R Dplyr Exercises

Programming for Data Science Bootcamp

Exercise 1

Install Tidyverse

library(tidyverse)

── Attaching core tidyverse packages ──────────────────────── tidyverse 2.0.0 ──
✔ dplyr     1.1.3     ✔ readr     2.1.4
✔ forcats   1.0.0     ✔ stringr   1.5.0
✔ ggplot2   3.4.4     ✔ tibble    3.2.1
✔ lubridate 1.9.3     ✔ tidyr     1.3.0
✔ purrr     1.0.2     
── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
✖ dplyr::filter() masks stats::filter()
✖ dplyr::lag()    masks stats::lag()
ℹ Use the conflicted package (<http://conflicted.r-lib.org/>) to force all conflicts to become errors

Exercise 2

We create a super simple tibble to use in the exercises.

You can find a text file called exercise-data.txt in the directory of today's module (M11).

Just cut and paste it into you code.

scores <- 
  tibble(
    name = c("mike", "carol", "greg", "marcia", "peter", "jan", "bobby", "cindy", "alice"),
    school = c("south", "south", "south", "south", "north", "north", "north", "south", "south"),
    teacher = c("johnson", "johnson", "johnson", "johnson",  "smith", "smith", "smith", "perry", "perry"),
    sex = c("male", "female", "male", "female", "male", "female", "male", "female", "female"),
    math_score = c(4, 3, 2, 4, 3, 4, 5, 4, 5),
    reading_score = c(1, 5, 2, 4, 5, 4, 1, 5, 4)
)

Exercise 3

View the tibble.

scores

A tibble: 9 × 6

name	school	teacher	sex	math_score	reading_score
<chr>	<chr>	<chr>	<chr>	<dbl>	<dbl>
mike	south	johnson	male	4	1
carol	south	johnson	female	3	5
greg	south	johnson	male	2	2
marcia	south	johnson	female	4	4
peter	north	smith	male	3	5
jan	north	smith	female	4	4
bobby	north	smith	male	5	1
cindy	south	perry	female	4	5
alice	south	perry	female	5	4

print(scores)

# A tibble: 9 × 6
  name   school teacher sex    math_score reading_score
  <chr>  <chr>  <chr>   <chr>       <dbl>         <dbl>
1 mike   south  johnson male            4             1
2 carol  south  johnson female          3             5
3 greg   south  johnson male            2             2
4 marcia south  johnson female          4             4
5 peter  north  smith   male            3             5
6 jan    north  smith   female          4             4
7 bobby  north  smith   male            5             1
8 cindy  south  perry   female          4             5
9 alice  south  perry   female          5             4

Exercise 4

Make sure you understand the difference between

• doing something and assigning it to a variable
• and just doing it without assigning it

when using %>%.

First, just get the first three rows (using slice()).

Then, assign the first three rows to a variable.

scores %>% 
  slice(1:3)

A tibble: 3 × 6

name	school	teacher	sex	math_score	reading_score
<chr>	<chr>	<chr>	<chr>	<dbl>	<dbl>
mike	south	johnson	male	4	1
carol	south	johnson	female	3	5
greg	south	johnson	male	2	2

Get the first 3 rows, and assign it to a new name scores_small.

scores_small <- scores %>% 
  slice(1:3)

See what's in scores_small.

scores_small

A tibble: 3 × 6

name	school	teacher	sex	math_score	reading_score
<chr>	<chr>	<chr>	<chr>	<dbl>	<dbl>
mike	south	johnson	male	4	1
carol	south	johnson	female	3	5
greg	south	johnson	male	2	2

Another option:

scores %>% head(3)

A tibble: 3 × 6

name	school	teacher	sex	math_score	reading_score
<chr>	<chr>	<chr>	<chr>	<dbl>	<dbl>
mike	south	johnson	male	4	1
carol	south	johnson	female	3	5
greg	south	johnson	male	2	2

And the old way:

scores[1:3,]

A tibble: 3 × 6

name	school	teacher	sex	math_score	reading_score
<chr>	<chr>	<chr>	<chr>	<dbl>	<dbl>
mike	south	johnson	male	4	1
carol	south	johnson	female	3	5
greg	south	johnson	male	2	2

Part 1: Arrange

Exercise 5

Sort the data by math_score from high to low.

