Mislabeled variable types cont.

• age is a character variable because there’s a five instead of 5.

• As seen below, we can use if_else(test, yes, no) to say if age is the character string "five", make it "5", and if not leave it as age.

students <- students |>
  janitor::clean_names() |>
  mutate(
    meal_plan = factor(meal_plan),
    age = parse_number(if_else(age == "five", "5", age))
  )

students

## # A tibble: 6 × 5
##   student_id full_name        favourite_food     meal_plan             age
##        <dbl> <chr>            <chr>              <fct>               <dbl>
## 1          1 Sunil Huffmann   Strawberry yoghurt Lunch only              4
## 2          2 Barclay Lynn     French fries       Lunch only              5
## 3          3 Jayendra Lyne    <NA>               Breakfast and lunch     7
## 4          4 Leon Rossini     Anchovies          Lunch only             NA
## 5          5 Chidiegwu Dunkel Pizza              Breakfast and lunch     5
## 6          6 Güvenç Attila    Ice cream          Lunch only              6