How to convert columns to multiple boolean columns with tidyverse

Question

I have a group of columns for each time and I want to convert it to a lot of boolean columns (one by category) with mutate() and across() like that:

data <- data.frame(category_t1 = c("A","B","C","C","A","B"),
                   category_t2 = c("A","C","B","B","B",NA),
                   category_t3 = c("C","C",NA,"B",NA,"A"))

data %>% mutate(across(starts_with("category"), 
                       ~case_when(.x == "A" ~ TRUE, !is.na(.x) ~ FALSE),
                       .names = "{str_replace(.col, 'category', 'A')}"),
                across(starts_with("category"), 
                       ~case_when(.x == "B" ~ TRUE, !is.na(.x) ~ FALSE),
                       .names = "{str_replace(.col, 'category', 'B')}"),
                across(starts_with("category"), 
                       ~case_when(.x == "C" ~ TRUE, !is.na(.x) ~ FALSE),
                       .names = "{str_replace(.col, 'category', 'C')}"))

Which makes:

category_t1 category_t2 category_t3  A_t1  A_t2  A_t3  B_t1  B_t2  B_t3  C_t1  C_t2
1         A           A           C  TRUE  TRUE FALSE FALSE FALSE FALSE FALSE FALSE
2         B           C           C FALSE FALSE FALSE  TRUE FALSE FALSE FALSE  TRUE
3         C           B        <NA> FALSE FALSE    NA FALSE  TRUE    NA  TRUE FALSE
4         C           B           B FALSE FALSE FALSE FALSE  TRUE  TRUE  TRUE FALSE
5         A           B        <NA>  TRUE FALSE    NA FALSE  TRUE    NA FALSE FALSE
6         B        <NA>           A FALSE    NA  TRUE  TRUE    NA FALSE FALSE    NA

It works but I would like to know if there is a better idea because here I am doing the same code 3 times instead of one big code (and imagine if I had 10 times to repeat it...). I though I could do it with map() but I didn't manage to make it work. I think there is a problem because of .names argument in across() that cannot connect with the string I use in case_when() .

I think maybe there is something to do in the ... argument, like:

data %>% mutate(across(starts_with("category"),
                       ~case_when(.x == mod ~ TRUE, !is.na(.x) ~ FALSE),
                       mod = levels(as.factor(data$category_t1)),
                       .names = "{str_replace(.col, 'category', mod)}"))

But of course that doesn't work here. Do you know how to do that?

Thanks a lot.

Answer 1

purrr 's map_dfc could match well with your current approach:

library(dplyr)
library(purrr)

bind_cols(data, 
          map_dfc(LETTERS[1:3], \(letter) { mutate(data,
                                                   across(starts_with("category"), 
                                                          ~ case_when(.x == letter ~ TRUE, !is.na(.x) ~ FALSE),
                                                   .names = paste0("{str_replace(.col, 'category', '", letter, "')}")),
                                                   .keep = "none") }
                  )
          )

Or skip the bind_cols and use .keep = ifelse(letter == "A", "all", "none") .

Output:

  category_t1 category_t2 category_t3  A_t1  A_t2  A_t3  B_t1  B_t2  B_t3  C_t1  C_t2  C_t3
1           A           A           C  TRUE  TRUE FALSE FALSE FALSE FALSE FALSE FALSE  TRUE
2           B           C           C FALSE FALSE FALSE  TRUE FALSE FALSE FALSE  TRUE  TRUE
3           C           B        <NA> FALSE FALSE    NA FALSE  TRUE    NA  TRUE FALSE    NA
4           C           B           B FALSE FALSE FALSE FALSE  TRUE  TRUE  TRUE FALSE FALSE
5           A           B        <NA>  TRUE FALSE    NA FALSE  TRUE    NA FALSE FALSE    NA
6           B        <NA>           A FALSE    NA  TRUE  TRUE    NA FALSE FALSE    NA FALSE

Answer 2

Not a tidyverse option (although pipe-compatible), it is very easily doable with package fastDummies :

fastDummies::dummy_cols(data, ignore_na = TRUE)

  category_t1 category_t2 category_t3 category_t1_A category_t1_B category_t1_C category_t2_A category_t2_B category_t2_C category_t3_A category_t3_B category_t3_C
1           A           A           C             1             0             0             1             0             0             0             0             1
2           B           C           C             0             1             0             0             0             1             0             0             1
3           C           B        <NA>             0             0             1             0             1             0            NA            NA            NA
4           C           B           B             0             0             1             0             1             0             0             1             0
5           A           B        <NA>             1             0             0             0             1             0            NA            NA            NA
6           B        <NA>           A             0             1             0            NA            NA            NA             1             0             0

Answer 3

A base solution with nested lapply() :

cbind(data, lapply(data, \(x) {
  lev <- levels(factor(x))
  sapply(setNames(lev, lev), \(y) x == y)
}))

  category_t1 category_t2 category_t3 category_t1.A category_t1.B category_t1.C category_t2.A category_t2.B category_t2.C category_t3.A category_t3.B category_t3.C
1           A           A           C          TRUE         FALSE         FALSE          TRUE         FALSE         FALSE         FALSE         FALSE          TRUE
2           B           C           C         FALSE          TRUE         FALSE         FALSE         FALSE          TRUE         FALSE         FALSE          TRUE
3           C           B        <NA>         FALSE         FALSE          TRUE         FALSE          TRUE         FALSE            NA            NA            NA
4           C           B           B         FALSE         FALSE          TRUE         FALSE          TRUE         FALSE         FALSE          TRUE         FALSE
5           A           B        <NA>          TRUE         FALSE         FALSE         FALSE          TRUE         FALSE            NA            NA            NA
6           B        <NA>           A         FALSE          TRUE         FALSE            NA            NA            NA          TRUE         FALSE         FALSE

Answer 4

We could use model.matrix from base R

m1 <- model.matrix( ~ 0 + ., replace(data, is.na(data), "NA"))> 0
cbind(data, m1[,-ncol(m1)])