Who had the best math score?

scores %>% 
  arrange(desc(math_score))

A tibble: 9 × 6

name	school	teacher	sex	math_score	reading_score
<chr>	<chr>	<chr>	<chr>	<dbl>	<dbl>
bobby	north	smith	male	5	1
alice	south	perry	female	5	4
mike	south	johnson	male	4	1
marcia	south	johnson	female	4	4
jan	north	smith	female	4	4
cindy	south	perry	female	4	5
carol	south	johnson	female	3	5
peter	north	smith	male	3	5
greg	south	johnson	male	2	2

Answer: Bobby and Alice both tied for the highest math score

Exercise 6

Sort the data by name in alphabetical order.

scores %>% 
  arrange(name)

A tibble: 9 × 6

name	school	teacher	sex	math_score	reading_score
<chr>	<chr>	<chr>	<chr>	<dbl>	<dbl>
alice	south	perry	female	5	4
bobby	north	smith	male	5	1
carol	south	johnson	female	3	5
cindy	south	perry	female	4	5
greg	south	johnson	male	2	2
jan	north	smith	female	4	4
marcia	south	johnson	female	4	4
mike	south	johnson	male	4	1
peter	north	smith	male	3	5

Exercise 7

Sort the data by sex so females show up first.

scores %>% 
  arrange(sex)

A tibble: 9 × 6

name	school	teacher	sex	math_score	reading_score
<chr>	<chr>	<chr>	<chr>	<dbl>	<dbl>
carol	south	johnson	female	3	5
marcia	south	johnson	female	4	4
jan	north	smith	female	4	4
cindy	south	perry	female	4	5
alice	south	perry	female	5	4
mike	south	johnson	male	4	1
greg	south	johnson	male	2	2
peter	north	smith	male	3	5
bobby	north	smith	male	5	1

Exercise 8

Sort the data by school, teacher, sex, math_score, and reading_score.

scores %>%
    arrange(school, teacher, sex, math_score, reading_score)

A tibble: 9 × 6

name	school	teacher	sex	math_score	reading_score
<chr>	<chr>	<chr>	<chr>	<dbl>	<dbl>
jan	north	smith	female	4	4
peter	north	smith	male	3	5
bobby	north	smith	male	5	1
carol	south	johnson	female	3	5
marcia	south	johnson	female	4	4
greg	south	johnson	male	2	2
mike	south	johnson	male	4	1
cindy	south	perry	female	4	5
alice	south	perry	female	5	4

Part 2: Select

Exercise 9

Select only name, math_score, and reading_score.

scores %>% 
  select(name, math_score, reading_score)

A tibble: 9 × 3

name	math_score	reading_score
<chr>	<dbl>	<dbl>
mike	4	1
carol	3	5
greg	2	2
marcia	4	4
peter	3	5
jan	4	4
bobby	5	1
cindy	4	5
alice	5	4

Exercise 10

Select all of the columns except the sex column.

scores %>% 
  select(-sex)

A tibble: 9 × 5

name	school	teacher	math_score	reading_score
<chr>	<chr>	<chr>	<dbl>	<dbl>
mike	south	johnson	4	1
carol	south	johnson	3	5
greg	south	johnson	2	2
marcia	south	johnson	4	4
peter	north	smith	3	5
jan	north	smith	4	4
bobby	north	smith	5	1
cindy	south	perry	4	5
alice	south	perry	5	4

Exercise 11

Select all of the columns except the math_score and reading_score columns.

scores %>% 
  select(-sex, -reading_score)

