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.

Practice

1. In R, which operator would you use to extract a single element from a list while removing the list structure?

easy

A. [[]]

B. []

C. $

D. :

Solution

1. Step 1: Understand list element extraction

   The [[]] operator extracts a single element from a list and returns it without the list structure.

2. Step 2: Compare with other operators

   The [] operator returns a subset as a list, $ is a shortcut for named elements but also returns the element itself, and : is a sequence operator unrelated to extraction.

3. Final Answer:

   [[]] -> Option A

4. Quick Check:

   Single element extraction = [[]] [OK]

Hint: Use double brackets [[]] to get single elements from lists [OK]

Common Mistakes:

• Using [] returns a list subset, not a single element
• Confusing $ with [[]] for unnamed elements
• Using : operator for extraction

2. Which of the following is the correct syntax to access the column named age from a data frame df using the $ operator?

easy

A. df[[age]]

B. df[age]

C. df$age

D. df$"age"

Solution

1. Step 1: Understand $ operator usage

   The $ operator accesses named elements directly by their name without quotes.

2. Step 2: Check syntax correctness

   df$age is correct. df[age] is incorrect because age is not quoted or defined as a variable. df[[age]] expects a variable or string inside double brackets. df$"age" is invalid syntax.

3. Final Answer:

   df$age -> Option C

4. Quick Check:

   Access named column with $ and name directly [OK]

Hint: Use df$columnName without quotes to access columns [OK]

Common Mistakes:

• Putting quotes inside $ operator like df$"age"
• Using single brackets without quotes
• Confusing [[]] with $ syntax

3. Given the list lst <- list(a = 1:3, b = 4:6), what is the output of lst["a"]?

medium

A. 1 2 3

B. 4 5 6

C. NULL

D. list(a = 1:3)

Solution

1. Step 1: Understand single bracket extraction on lists

   Using single brackets [] with a name returns a sublist containing that element, preserving the list structure.

2. Step 2: Apply to given list

   lst["a"] returns a list with element named 'a' containing 1:3, not just the vector 1 2 3.

3. Final Answer:

   list(a = 1:3) -> Option D

4. Quick Check:

   Single bracket returns sublist = list(a = 1:3) [OK]

Hint: Single brackets [] return sublist, double brackets [[]] extract element [OK]

Common Mistakes:

• Expecting [] to return just the vector inside
• Confusing [] and [[]] behavior
• Assuming output is NULL or wrong element

4. Consider the code:

df <- data.frame(x = 1:3, y = 4:6)
value <- df[["z"]]

What is the error and how to fix it?

medium

A. Syntax error; fix by adding quotes around z

B. Error: subscript out of bounds; fix by using an existing column name

C. No error; value will be NULL

D. Error: object 'z' not found; fix by defining z first

Solution

1. Step 1: Identify the error cause

   The data frame df has columns 'x' and 'y' but no 'z'. Accessing df[["z"]] causes a subscript out of bounds error.

2. Step 2: Fix the error

   Use a valid column name like 'x' or 'y' inside double brackets to avoid the error.

3. Final Answer:

   Error: subscript out of bounds; fix by using an existing column name -> Option B

4. Quick Check:

   Accessing non-existent column causes subscript error [OK]

Hint: Check column names before using [[]] to avoid subscript errors [OK]

Common Mistakes:

• Assuming [[]] returns NULL if column missing
• Confusing syntax error with runtime error
• Not verifying column existence before access

5. You have a list lst <- list(name = "Alice", scores = c(90, 85, 88)). Which code correctly extracts the second score?

hard

A. lst$scores[2]

B. lst["scores"][2]

C. lst$scores["2"]

D. lst[[scores]][2]

Solution

1. Step 1: Understand list element access

   lst$scores accesses the vector of scores directly.

2. Step 2: Extract the second element from the vector

   Using [2] on the vector gets the second score. So lst$scores[2] works correctly.

3. Step 3: Check other options

   lst["scores"][2] uses single brackets which returns a list, and indexing a length-1 list with [2] causes subscript out of bounds. lst$scores["2"] uses character index on unnamed vector, returns NA. lst[[scores]][2] is invalid because scores is not quoted.

4. Final Answer:

   lst$scores[2] -> Option A

5. Quick Check:

   Use $ to get vector, then [2] for element [OK]

Hint: Use $ to get vector, then [index] for element inside [OK]

Common Mistakes:

• Using single brackets [] on list then [2] on sublist (out of bounds)
• Using character index ["2"] on unnamed vector (returns NA)
• Not quoting names inside [[]]