Pipe chaining operations in R Programming - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of pipe chaining with dplyr filter and mutate

What is the output of this R code using pipe chaining?

R Programming

library(dplyr)
data <- tibble(x = 1:5, y = c(2, 4, 6, 8, 10))
result <- data %>% filter(x > 2) %>% mutate(z = y / x) %>% pull(z)
print(result)

A [1] 2 2 2

B [1] 3 2 2

C [1] 2 2 3

D [1] 2 3 2

💡 Hint

Remember filter keeps rows where x > 2, then z is y / x.

The filter keeps rows where x is 3, 4, 5. Then z is y divided by x: 6/3=2, 8/4=2, 10/5=2.

🧠 Conceptual

intermediate

Understanding pipe chaining with base R and magrittr

Which option correctly describes what this pipe chain does?

R Programming

library(magrittr)
result <- 1:5 %>% sum() %>% sqrt()

A Calculates the sum of numbers 1 to 5 and then squares the result

B Calculates the square root of the sum of numbers 1 to 5

C Calculates the sum of the square roots of numbers 1 to 5

D Calculates the square root of each number from 1 to 5 and sums them

💡 Hint

Remember pipe passes the result of one function as input to the next.

The pipe sends 1:5 to sum(), which returns 15, then sqrt(15) is calculated.

🔧 Debug

advanced

Identify the error in this pipe chain

What error does this R code produce?

R Programming

library(dplyr)
data <- tibble(a = 1:3, b = 4:6)
result <- data %>% select(a) %>% filter(b > 4)

A Error: unexpected symbol in filter

B No error, returns filtered tibble

C Error: object 'b' not found

D Error: select() expects at least one column

💡 Hint

Check which columns remain after select(a).

After selecting only column 'a', column 'b' is removed, so filter(b > 4) fails with 'object b not found'.

📝 Syntax

advanced

Which pipe chaining syntax is correct?

Which option shows the correct pipe chaining syntax in R?

A data %>% filter(x > 1) %>% mutate(y = x * 2)

B data |> filter(x > 1) %>% mutate(y = x * 2)

C data %>% filter(x > 1) |> mutate(y = x * 2)

D data |> filter(x > 1) |> mutate(y = x * 2)

💡 Hint

Check if the pipe operators are used consistently.

Option A uses the magrittr pipe operator consistently. Mixing %>% and |> in one chain is invalid.

🚀 Application

expert

Result of complex pipe chaining with multiple functions

What is the value of final after running this code?

R Programming

library(dplyr)
library(magrittr)
data <- tibble(val = c(2, 4, 6, 8))
final <- data %>%
  mutate(sq = val^2) %>%
  filter(sq %% 8 == 0) %>%
  summarise(total = sum(sq)) %>%
  pull(total)
print(final)

A 100

B 64

C 120

D 80

💡 Hint

Calculate squares, filter those divisible by 8, then sum.

Squares: 4, 16, 36, 64. Divisible by 8: 16, 64. Sum: 80.

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