Pipe chaining operations in R Programming - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to chain the operations using the pipe operator.

R Programming

library(dplyr)
data <- data.frame(x = 1:5, y = 6:10)
result <- data [1] filter(x > 2)

Drag options to blanks, or click blank then click option'

A%$%

B%>%

C%T>%

D%<>%

Attempts:

3 left

💡 Hint

Common Mistakes

Using other pipe operators like %$% or %T>% which have different purposes.

The pipe operator %>% passes the data to the next function.

2fill in blank

medium

Complete the code to select column 'y' after filtering.

R Programming

library(dplyr)
data <- data.frame(x = 1:5, y = 6:10)
result <- data %>% filter(x > 2) [1] select(y)

Drag options to blanks, or click blank then click option'

A%>%

B%$%

C%T>%

D%<>%

Attempts:

3 left

💡 Hint

Common Mistakes

Using different pipe operators that do not chain functions properly.

Use %>% again to continue chaining operations.

3fill in blank

hard

Fix the error in the pipe chain to calculate the mean of column 'y'.

R Programming

library(dplyr)
data <- data.frame(x = 1:5, y = 6:10)
mean_y <- data %>% filter(x > 2) [1] summarise(mean_y = mean(y))

Drag options to blanks, or click blank then click option'

A%T>%

B%$%

C%>%

D%<>%

Attempts:

3 left

💡 Hint

Common Mistakes

Using %$% which exposes variables but does not chain properly here.

The pipe operator %>% correctly passes the filtered data to summarise.

4fill in blank

hard

Fill both blanks to create a pipe chain that filters and then mutates a new column.

R Programming

library(dplyr)
data <- data.frame(x = 1:5, y = 6:10)
result <- data [1] filter(x >= 3) [2] mutate(z = y * 2)

Drag options to blanks, or click blank then click option'

A%>%

B%T>%

C%<>%

D%$%

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing different pipe operators in the same chain.

Use %>% to chain both filter and mutate operations.

5fill in blank

hard

Fill all three blanks to chain filter, mutate, and then select columns.

R Programming

library(dplyr)
data <- data.frame(x = 1:5, y = 6:10)
result <- data [1] filter(x < 5) [2] mutate(z = y + 1) [3] select(x, z)

Drag options to blanks, or click blank then click option'

A%$%

B%>%

C%<>%

D%T>%

Attempts:

3 left

💡 Hint

Common Mistakes

Using different pipe operators that break the chain.

Use %>% consistently to chain all operations in order.

Practice

1.

What does the pipe operator %>% do in R?

easy

A. It creates a new variable to store intermediate results.

B. It comments out the rest of the line in the code.

C. It passes the output of one function as input to the next function.

D. It repeats the previous function multiple times.

Solution

1. Step 1: Understand the pipe operator purpose

   The pipe operator %>% takes the result from the left side and passes it as the first argument to the function on the right side.

2. Step 2: Compare options with pipe behavior

   Only It passes the output of one function as input to the next function. correctly describes this behavior; other options describe unrelated actions.

3. Final Answer:

   It passes the output of one function as input to the next function. -> Option C

4. Quick Check:

   Pipe operator = passes output to next function [OK]

Hint: Pipe means pass output directly to next function [OK]

Common Mistakes:

• Thinking pipe creates new variables
• Confusing pipe with comment symbol
• Assuming pipe repeats functions

2.

Which of the following is the correct syntax to use pipe chaining in R?

library(dplyr)
data <- data.frame(x = 1:3, y = 4:6)
data %>% ? %>% head()

easy

A. filter[x > 1]

B. filter(x > 1)

C. filter{ x > 1 }

D. filter x > 1

Solution

1. Step 1: Identify correct function call syntax in pipe

   In R, functions are called with parentheses enclosing arguments, e.g., filter(x > 1).

2. Step 2: Check options for correct syntax

   Only filter(x > 1) uses parentheses correctly; others use invalid syntax for function calls.

3. Final Answer:

   filter(x > 1) -> Option B

4. Quick Check:

   Function calls need parentheses [OK]

Hint: Use parentheses for function arguments in pipe [OK]

Common Mistakes:

• Omitting parentheses in function calls
• Using braces or brackets instead of parentheses
• Writing function and arguments without separator

3.

What is the output of this R code?

library(dplyr)
data <- tibble(x = 1:5, y = 6:10)
data %>% filter(x > 2) %>% mutate(z = x + y) %>% pull(z)

medium

A. [1] 11 13 15

B. [1] 3 4 5

C. [1] 7 8 9

D. [1] 6 7 8

Solution

1. Step 1: Filter rows where x > 2

   Original data: x = 1,2,3,4,5; y = 6,7,8,9,10. filter(x > 2) keeps x=3 (y=8), x=4 (y=9), x=5 (y=10).

2. Step 2: Mutate z = x + y and pull(z)

   z values: 3+8=11, 4+9=13, 5+10=15. pull(z) outputs [1] 11 13 15.

3. Final Answer:

   [1] 11 13 15 -> Option A

4. Quick Check:

   filter(x > 2) %>% mutate(z = x + y) %>% pull(z) = [1] 11 13 15 [OK]

Hint: Follow pipe steps left to right carefully [OK]

Common Mistakes:

• Not applying filter before mutate
• Mixing up x and y values
• Ignoring pull() extracts vector

4.

Find the error in this pipe chain and select the fix:

library(dplyr)
data <- tibble(a = 1:4, b = 5:8)
data %>% filter(a > 2) %>% mutate(c = a + b) %>% select(c, d)

medium

A. Remove mutate(c = a + b)

B. Add filter(b > 5) before mutate

C. Replace %>% with +

D. Change select(c, d) to select(c)

Solution

1. Step 1: Identify error in select()

   The code tries to select columns c and d, but d does not exist in the data.

2. Step 2: Fix select() to only include existing columns

   Changing select(c, d) to select(c) removes the error by selecting only the existing column c.

3. Final Answer:

   Change select(c, d) to select(c) -> Option D

4. Quick Check:

   Select only existing columns to avoid error [OK]

Hint: Select only columns present in data [OK]

Common Mistakes:

• Selecting non-existent columns
• Using wrong operators instead of pipe
• Removing needed mutate step

5.

Using pipe chaining, how can you transform this data frame df to get the sum of squares of even numbers in column num?

df <- data.frame(num = 1:6)

Choose the correct pipe chain.

hard

A. df %>% filter(num %% 2 == 0) %>% mutate(sq = num^2) %>% summarise(total = sum(sq))

B. df %>% mutate(sq = num^2) %>% filter(num %% 2 == 0) %>% sum(sq)

C. df %>% filter(num %% 2 == 0) %>% summarise(total = sum(num))

D. df %>% summarise(total = sum(num^2)) %>% filter(num %% 2 == 0)

Solution

1. Step 1: Filter even numbers from column num

   Use filter(num %% 2 == 0) to keep only even numbers.

2. Step 2: Create a new column sq with squares of num

   Use mutate(sq = num^2) to add squared values.

3. Step 3: Summarise total sum of sq column

   Use summarise(total = sum(sq)) to get the sum of squares.

4. Final Answer:

   df %>% filter(num %% 2 == 0) %>% mutate(sq = num^2) %>% summarise(total = sum(sq)) -> Option A

5. Quick Check:

   Filter -> mutate -> summarise = sum squares of evens [OK]

Hint: Filter first, then mutate, then summarise [OK]

Common Mistakes:

• Summarising before filtering
• Using sum() incorrectly inside pipe
• Mutating after summarise