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.