Nested lists in R Programming - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a nested list with two elements.

R Programming

nested_list <- list(1, [1])

Drag options to blanks, or click blank then click option'

Amatrix(2, 3)

B2:3

Clist(2, 3)

Dc(2, 3)

Attempts:

3 left

💡 Hint

Common Mistakes

Using c(2, 3) creates a vector, not a nested list.

Using 2:3 creates a sequence, not a nested list.

The list() function creates a nested list. Using list(2, 3) as the second element nests another list inside the main list.

2fill in blank

medium

Complete the code to access the second element of the nested list.

R Programming

nested_list <- list(1, list(2, 3))
second_element <- nested_list[[[1]]]

Drag options to blanks, or click blank then click option'

A1

B2

C3

D4

Attempts:

3 left

💡 Hint

Common Mistakes

Using single brackets [ ] returns a sublist, not the element itself.

Using the wrong index number.

In R, double square brackets [[ ]] access elements of a list. The second element is at position 2.

3fill in blank

hard

Fix the error in the code to access the number 3 inside the nested list.

R Programming

nested_list <- list(1, list(2, 3))
value <- nested_list[[2]][[[1]]]

Drag options to blanks, or click blank then click option'

A2

B1

C3

D4

Attempts:

3 left

💡 Hint

Common Mistakes

Using [[1]] returns 2, not 3.

Using single brackets [ ] instead of double brackets.

The inner list is at position 2 inside the second element. So to get the number 3, use [[2]] on the inner list.

4fill in blank

hard

Fill both blanks to create a nested list and access the number 4 inside it.

R Programming

nested_list <- list(1, list(3, [1]))
value <- nested_list[[2]][[[2]]]

Drag options to blanks, or click blank then click option'

A4

B1

C2

D3

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up the positions of elements in the inner list.

Using wrong indices to access elements.

The inner list contains 3 and 4. To access 4, put it as the second element in the inner list and access it with [[2]][[2]].

5fill in blank

hard

Fill all three blanks to create a nested list and access the number 6 inside it.

R Programming

nested_list <- list([1], list([2], 6))
value <- nested_list[[[3]]][[2]]

Drag options to blanks, or click blank then click option'

A5

B4

C1

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong indices to access elements.

Confusing the order of elements in the lists.

The outer list has 5 and an inner list with 4 and 6. To access 6, get the second element of the outer list with [[2]] and then the second element inside it.

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