A tibble: 9 × 4

name	school	teacher	math_score
<chr>	<chr>	<chr>	<dbl>
mike	south	johnson	4
carol	south	johnson	3
greg	south	johnson	2
marcia	south	johnson	4
peter	north	smith	3
jan	north	smith	4
bobby	north	smith	5
cindy	south	perry	4
alice	south	perry	5

Exercise 12

Keep all of the columns but rearrange them so that sex is the first column.

scores %>% 
  select(sex, everything())

A tibble: 9 × 6

sex	name	school	teacher	math_score	reading_score
<chr>	<chr>	<chr>	<chr>	<dbl>	<dbl>
male	mike	south	johnson	4	1
female	carol	south	johnson	3	5
male	greg	south	johnson	2	2
female	marcia	south	johnson	4	4
male	peter	north	smith	3	5
female	jan	north	smith	4	4
male	bobby	north	smith	5	1
female	cindy	south	perry	4	5
female	alice	south	perry	5	4

Part 3: Filter

Exercise 13

Filter by students who are male and went to south.

Option 1

scores %>% 
  filter(sex == "male" & school == "south")

A tibble: 2 × 6

name	school	teacher	sex	math_score	reading_score
<chr>	<chr>	<chr>	<chr>	<dbl>	<dbl>
mike	south	johnson	male	4	1
greg	south	johnson	male	2	2

Option 2

Using a comma , instead of &.

scores %>% 
  filter(sex == "male", school == "south")

A tibble: 2 × 6

name	school	teacher	sex	math_score	reading_score
<chr>	<chr>	<chr>	<chr>	<dbl>	<dbl>
mike	south	johnson	male	4	1
greg	south	johnson	male	2	2

Exercise 14

Filter by students who did above average in math.

scores %>%
    filter(math_score > mean(math_score))

A tibble: 6 × 6

name	school	teacher	sex	math_score	reading_score
<chr>	<chr>	<chr>	<chr>	<dbl>	<dbl>
mike	south	johnson	male	4	1
marcia	south	johnson	female	4	4
jan	north	smith	female	4	4
bobby	north	smith	male	5	1
cindy	south	perry	female	4	5
alice	south	perry	female	5	4

Exercise 15

Use filter() to figure out how many students had a math score of $4$ or more and a reading score of $3$ or more.

scores %>%
    filter(math_score >= 4 & reading_score >= 3)

A tibble: 4 × 6

name	school	teacher	sex	math_score	reading_score
<chr>	<chr>	<chr>	<chr>	<dbl>	<dbl>
marcia	south	johnson	female	4	4
jan	north	smith	female	4	4
cindy	south	perry	female	4	5
alice	south	perry	female	5	4

Exercise 16

Filter by students who got a $3$ or worse in either math or reading.

scores %>%
    filter(math_score <= 3 | reading_score <= 3)

A tibble: 5 × 6

name	school	teacher	sex	math_score	reading_score
<chr>	<chr>	<chr>	<chr>	<dbl>	<dbl>
mike	south	johnson	male	4	1
carol	south	johnson	female	3	5
greg	south	johnson	male	2	2
peter	north	smith	male	3	5
bobby	north	smith	male	5	1

Exercise 17

Filter by students who got a reading score of $2$, $3$, or $4$.

scores %>%
    filter(reading_score == 2 | reading_score == 3 | reading_score == 4)

A tibble: 4 × 6

name	school	teacher	sex	math_score	reading_score
<chr>	<chr>	<chr>	<chr>	<dbl>	<dbl>
greg	south	johnson	male	2	2
marcia	south	johnson	female	4	4
jan	north	smith	female	4	4
alice	south	perry	female	5	4

scores %>%
    filter(reading_score >= 2 & reading_score <= 4)

A tibble: 4 × 6

name	school	teacher	sex	math_score	reading_score
<chr>	<chr>	<chr>	<chr>	<dbl>	<dbl>
greg	south	johnson	male	2	2
marcia	south	johnson	female	4	4
jan	north	smith	female	4	4
alice	south	perry	female	5	4

scores %>%
    filter(reading_score %in% 2:4)

A tibble: 4 × 6

name	school	teacher	sex	math_score	reading_score
<chr>	<chr>	<chr>	<chr>	<dbl>	<dbl>
greg	south	johnson	male	2	2
marcia	south	johnson	female	4	4
jan	north	smith	female	4	4
alice	south	perry	female	5	4

Exercise 18

Filter by students who have a name that starts with 'm'.

