How do I count the percentage makeup of TRUES in a table?

This is the code used to derive the first table in my question.

JH %>% group_by(ATT_ID, CAR=="B") %>%
summarize(count = n(), .groups = "drop")

ATT_ID	CAR == "B"	Count
ONE	FALSE	1
TWO	TRUE	1
THREE	TRUE	3
THREE	FALSE	5
FOUR	FALSE	2
FIVE	TRUE	4
SIX	TRUE	8
SIX	FALSE	4

How can I get the table above to look like:

ATT_ID	Percentage of "B"
ONE	0%
TWO	100%
THREE	37.5%
FOUR	0%
FIVE	100%
SIX	67%

Notice how some ID's are seen twice so as to show the presence of both FALSE & TRUE whereas other ID's appear once to showcase the presence of only one or the other.

Thank you

Solution

Grouped by 'ATT_ID', get the sum of Count where CAR=="B" is TRUE and divide by the sum of full Count

library(dplyr)
df1 %>%
   group_by(ATT_ID = factor(ATT_ID, levels = unique(ATT_ID))) %>%
  summarise(Percentage_of_B = paste0(round(
      sum(Count[`CAR == "B"`])/sum(Count) * 100, 1), "%"))

-output

# A tibble: 6 × 2
  ATT_ID Percentage_of_B
  <fct>  <chr>          
1 ONE    0%             
2 TWO    100%           
3 THREE  37.5%          
4 FOUR   0%             
5 FIVE   100%           
6 SIX    66.7%

data

df1 <- structure(list(ATT_ID = c("ONE", "TWO", "THREE", "THREE", "FOUR", 
"FIVE", "SIX", "SIX"), `CAR == "B"` = c(FALSE, TRUE, TRUE, FALSE, 
FALSE, TRUE, TRUE, FALSE), Count = c(1L, 1L, 3L, 5L, 2L, 4L, 
8L, 4L)), class = "data.frame", row.names = c(NA, -8L))