Nested lists in R Programming - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of nested list indexing

What is the output of the following R code?

R Programming

nested_list <- list(a = list(1, 2, 3), b = list(4, 5, 6))
result <- nested_list$a[[2]]
print(result)

A [1] 3

B [1] 2

C [1] 1

D NULL

💡 Hint

Remember that [[ ]] extracts the element itself, not a sublist.

nested_list$a is a list of three elements: 1, 2, 3. Using [[2]] extracts the second element, which is 2.

❓ Predict Output

intermediate

Length of nested list element

What is the value of length(nested_list$b) after running this code?

R Programming

nested_list <- list(a = list(1, 2, 3), b = list(4, 5, 6))
length(nested_list$b)

A 3

B 1

C 6

D NULL

💡 Hint

length() returns the number of elements in the list, not the sum of elements.

nested_list$b is a list with three elements: 4, 5, and 6, so length is 3.

❓ Predict Output

advanced

Output of nested list modification

What is the output of this R code?

R Programming

nested_list <- list(a = list(1, 2), b = list(3, 4))
nested_list$a[[2]] <- 10
print(nested_list$a)

A Error: object of type 'closure' is not subsettable

B

[[1]] 1
[[2]] 2

C [1] 1 10

D

[[1]] 1
[[2]] 10

💡 Hint

Using [[ ]] allows you to replace the element inside the nested list.

The second element of nested_list$a is replaced by 10. Printing nested_list$a shows the list with elements 1 and 10.

❓ Predict Output

advanced

Accessing deeply nested list element

What is the output of this code?

R Programming

nested_list <- list(x = list(y = list(z = 42)))
print(nested_list$x$y$z)

A [1] 42

B NULL

C Error: object 'z' not found

D [1] "42"

💡 Hint

Use $ to access named elements inside nested lists.

nested_list$x$y$z accesses the value 42 stored deeply inside the nested list.

🧠 Conceptual

expert

Number of elements in a nested list

Given the nested list below, how many top-level elements does it have?

R Programming

nested_list <- list(alpha = list(1, 2), beta = list(3, 4), gamma = 5)

A 5

B 2

C 3

D 1

💡 Hint

Count only the elements directly inside nested_list, not inside sublists.

nested_list has three top-level elements named alpha, beta, and gamma.

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