Hint: type ?substr in the console and then scroll to the bottom of the help file to see useful examples.

# ?substr

scores %>%
    filter(substr(name, 1, 1) == 'm')

A tibble: 2 × 6

name	school	teacher	sex	math_score	reading_score
<chr>	<chr>	<chr>	<chr>	<dbl>	<dbl>
mike	south	johnson	male	4	1
marcia	south	johnson	female	4	4

Part 4: Filter with groups

Exercise 19

Filter to teachers whose best math student got a score of $5$.

scores %>% 
  group_by(teacher) %>% 
  filter(max(math_score) == 5)

A grouped_df: 5 × 6

name	school	teacher	sex	math_score	reading_score
<chr>	<chr>	<chr>	<chr>	<dbl>	<dbl>
peter	north	smith	male	3	5
jan	north	smith	female	4	4
bobby	north	smith	male	5	1
cindy	south	perry	female	4	5
alice	south	perry	female	5	4

Note that all students from both teachers are shown.

Exercise 20

Filter to the sex with a mean math score of $4$.

scores %>%
    group_by(sex) %>%
    filter(mean(math_score) == 4)

A grouped_df: 5 × 6

name	school	teacher	sex	math_score	reading_score
<chr>	<chr>	<chr>	<chr>	<dbl>	<dbl>
carol	south	johnson	female	3	5
marcia	south	johnson	female	4	4
jan	north	smith	female	4	4
cindy	south	perry	female	4	5
alice	south	perry	female	5	4

Part 5: Mutate

Exercise 21

Set the math and reading scores to $10$ times their original values.

scores %>% 
  mutate(
      math_score =  math_score * 10, 
      reading_score = reading_score * 10
  )

A tibble: 9 × 6

name	school	teacher	sex	math_score	reading_score
<chr>	<chr>	<chr>	<chr>	<dbl>	<dbl>
mike	south	johnson	male	40	10
carol	south	johnson	female	30	50
greg	south	johnson	male	20	20
marcia	south	johnson	female	40	40
peter	north	smith	male	30	50
jan	north	smith	female	40	40
bobby	north	smith	male	50	10
cindy	south	perry	female	40	50
alice	south	perry	female	50	40

Exercise 22

Create a new column called math_reading_avg which is the average of a student's math and reading scores.

That is, combine the two scores and take their average.

scores %>% 
    mutate(math_reading_avg = (math_score + reading_score) / 2)

A tibble: 9 × 7

name	school	teacher	sex	math_score	reading_score	math_reading_avg
<chr>	<chr>	<chr>	<chr>	<dbl>	<dbl>	<dbl>
mike	south	johnson	male	4	1	2.5
carol	south	johnson	female	3	5	4.0
greg	south	johnson	male	2	2	2.0
marcia	south	johnson	female	4	4	4.0
peter	north	smith	male	3	5	4.0
jan	north	smith	female	4	4	4.0
bobby	north	smith	male	5	1	3.0
cindy	south	perry	female	4	5	4.5
alice	south	perry	female	5	4	4.5

Exercise 23

Create a new column high_math_achiever that is an indicator of if a student got a 4 or better on their math score.

scores %>% 
  mutate(high_math_achiever = math_score >= 4)

