Logical indexing in R Programming - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of logical indexing with NA values

What is the output of the following R code?

R Programming

x <- c(2, NA, 5, 8, NA)
result <- x[!is.na(x) & x > 4]
print(result)

A [1] 5 8

B [1] 2 5 8

C [1] NA NA

D [1] 5 8 NA

💡 Hint

Remember that logical indexing excludes NA when using !is.na().

The code selects elements of x that are not NA and greater than 4. Only 5 and 8 satisfy both conditions.

❓ Predict Output

intermediate

Logical indexing with multiple conditions

What does this R code print?

R Programming

v <- c(10, 15, 20, 25, 30)
subset <- v[v %% 10 == 0 & v > 15]
print(subset)

A [1] 10 20 30

B [1] 20 30

C [1] 20

D [1] 15 25

💡 Hint

Check which numbers are multiples of 10 and greater than 15.

The code selects numbers divisible by 10 and greater than 15. These are 20 and 30.

🔧 Debug

advanced

Identify the error in logical indexing

What error does this R code produce?

R Programming

data <- c(1, 2, 3, 4, 5)
index <- c(TRUE, FALSE, TRUE, FALSE, TRUE, TRUE)
result <- data[index]
print(result)

A [1] 1 2

B Error: subscript out of bounds

C [1] 1 3 5

D Error: logical subscript too long

💡 Hint

Check the length of the logical vector compared to the data vector.

The logical index vector is longer than the data vector, causing the error 'logical subscript too long'.

❓ Predict Output

advanced

Logical indexing with named vectors

What is the output of this R code?

R Programming

scores <- c(John=85, Mary=92, Paul=78, Lisa=90)
high_scores <- scores[scores >= 90]
print(names(high_scores))

A [1] "Mary" "Lisa"

B NULL

C [1] "John" "Mary"

D [1] 92 90

💡 Hint

Logical indexing keeps names of selected elements.

The code selects scores 90 or above and prints their names, which are Mary and Lisa.

❓ Predict Output

expert

Complex logical indexing with NA and recycling

What is the output of this R code?

R Programming

x <- c(NA, 3, NA, 7, 5)
idx <- c(TRUE, FALSE)
result <- x[!is.na(x) & idx]
print(result)

A [1] 3 7 5

B [1] 3 5

C [1] 5

D Error: logical subscript too long

💡 Hint

Logical vector idx is recycled to match length of x.

The logical vector idx is recycled to length 5: TRUE, FALSE, TRUE, FALSE, TRUE. Combined with !is.na(x), positions 2, 4 and 5 are not NA. But idx at position 2 is FALSE, position 4 is FALSE, so only position 5 is selected.

Practice

1. What does logical indexing in R do when applied to a vector like v[c(TRUE, FALSE, TRUE)]?

easy

A. Selects elements of v where the index is TRUE

B. Selects elements of v where the index is FALSE

C. Returns the length of v

D. Returns an error because logical vectors cannot index

Solution

1. Step 1: Understand logical indexing

   Logical indexing uses a TRUE/FALSE vector to pick elements from the original vector where TRUE is present.

2. Step 2: Apply logical vector to v

   Elements at positions with TRUE in the logical vector are selected; others are skipped.

3. Final Answer:

   Selects elements of v where the index is TRUE -> Option A

4. Quick Check:

   TRUE picks elements [OK]

Hint: TRUE means pick element, FALSE means skip [OK]

Common Mistakes:

• Thinking FALSE picks elements
• Confusing logical indexing with numeric indexing
• Assuming logical vector length doesn't matter

2. Which of the following is the correct syntax to select elements of vector x where values are greater than 5 using logical indexing?

easy

A. x[> 5]

B. x > 5[x]

C. x[x > 5]

D. x[5 > x]

Solution

1. Step 1: Understand logical condition syntax

   The condition x > 5 creates a logical vector TRUE where elements are greater than 5.

2. Step 2: Use logical vector to index x

   Using x[x > 5] selects elements of x where the condition is TRUE.

3. Final Answer:

   x[x > 5] -> Option C

4. Quick Check:

   Correct logical indexing syntax [OK]

Hint: Put condition inside square brackets after vector [OK]

Common Mistakes:

• Placing condition outside brackets
• Using invalid syntax like x[> 5]
• Swapping operands in condition

3. What is the output of the following R code?

v <- c(10, 3, 7, 2, 9)
v[v >= 7]

medium

A. [1] 10 3 7 2 9

B. [1] 3 2

C. Error: invalid logical indexing

D. [1] 10 7 9

Solution

1. Step 1: Evaluate the logical condition v >= 7

   This creates a logical vector: TRUE, FALSE, TRUE, FALSE, TRUE for elements 10, 3, 7, 2, 9 respectively.

2. Step 2: Select elements where condition is TRUE

   Elements at TRUE positions are 10, 7, and 9.

3. Final Answer:

   [1] 10 7 9 -> Option D

4. Quick Check:

   Only values >= 7 selected [OK]

Hint: Check which elements satisfy condition, pick those [OK]

Common Mistakes:

• Including elements less than 7
• Confusing >= with >
• Expecting an error due to indexing

4. Identify the error in this R code snippet:

nums <- c(4, 8, 15, 16, 23, 42)
nums[nums > 10 & TRUE]

medium

A. No error, code runs correctly

B. Logical vector length mismatch

C. Incorrect use of & with TRUE

D. Syntax error due to missing parentheses

Solution

1. Step 1: Analyze logical condition nums > 10 & TRUE

   The expression nums > 10 returns a logical vector; AND with TRUE keeps it unchanged.

2. Step 2: Check for errors

   The logical vector length matches nums, and syntax is correct, so no error occurs.

3. Final Answer:

   No error, code runs correctly -> Option A

4. Quick Check:

   Logical AND with TRUE keeps vector same [OK]

Hint: AND with TRUE keeps condition unchanged [OK]

Common Mistakes:

• Thinking & TRUE causes error
• Expecting length mismatch error
• Missing that & works element-wise

5. Given a vector temps <- c(23, NA, 19, 25, NA, 21), how can you use logical indexing to select only the non-missing values?

hard

A. temps[is.na(temps)]

B. temps[!is.na(temps)]

C. temps[temps != NA]

D. temps[temps > NA]

Solution

1. Step 1: Understand how to detect missing values

   The function is.na(temps) returns TRUE for missing values (NA) and FALSE otherwise.

2. Step 2: Use logical NOT to select non-missing

   Using !is.na(temps) creates a logical vector TRUE where values are not missing.

3. Step 3: Index temps with this logical vector

   This selects only the non-NA values from temps.

4. Final Answer:

   temps[!is.na(temps)] -> Option B

5. Quick Check:

   Use !is.na() to filter out NA [OK]

Hint: Use !is.na() to pick non-missing values [OK]

Common Mistakes:

• Using temps != NA which is invalid
• Selecting is.na(temps) to get missing values
• Trying to compare with NA using > or <