Why does my dplyr percentile calculation not work with tidy evaluation? - r

Question

I have a tibble with student test data, and I wish to convert these to percentiles using dplyr. For the sake of having a minimal example, imagine the following setup of three students.
require(tidyverse)
tbl <- tibble(Name = c("Alice", "Bob", "Cat"), Test = c(16, 13, 15))
The following code works and yields the desired output.
tbl %>% mutate(TestPercentile = cume_dist(Test) * 100)
# A tibble: 3 x 3
Name Test TestPercentile
<chr> <dbl> <dbl>
1 Alice 16 100
2 Bob 13 33.3
3 Cat 15 66.7
However, I actually want to do it programmatically because there are many such columns.
colname <- "Test"
percname <- str_c(colname, "Percentile")
tbl %>% mutate({{percname}} := cume_dist({{colname}}) * 100)
# A tibble: 3 x 3
Name Test TestPercentile
<chr> <dbl> <dbl>
1 Alice 16 100
2 Bob 13 100
3 Cat 15 100
Why does cume_dist make the percentile 100 for all students when I try to use tidy evaluation like this? (And ideally, if I can be permitted a second question, how can I fix it?)

Answer 1

If by programmatically you mean you want to write your own function, you can do it like this:
calculate_percentile <- function(data, colname) {
data %>%
mutate("{{colname}}Percentile" := cume_dist({{colname}} * 100))
}
tbl %>%
calculate_percentile(Test)
# A tibble: 3 x 3
Name Test TestPercentile
<chr> <dbl> <dbl>
1 Alice 16 1
2 Bob 13 0.333
3 Cat 15 0.667
Edit for multiple columns
New Data
tbl <- tibble(Name = c("Alice", "Bob", "Cat"), Test = c(16, 13, 15), Test_math = c(16, 30, 55), Test_music = c(3, 78, 34))
calculate_percentile <- function(data, colnames) {
data %>%
mutate(across({{colnames}}, ~cume_dist(.) * 100, .names = "{col}Percentile"))
}
test_columns <- c("Test_math", "Test_music")
tbl %>%
calculate_percentile(test_columns)
# A tibble: 3 x 6
Name Test Test_math Test_music Test_mathPercentile Test_musicPercentile
<chr> <dbl> <dbl> <dbl> <dbl> <dbl>
1 Alice 16 16 3 33.3 33.3
2 Bob 13 30 78 66.7 100
3 Cat 15 55 34 100 66.7
Why does your solution not work? Because your solution applies cume_dist literally to the string "test":
tbl %>% mutate({{percname}} := print({{colname}}))
[1] "Test"
# A tibble: 3 x 5
Name Test Test_math Test_music TestPercentile
<chr> <dbl> <dbl> <dbl> <chr>
1 Alice 16 16 3 Test
2 Bob 13 30 78 Test
3 Cat 15 55 34 Test
Why does this give a TestPercentile value of 100? Because cume_dist of "test" is 1:
cume_dist("test")
#[1] 1
So we need R to tell not to evaluate the string "test" per se but to look for a variable with this name, which we can do like this:
tbl %>% mutate({{percname}} := cume_dist(!!parse_quo(colname, env = global_env())) * 100)
# A tibble: 3 x 5
Name Test Test_math Test_music TestPercentile
<chr> <dbl> <dbl> <dbl> <dbl>
1 Alice 16 16 3 100
2 Bob 13 30 78 33.3
3 Cat 15 55 34 66.7
#Check that this uses the values of "Test" and not "Test" per se:
tbl %>% mutate({{percname}} := print(!!parse_quo(colname, env = global_env())))
[1] 16 13 15
# A tibble: 3 x 5
Name Test Test_math Test_music TestPercentile
<chr> <dbl> <dbl> <dbl> <dbl>
1 Alice 16 16 3 16
2 Bob 13 30 78 13
3 Cat 15 55 34 15

Answer 2

Passing column name as string :
library(dplyr)
library(rlang)
return_percentile <- function(data, colname) {
percname <- paste0(colname, "Percentile")
data %>% mutate({{percname}} := cume_dist(!!sym(colname)) * 100)
}
tbl %>% return_percentile("Test")
# A tibble: 3 x 3
# Name Test TestPercentile
# <chr> <dbl> <dbl>
#1 Alice 16 100
#2 Bob 13 33.3
#3 Cat 15 66.7
Passing column name unquoted :
return_percentile <- function(data, colname) {
percname <- paste0(deparse(substitute(colname)), "Percentile")
data %>% mutate({{percname}} := cume_dist({{colname}}) * 100)
}
tbl %>% return_percentile(Test)
# A tibble: 3 x 3
# Name Test TestPercentile
# <chr> <dbl> <dbl>
#1 Alice 16 100
#2 Bob 13 33.3
#3 Cat 15 66.7