Pipe chaining operations in R Programming - Mini Project: Build & Apply

Pipe chaining operations

📖 Scenario: You work in a small bakery and keep track of daily sales in a list. You want to find out which items sold more than 10 units and then get their names in uppercase to display on a special board.

🎯 Goal: Use pipe chaining operations to filter items with sales greater than 10 and convert their names to uppercase.

📋 What You'll Learn

Create a named vector called sales with exact entries: Bread = 15, Cookies = 8, Cake = 12, Pie = 5

Create a variable called threshold and set it to 10

Use pipe chaining with %>% to filter sales for items with sales greater than threshold and then convert the item names to uppercase

Print the final result

💡 Why This Matters

🌍 Real World

Filtering and transforming sales data helps businesses quickly identify top-selling products and display them clearly for customers or staff.

💼 Career

Data analysts and business intelligence professionals often use pipe chaining in R to write clean, readable code for data filtering and transformation.

1

Create the sales data

Create a named vector called sales with these exact entries: Bread = 15, Cookies = 8, Cake = 12, Pie = 5

R Programming

# Create the sales vector with exact names and values
# Your code here

Hint

Use c() to create a named vector with names and values separated by =.

2

Set the sales threshold

Create a variable called threshold and set it to 10

R Programming

sales <- c(Bread = 15, Cookies = 8, Cake = 12, Pie = 5)
# Create the threshold variable and set it to 10
# Your code here

Hint

Just assign the number 10 to the variable threshold.

3

Filter and transform with pipe chaining

Use pipe chaining with %>% to filter sales for items with sales greater than threshold and then convert the item names to uppercase. Store the result in a variable called top_items. Use magrittr::extract() or base R to filter and names() with toupper() to convert names.

R Programming

sales  threshold and convert names to uppercase
# Store result in top_items
# Your code here

Hint

Use %>% to chain filtering and renaming. Filtering keeps sales above threshold. Then change names to uppercase with toupper().

4

Print the final result

Write print(top_items) to display the filtered and renamed sales vector.

R Programming

library(magrittr)
top_items %
  magrittr::extract(. > threshold) %>%
  { names(.) <- toupper(names(.)); . }

# Print the top_items vector
# Your code here

Hint

Use print(top_items) to show the final filtered and renamed vector.

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