A tibble: 9 × 7

name	school	teacher	sex	math_score	reading_score	high_math_achiever
<chr>	<chr>	<chr>	<chr>	<dbl>	<dbl>	<lgl>
mike	south	johnson	male	4	1	TRUE
carol	south	johnson	female	3	5	FALSE
greg	south	johnson	male	2	2	FALSE
marcia	south	johnson	female	4	4	TRUE
peter	north	smith	male	3	5	FALSE
jan	north	smith	female	4	4	TRUE
bobby	north	smith	male	5	1	TRUE
cindy	south	perry	female	4	5	TRUE
alice	south	perry	female	5	4	TRUE

Exercise 24

Create a new column reading_score_centered that is a student's reading score minus the mean of all students' reading scores.

scores %>% 
  mutate(reading_score_centered = reading_score - mean(reading_score))

A tibble: 9 × 7

name	school	teacher	sex	math_score	reading_score	reading_score_centered
<chr>	<chr>	<chr>	<chr>	<dbl>	<dbl>	<dbl>
mike	south	johnson	male	4	1	-2.4444444
carol	south	johnson	female	3	5	1.5555556
greg	south	johnson	male	2	2	-1.4444444
marcia	south	johnson	female	4	4	0.5555556
peter	north	smith	male	3	5	1.5555556
jan	north	smith	female	4	4	0.5555556
bobby	north	smith	male	5	1	-2.4444444
cindy	south	perry	female	4	5	1.5555556
alice	south	perry	female	5	4	0.5555556

Note how this may be confusing: reading_score is used twice in the mutate() arugment.

In the first case, it refers to value for the current observation.

In the second case, it refers to the all the values.

Exercise 25

Create a new column called science_score that adds the square of math_score to reading_score.

Display the result by sorting on the new column in descending order.

scores %>% 
    mutate(science_score = math_score**2 + reading_score) %>%
    arrange(desc(science_score))

A tibble: 9 × 7

name	school	teacher	sex	math_score	reading_score	science_score
<chr>	<chr>	<chr>	<chr>	<dbl>	<dbl>	<dbl>
alice	south	perry	female	5	4	29
bobby	north	smith	male	5	1	26
cindy	south	perry	female	4	5	21
marcia	south	johnson	female	4	4	20
jan	north	smith	female	4	4	20
mike	south	johnson	male	4	1	17
carol	south	johnson	female	3	5	14
peter	north	smith	male	3	5	14
greg	south	johnson	male	2	2	6

Part 6: Mutate with groups

Exercise 26

Create a math_score_centered_by_sex column that subtracts a student's math_score from the mean math_score by sex.

Display the results in descending order on the new column.

scores %>% 
  group_by(sex) %>% 
  mutate(math_score_centered_by_sex = math_score - mean(math_score)) %>%
  arrange(desc(math_score_centered_by_sex))

A grouped_df: 9 × 7

name	school	teacher	sex	math_score	reading_score	math_score_centered_by_sex
<chr>	<chr>	<chr>	<chr>	<dbl>	<dbl>	<dbl>
bobby	north	smith	male	5	1	1.5
alice	south	perry	female	5	4	1.0
mike	south	johnson	male	4	1	0.5
marcia	south	johnson	female	4	4	0.0
jan	north	smith	female	4	4	0.0
cindy	south	perry	female	4	5	0.0
peter	north	smith	male	3	5	-0.5
carol	south	johnson	female	3	5	-1.0
greg	south	johnson	male	2	2	-1.5

Question: If Mike and Cindy both got a $4$ for their math score, then
why does Mike have a higher math_score_centered_by_sex score?

Answer: He is higher relative to the other males, which are lower overall.

