Accessing elements ([], [[]], $) in R Programming - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of list element access with single brackets

What is the output of this R code?

my_list <- list(a = 1:3, b = 4:6)
result <- my_list["a"]
print(result)

R Programming

my_list <- list(a = 1:3, b = 4:6)
result <- my_list["a"]
print(result)

A Vector 1 2 3

B List of 1 element named 'a' containing 1 2 3

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

D NULL

💡 Hint

Remember that single brackets return a sublist, not the element itself.

Using single brackets ["a"] on a list returns a sublist containing the element named 'a'. It keeps the list structure.

❓ Predict Output

intermediate

Output of list element access with double brackets

What is the output of this R code?

my_list <- list(a = 1:3, b = 4:6)
result <- my_list[["a"]]
print(result)

R Programming

my_list <- list(a = 1:3, b = 4:6)
result <- my_list[["a"]]
print(result)

A Vector 1 2 3

B List of 1 element named 'a' containing 1 2 3

C Error: subscript out of bounds

D NULL

💡 Hint

Double brackets extract the element itself, not a sublist.

Using double brackets [["a"]] extracts the element named 'a' from the list, returning the vector 1 2 3 directly.

❓ Predict Output

advanced

Accessing data frame columns with $ operator

What is the output of this R code?

df <- data.frame(x = 1:3, y = 4:6)
result <- df$y
print(result)

R Programming

df <- data.frame(x = 1:3, y = 4:6)
result <- df$y
print(result)

A List with element y containing 4 5 6

B Data frame with column y only

C Error: object 'y' not found

D Vector 4 5 6

💡 Hint

The $ operator extracts a column as a vector.

Using df$y extracts the column 'y' from the data frame as a vector 4 5 6.

❓ Predict Output

advanced

Difference between single and double brackets on nested list

What is the output of this R code?

nested_list <- list(a = list(b = 10))
result1 <- nested_list["a"]
result2 <- nested_list[["a"]]
print(result1)
print(result2)

R Programming

nested_list <- list(a = list(b = 10))
result1 <- nested_list["a"]
result2 <- nested_list[["a"]]
print(result1)
print(result2)

A result1 is a list with element 'a' containing list(b=10); result2 is list(b=10)

B result1 is list(b=10); result2 is a list with element 'a' containing list(b=10)

C Both result1 and result2 are list(b=10)

D Both result1 and result2 are lists with element 'a' containing list(b=10)

💡 Hint

Single brackets keep the outer list structure; double brackets extract the inner element.

result1 keeps the outer list with element 'a' which itself is a list; result2 extracts the inner list directly.

🧠 Conceptual

expert

Why does my_list$a and my_list[["a"]] differ from my_list["a"]?

Consider this R list:

my_list <- list(a = 1:3, b = 4:6)

Which statement best explains the difference between my_list$a, my_list[["a"]], and my_list["a"]?

A All three return the same vector 1 2 3 but with different names.

B <code>my_list$a</code> returns NULL, <code>my_list[["a"]]</code> returns a list, and <code>my_list["a"]</code> returns the vector 1 2 3.

C <code>my_list$a</code> and <code>my_list[["a"]]</code> return the element itself (vector 1 2 3), while <code>my_list["a"]</code> returns a sublist containing that element.

D <code>my_list$a</code> returns a list, <code>my_list[["a"]]</code> returns NULL, and <code>my_list["a"]</code> returns an error.

💡 Hint

Think about how $ and [[ ]] extract elements versus [ ] returning sublists.

The $ operator and double brackets [[ ]] extract the element itself, while single brackets [ ] return a sublist preserving the list structure.