Nested lists in R Programming - Mini Project: Build & Apply

Working with Nested Lists in R

📖 Scenario: Imagine you are organizing a small library. Each shelf holds a list of books, and each book has a title and an author. You want to store this information in a nested list in R.

🎯 Goal: Create a nested list in R where each shelf is a list of books, and each book is represented by a list containing its title and author. Then, extract and print the title of the first book on the second shelf.

📋 What You'll Learn

Create a nested list called library with two shelves.

Each shelf should be a list of two books.

Each book should be a list with elements title and author.

Create a variable called first_book_second_shelf to store the title of the first book on the second shelf.

Print the value of first_book_second_shelf.

💡 Why This Matters

🌍 Real World

Nested lists help organize complex data like library catalogs, menus, or survey responses.

💼 Career

Understanding nested data structures is important for data analysis, software development, and managing hierarchical information.

1

Create the nested list library

Create a nested list called library with two shelves. Each shelf should be a list of two books. Each book should be a list with elements title and author. Use these exact values:

First shelf:
- Book 1: title = "The Hobbit", author = "J.R.R. Tolkien"
- Book 2: title = "1984", author = "George Orwell"

Second shelf:
- Book 1: title = "Pride and Prejudice", author = "Jane Austen"
- Book 2: title = "To Kill a Mockingbird", author = "Harper Lee"

R Programming

# Create the nested list 'library' with the two shelves and books
# Your code here

Hint

Use list() to create lists inside lists. Each shelf is a list of two book lists.

2

Create variable first_book_second_shelf

Create a variable called first_book_second_shelf and assign it the title of the first book on the second shelf from the library nested list.

R Programming

library <- list(
  shelf1 = list(
    list(title = "The Hobbit", author = "J.R.R. Tolkien"),
    list(title = "1984", author = "George Orwell")
  ),
  shelf2 = list(
    list(title = "Pride and Prejudice", author = "Jane Austen"),
    list(title = "To Kill a Mockingbird", author = "Harper Lee")
  )
)
# Assign the title of the first book on the second shelf to 'first_book_second_shelf'
# Your code here

Hint

Use $ and double square brackets [[ ]] to access nested list elements.

3

Print the title stored in first_book_second_shelf

Write a print() statement to display the value of the variable first_book_second_shelf.

R Programming

library <- list(
  shelf1 = list(
    list(title = "The Hobbit", author = "J.R.R. Tolkien"),
    list(title = "1984", author = "George Orwell")
  ),
  shelf2 = list(
    list(title = "Pride and Prejudice", author = "Jane Austen"),
    list(title = "To Kill a Mockingbird", author = "Harper Lee")
  )
)
first_book_second_shelf <- library$shelf2[[1]]$title
# Print the title of the first book on the second shelf
# Your code here

Hint

Use print() to show the value stored in a variable.

4

Final Output Check

Run the program to print the title of the first book on the second shelf.

R Programming

library <- list(
  shelf1 = list(
    list(title = "The Hobbit", author = "J.R.R. Tolkien"),
    list(title = "1984", author = "George Orwell")
  ),
  shelf2 = list(
    list(title = "Pride and Prejudice", author = "Jane Austen"),
    list(title = "To Kill a Mockingbird", author = "Harper Lee")
  )
)
first_book_second_shelf <- library$shelf2[[1]]$title
print(first_book_second_shelf)
# Run the program to see the output

Hint

The output should be the title of the first book on the second shelf.

Practice

1. What is a nested list in R?

easy

A. A list with only numeric elements

B. A list that contains other lists as elements

C. A vector with mixed data types

D. A matrix with rows and columns

Solution

1. Step 1: Understand list structure in R

   Lists in R can hold different types of elements, including other lists.

2. Step 2: Define nested list

   A nested list is a list where some elements are themselves lists, allowing complex data grouping.