Exercise 27

Create a reading_score_centered_by_teacher column.

scores %>% 
  group_by(teacher) %>% 
  mutate(math_score_centered_by_teacher = math_score - mean(math_score))

A grouped_df: 9 × 7

name	school	teacher	sex	math_score	reading_score	math_score_centered_by_teacher
<chr>	<chr>	<chr>	<chr>	<dbl>	<dbl>	<dbl>
mike	south	johnson	male	4	1	0.75
carol	south	johnson	female	3	5	-0.25
greg	south	johnson	male	2	2	-1.25
marcia	south	johnson	female	4	4	0.75
peter	north	smith	male	3	5	-1.00
jan	north	smith	female	4	4	0.00
bobby	north	smith	male	5	1	1.00
cindy	south	perry	female	4	5	-0.50
alice	south	perry	female	5	4	0.50

Exercise 28

Make a number_of_students_in_class column that is number of students in a teacher’s class.

For example, it should be $4$ for mike and $3$ for peter.

scores %>% 
  group_by(teacher) %>% 
  mutate(number_of_students_in_class = n())

A grouped_df: 9 × 7

name	school	teacher	sex	math_score	reading_score	number_of_students_in_class
<chr>	<chr>	<chr>	<chr>	<dbl>	<dbl>	<int>
mike	south	johnson	male	4	1	4
carol	south	johnson	female	3	5	4
greg	south	johnson	male	2	2	4
marcia	south	johnson	female	4	4	4
peter	north	smith	male	3	5	3
jan	north	smith	female	4	4	3
bobby	north	smith	male	5	1	3
cindy	south	perry	female	4	5	2
alice	south	perry	female	5	4	2

Part 7: Summarize

Exercise 29

Use summarize() command to find the mean math score for all students.

scores %>% 
  summarize(math_score_mean = mean(math_score))

A tibble: 1 × 1

math_score_mean
<dbl>
3.777778

Note how summarize() creates a new column, like mutate().

It does this because it creates a new table.

Exercise 30

Find the mean reading score for all students.

scores %>% 
  summarize(math_score_mean = mean(reading_score))

A tibble: 1 × 1

math_score_mean
<dbl>
3.444444

Exercise 31

Find the mean for math scores and reading scores added together.

Call the column both_score_mean.

scores %>% 
  summarize(both_score_mean = mean((math_score + reading_score)/2))

A tibble: 1 × 1

both_score_mean
<dbl>
3.611111

Part 8: Summarize with groups

Exercise 32

Find the minimum math score for each school.

scores %>% 
  group_by(school) %>% 
  summarize(min_math_score = min(math_score))

A tibble: 2 × 2

school	min_math_score
<chr>	<dbl>
north	3
south	2

Exercise 33

Find the maximum math score for each teacher.

scores %>% 
  group_by(teacher) %>% 
  summarize(max_math_score = max(math_score))

A tibble: 3 × 2

teacher	max_math_score
<chr>	<dbl>
johnson	4
perry	5
smith	5

Excercise 11.19 Redone

scores %>% 
    group_by(teacher) %>% 
    summarize(max_math = max(math_score)) %>%
    filter(max_math == 5) %>%
    select(teacher)

A tibble: 2 × 1

teacher
<chr>
perry
smith

Exercise 34

If we grouped by sex, and then summarized with the minimum reading score, how many rows would the resulting data frame have?

$2$, because there are $2$ values for sex in the table.

Group by with summarization operations yield dataframes with the same number of rows as distinct values in the grouping.

This is an example of the Split-Apply-Combine pattern.

• To group is to split by distinct value of some feature of set of features
• To summarize to apply some aggregate function to each group
• To combine means to bind the resulting single row tables into one table, with one row for each group

See this article by Brian S. Yandell, a professor of statistics at Wisconsin, for a deeper dive into this topic.

scores %>% 
  group_by(sex) %>%
  summarize(min_reading_score = min(reading_score))

A tibble: 2 × 2

sex	min_reading_score
<chr>	<dbl>
female	4
male	1

Exercise 35

Create a data frame with the mean and median reading score by sex, as well as the number of students of that sex.

scores %>% 
  group_by(sex) %>% 
  summarize(mean_reading_score = mean(reading_score), 
            median_reading_score = median(reading_score), 
            n = n())

A tibble: 2 × 4

sex	mean_reading_score	median_reading_score	n
<chr>	<dbl>	<dbl>	<int>
female	4.40	4.0	5
male	2.25	1.5	4

Part 9: Combining verbs

Exercise 36

Select just the name and math_score columns.

