Accessing elements ([], [[]], $) in R Programming - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to access the first element of the vector v.

R Programming

v <- c(10, 20, 30)
first_element <- v[1]1

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

A{

B[[

C$

D[

Attempts:

3 left

💡 Hint

Common Mistakes

Using double brackets [[ ]] which are for lists or data frames.

Using $ which is for named list or data frame elements.

In R, single square brackets [] are used to access elements of a vector by index.

2fill in blank

medium

Complete the code to extract the first element from the list lst.

R Programming

lst <- list(a = 5, b = 10)
first_element <- lst[1]1

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

A[[

B{

C$

D[

Attempts:

3 left

💡 Hint

Common Mistakes

Using single brackets which return a list, not the element.

Using $ without the element name.

Double square brackets [[ ]] extract the element itself from a list, not a sublist.

3fill in blank

hard

Fix the error in the code to access the element named 'b' in the list lst.

R Programming

lst <- list(a = 5, b = 10)
value <- lst[1]b

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

A$

B[[b]]

C[b]

D[['b']]

Attempts:

3 left

💡 Hint

Common Mistakes

Using brackets with unquoted names which causes an error.

Using double brackets without quotes around the name.

The dollar sign $ is used to access named elements in a list by name without quotes.

4fill in blank

hard

Fill both blanks to create a named vector and access the element named 'second'.

R Programming

vec <- c(first = 1, second = 2, third = 3)
value <- vec[1]"second"[2]

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

A[

B[[

C]

Attempts:

3 left

💡 Hint

Common Mistakes

Using double brackets which are not valid for vectors.

Forgetting to close the brackets.

Single brackets [] with quotes access named elements in a vector and return a vector of length 1.

5fill in blank

hard

Complete the code to create a list with named elements and access the element named 'age'.

R Programming

person <- list(name = "Alice", age = 30, city = "NY")
age_value <- person[1]"age"

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

A[

B[[

C]

Attempts:

3 left

💡 Hint

Common Mistakes

Using single brackets which return a sublist, not the element.

Not quoting the name inside the brackets.

Use double brackets [[ ]] with quotes to extract the element itself from a list by name.

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 [[]]