3. Final Answer:

   A list that contains other lists as elements -> Option B

4. Quick Check:

   Nested list = list inside list [OK]

Hint: Nested lists hold lists inside lists for complex data [OK]

Common Mistakes:

• Confusing nested lists with vectors
• Thinking nested lists must be numeric
• Mixing up matrices and lists

2. Which of the following is the correct way to create a nested list in R?

easy

A. nested <- list(a = 1, b = list(c = 2, d = 3))

B. nested <- c(a = 1, b = c(c = 2, d = 3))

C. nested <- matrix(c(1, 2, 3, 4), nrow=2)

D. nested <- data.frame(a = 1, b = 2)

Solution

1. Step 1: Identify list creation syntax

   Use list() to create lists; nested lists are lists inside list() calls.

2. Step 2: Check each option

   nested <- list(a = 1, b = list(c = 2, d = 3)) uses list() with a nested list inside, which is correct syntax for nested lists.

3. Final Answer:

   nested <- list(a = 1, b = list(c = 2, d = 3)) -> Option A

4. Quick Check:

   list() with list() inside = nested list [OK]

Hint: Use list() inside list() to nest lists [OK]

Common Mistakes:

• Using c() instead of list() for nested lists
• Confusing data.frame or matrix with lists
• Missing nested list syntax

3. What is the output of the following R code?

nested <- list(x = 10, y = list(a = 5, b = 15))
nested$y$a

medium

A. 10

B. NULL

C. 15

D. 5

Solution

1. Step 1: Understand list structure

   nested is a list with element y, which is itself a list containing a=5 and b=15.

2. Step 2: Access nested element

   nested$y$a accesses element 'a' inside the nested list 'y', which is 5.

3. Final Answer:

   5 -> Option D

4. Quick Check:

   nested$y$a = 5 [OK]

Hint: Use $ to access nested list elements stepwise [OK]

Common Mistakes:

• Accessing nested elements with single brackets
• Confusing element names
• Expecting the outer element value

4. Identify the error in this R code for nested lists:

nested <- list(a = 1, b = list(c = 2, d = 3))
value <- nested$b[c]

medium

A. Missing assignment operator <-

B. Incorrect list creation syntax

C. Using single brackets instead of double brackets to access list element

D. Using $ instead of [[ ]] to access nested list

Solution

1. Step 1: Understand list element access

   To access a single element by name in a list, use double brackets [[ ]] or $; single brackets [ ] return a sublist.

2. Step 2: Analyze the code

   nested$b[c] uses single brackets with a character index, which returns a sublist, not the element value.

3. Final Answer:

   Using single brackets instead of double brackets to access list element -> Option C

4. Quick Check:

   Use [[ ]] or $ for single element access [OK]

Hint: Use [[ ]] or $ to get single list elements, not [ ] [OK]

Common Mistakes:

• Using [ ] returns sublist, not element
• Confusing list and vector indexing
• Forgetting to use quotes for names

5. Given a nested list data <- list(group1 = list(scores = c(10, 20)), group2 = list(scores = c(15, 25))), how do you create a new list with the average score of each group?

hard

A. averages <- list(group1 = mean(data$group1$scores), group2 = mean(data$group2$scores))

B. averages <- list(group1 = sum(data$group1$scores), group2 = sum(data$group2$scores))

C. averages <- list(group1 = data$group1$scores[1], group2 = data$group2$scores[1])

D. averages <- list(group1 = length(data$group1$scores), group2 = length(data$group2$scores))

Solution

1. Step 1: Access nested scores vectors

   Use data$group1$scores and data$group2$scores to get numeric vectors for each group.

2. Step 2: Calculate mean for each group

   Apply mean() function to each scores vector to get average scores.

3. Step 3: Create new list with averages

   Assign the calculated means to a new list with same group names.

4. Final Answer:

   averages <- list(group1 = mean(data$group1$scores), group2 = mean(data$group2$scores)) -> Option A

5. Quick Check:

   mean() on nested scores = averages list [OK]

Hint: Use mean() on nested vectors accessed by $ [OK]

Common Mistakes:

• Using sum() instead of mean()
• Accessing scores incorrectly
• Confusing length with average