Then create a new column math_score_ec that is a students math score plus 5 extra credit points.

Finally, arrange the data frame by math_score_ec from low to high.

scores %>% 
  select(name, math_score) %>% 
  mutate(math_score_ec = math_score + 5) %>% 
  arrange(math_score_ec)

A tibble: 9 × 3

name	math_score	math_score_ec
<chr>	<dbl>	<dbl>
greg	2	7
carol	3	8
peter	3	8
mike	4	9
marcia	4	9
jan	4	9
cindy	4	9
bobby	5	10
alice	5	10

Exercise 37

Select every column except the teacher column.

Create a new variabled called mean_score that is the mean of a student’s math and reading score.

Finally, arrange the data frame by mean_score from low to high.

scores %>% 
  select(-teacher) %>% 
  mutate(mean_score = (math_score + reading_score)/2) %>% 
  arrange(mean_score)

A tibble: 9 × 6

name	school	sex	math_score	reading_score	mean_score
<chr>	<chr>	<chr>	<dbl>	<dbl>	<dbl>
greg	south	male	2	2	2.0
mike	south	male	4	1	2.5
bobby	north	male	5	1	3.0
carol	south	female	3	5	4.0
marcia	south	female	4	4	4.0
peter	north	male	3	5	4.0
jan	north	female	4	4	4.0
cindy	south	female	4	5	4.5
alice	south	female	5	4	4.5

Exercise 38

Remove any students with 'smith' as a teacher, then find the mean math_score by sex.

scores %>% 
  filter(teacher != "smith") %>% 
  group_by(sex) %>% 
  summarize(mean_math_score = mean(math_score))

A tibble: 2 × 2

sex	mean_math_score
<chr>	<dbl>
female	4
male	3

Exercise 39

Find the min, max, and median reading_score for female students at south school.

scores %>% 
  filter(sex == "female", school == "south") %>% 
  summarize(min_reading_score = min(reading_score),
            max_reading_score = max(reading_score),
            median_reading_score = median(reading_score))

A tibble: 1 × 3

min_reading_score	max_reading_score	median_reading_score
<dbl>	<dbl>	<dbl>
4	5	4.5

Exercise 40

Inspect each of the following code blocks. They both do about the same thing. Which one do you think is preffered from a computer efficiency standpoint?

Code block 1

scores %>% 
  group_by(school, teacher) %>% 
  summarize(max_math_score = max(math_score)) %>% 
  filter(school == "south")

`summarise()` has grouped output by 'school'. You can override using the
`.groups` argument.

A grouped_df: 2 × 3

school	teacher	max_math_score
<chr>	<chr>	<dbl>
south	johnson	4
south	perry	5

Code block 2

scores %>% 
  filter(school == "south") %>% 
  group_by(teacher) %>% 
  summarize(max_math_score = max(math_score))

A tibble: 2 × 2

teacher	max_math_score
<chr>	<dbl>
johnson	4
perry	5

Answer: They both get the max math score by teacher for teachers at south school. The first block calculates the max_math_score for both north and south and then filters out north after that calculation. The second block filters out north right away. This is preferred because it prevents the computer from making unnecessary calculations.

Regarding the error, we could have done this in the first code block:

scores %>% 
  group_by(school, teacher) %>% 
  summarize(max_math_score = max(math_score), .groups = 'keep') %>% 
  filter(school == "south")

A grouped_df: 2 × 3

school	teacher	max_math_score
<chr>	<chr>	<dbl>
south	johnson	4
south	perry	5

Credits

This notebook is adapted largely from Stenhaug's "The 5 verbs of dplyr"