Activity: Data wrangling across columns and functions

Student grades

You are a TA for a statistics course. The instructor of the course is interested in assessing how students performed on each assignment in the class.

You are provided with a CSV file (student_grades.csv), from Canvas, containing the grades for each student on each assignment in the course. Here are the instructions the professor gives you:

• There are 6 homeworks, 2 midterms, a final exam, and a project
• Each homework is scored out of 10. All other assignments are scored out of 100
• If a student did not submit an assignment, it is marked as NA in the CSV file. These missing assignments should receive a score of 0

The data can be imported into R with the following code:

library(tidyverse)

student_grades <- read_csv("https://sta279-f25.github.io/data/student_grades.csv")

Questions

Your answers to these questions should involve functions like starts_with, where, across, etc. You should not list all of the homework columns explicitly, e.g.

1. What was the average exam score for each midterm?

Solution:

student_grades |>
  summarize(across(starts_with("midterm"), mean))

# A tibble: 1 × 2
  midterm_1 midterm_2
      <dbl>     <dbl>
1      71.8      71.2

2. What fraction of students failed each midterm (a grade less than 60%)?

Solution:

One option:

failure_rate <- function(x){
  mean(x < 60)
}

student_grades |>
  summarize(across(starts_with("midterm"), failure_rate))

# A tibble: 1 × 2
  midterm_1 midterm_2
      <dbl>     <dbl>
1     0.167       0.2

Another option:

student_grades |>
  summarize(across(starts_with("midterm"), function(x) mean(x < 60)))

# A tibble: 1 × 2
  midterm_1 midterm_2
      <dbl>     <dbl>
1     0.167       0.2

3. What was the average score for each homework, if you ignore missing submissions?

Solution:

One option:

mean_no_na <- function (x) {
  mean(x, na.rm=T)
}

student_grades |>
  summarize(across(starts_with("hw"), mean_no_na))

# A tibble: 1 × 6
   hw_1  hw_2  hw_3  hw_4  hw_5  hw_6
  <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1  7.18  6.93  7.21  6.87  6.63  7.47

Another option:

student_grades |>
  summarize(across(starts_with("hw"), function(x) mean(x, na.rm=T)))

# A tibble: 1 × 6
   hw_1  hw_2  hw_3  hw_4  hw_5  hw_6
  <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1  7.18  6.93  7.21  6.87  6.63  7.47

4. What was the average score for each homework, if you treat missing assignment as 0? Hint: One approach could use the replace_na function

Solution:

One option:

mean_missing <- function (x) {
  x[is.na(x)] <- 0
  mean(x)
}

student_grades |>
  summarize(across(starts_with("hw"), mean_missing))

# A tibble: 1 × 6
   hw_1  hw_2  hw_3  hw_4  hw_5  hw_6
  <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1   6.7  6.93  6.97  6.87  5.97  7.47

Another option:

student_grades |>
  summarize(across(starts_with("hw"), 
                   function(x) mean(replace_na(x, 0))))

# A tibble: 1 × 6
   hw_1  hw_2  hw_3  hw_4  hw_5  hw_6
  <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1   6.7  6.93  6.97  6.87  5.97  7.47

Another option:

student_grades |>
  mutate(across(everything(), function(x) replace_na(x, 0))) |>
  summarize(across(starts_with("hw"), mean))

# A tibble: 1 × 6
   hw_1  hw_2  hw_3  hw_4  hw_5  hw_6
  <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1   6.7  6.93  6.97  6.87  5.97  7.47

5. For each homework assignment, how many students failed to submit?

Solution:

One option:

count_na <- function (x) {
  sum(is.na(x))
}

student_grades |>
  summarize(across(starts_with("hw"), count_na))

# A tibble: 1 × 6
   hw_1  hw_2  hw_3  hw_4  hw_5  hw_6
  <int> <int> <int> <int> <int> <int>
1     2     0     1     0     3     0

Another option:

student_grades |>
  summarize(across(starts_with("hw"), function(x) sum(is.na(x))))

# A tibble: 1 × 6
   hw_1  hw_2  hw_3  hw_4  hw_5  hw_6
  <int> <int> <int> <int> <int> <int>
1     2     0     1     0     3     0