Named list elements in R Programming - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of accessing named list elements

What is the output of the following R code?

R Programming

my_list <- list(a = 10, b = 20, c = 30)
result <- my_list$b
print(result)

A [1] 30

B [1] 10

C [1] 20

D NULL

💡 Hint

Remember that named list elements can be accessed by their names using the $ operator.

The list has elements named 'a', 'b', and 'c' with values 10, 20, and 30 respectively. Accessing my_list$b returns the value 20.

❓ Predict Output

intermediate

Output of modifying named list elements

What will be the output after running this R code?

R Programming

my_list <- list(x = 5, y = 15)
my_list$y <- my_list$y + 10
print(my_list$y)

A Error: object 'y' not found

B [1] 15

C [1] 10

D [1] 25

💡 Hint

Think about how the value of 'y' is updated by adding 10.

Initially, my_list$y is 15. Adding 10 updates it to 25, so printing my_list$y outputs 25.

🔧 Debug

advanced

Identify the error in accessing named list elements

What error does this R code produce?

R Programming

my_list <- list(a = 1, b = 2)
print(my_list[['c']])

A list()

B Error: object 'c' not found

C [1] NA

D NULL

💡 Hint

Accessing a non-existent name with double brackets produces an error.

Using double brackets with a non-existent name produces an error: subscript out of bounds.

❓ Predict Output

advanced

Output of named list element extraction with double brackets

What is the output of this R code?

R Programming

my_list <- list(a = 100, b = 200)
result <- my_list[['a']]
print(result)

A [1] 100

B [1] 200

C list(a = 100)

D Error: subscript out of bounds

💡 Hint

Double brackets extract the element itself, not a sublist.

Using double brackets with a name extracts the element value directly, so my_list[['a']] returns 100.

🧠 Conceptual

expert

Number of elements in a named list after modification

Consider the following R code. How many elements does my_list have after execution?

R Programming

my_list <- list(one = 1, two = 2)
my_list$three <- 3
my_list$two <- NULL

A 2

B 3

C 1

D 0

💡 Hint

Assigning NULL to a list element removes it.

Initially, my_list has 2 elements. Adding 'three' makes 3 elements. Setting 'two' to NULL removes it, so 2 elements remain.