3 =c(0.1,0.03,0.04),
                     exp_4 =c(0.1,0.03,0.04))


ori_df %>% mutate(across(contains("base"),~ .* exp_1,.names ="{sub('base','amount_exp_1',{.col})}")) %>% 
  mutate(across(contains("base"),~ .* exp_2,.names ="{sub('base','amount_exp_2',{.col})}")) %>% 
  mutate(across(contains("base"),~ .* exp_3,.names ="{sub('base','amount_exp_3',{.col})}")) %>% 
  mutate(across(contains("base"),~ .* exp_4,.names ="{sub('base','amount_exp_4',{.col})}"))

Answer 1

across() has a .fns argument that allows you to specify multiple functions in a single across call that will be applied to each column. I had to change the sub() in names for some reason to make it work. The order of columns is slightly different but I hope this is not a big problem.

ori_df %>% 
  mutate(across(contains("base"), 
        .fns = list(amount_exp_1 = ~ .* exp_1,
                    amount_exp_2 = ~ .* exp_2, 
                    amount_exp_3 = ~ .* exp_3, 
                    amount_exp_4 = ~ .* exp_4),
        .names = "{.fn}{sub('base','',{.col})}"))

Answer 2

Here is another solution with tidyr::unnest. This way, you don't need to specify the numbers for exp columns. However, one caveat is that this will result in slightly different column names than OP's example. i.e. amount_1_exp_1, amount_1_exp_2, ... as opposed to amount_exp_1_1, amount_exp_1_2... If you want to keep the names in the original format, you could change them with rename_with.

library(dplyr)
ori_df |> 
  mutate(across(
    contains("base"), 
    ~ .x * across(contains("exp")),
    .names="{sub('base','amount',{.col})}"
  )) |> 
  tidyr::unnest(cols=contains("amount"),
                names_sep = "_") 

# output
# A tibble: 3 × 19
  base_1 base_2 base_3 exp_1 exp_2 exp_3 exp_4 amount_1_exp_1 amount_1_exp_2 amount_1_exp_3
   <int>  <int>  <int> <dbl> <dbl> <dbl> <dbl>          <dbl>          <dbl>          <dbl>
1      1      7      3   0.4  0.1   0.1   0.1             0.4            0.1           0.1 
2      2      8      2   0.1  0.05  0.03  0.03            0.2            0.1           0.06
3      3      9      1   0.7  0.6   0.04  0.04            2.1            1.8           0.12
# ℹ 9 more variables: amount_1_exp_4 <dbl>, amount_2_exp_1 <dbl>, amount_2_exp_2 <dbl>,
#   amount_2_exp_3 <dbl>, amount_2_exp_4 <dbl>, amount_3_exp_1 <dbl>, amount_3_exp_2 <dbl>,
#   amount_3_exp_3 <dbl>, amount_3_exp_4 <dbl>

(Adding a call to rename_with)

... |> 
  rename_with(.cols = contains("amount"),
              .fn = ~ sub("^amount_(\\d)_exp_(\\d)",
                          "amount_exp_\\1_\\2",
                          .x))

Answer 3

Find another code ， just mark it

1:4 %>% 
  map(~ {
    exp_col <- paste0("exp_", .x)
    new_prefix <- paste0("amount_", exp_col, "_")
    ori_df %>% 
      mutate(across(contains("base"), 
                   ~ . * get(exp_col),
                   .names = "{sub('base', new_prefix, .col)}"))
  }) %>% 
  reduce(left_join, by = names(ori_df))