grep and grepl in R Programming - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of grep with value = TRUE

What is the output of this R code?

vec <- c("apple", "banana", "cherry", "date")
grep("a", vec, value = TRUE)

R Programming

vec <- c("apple", "banana", "cherry", "date")
grep("a", vec, value = TRUE)

A [1] "apple" "banana" "date"

B [1] 1 2 4

C [1] "apple" "banana" "cherry" "date"

D [1] 1 2 3 4

💡 Hint

Remember, value = TRUE returns the matching elements, not their positions.

grep with value = TRUE returns the elements of the vector that contain the pattern. Here, "apple", "banana", and "date" contain the letter 'a'.

❓ Predict Output

intermediate

Output of grepl with logical vector

What is the output of this R code?

vec <- c("cat", "dog", "bird", "cow")
grepl("o", vec)

R Programming

vec <- c("cat", "dog", "bird", "cow")
grepl("o", vec)

A [1] FALSE FALSE FALSE TRUE

B [1] TRUE TRUE FALSE TRUE

C [1] TRUE FALSE FALSE TRUE

D [1] FALSE TRUE FALSE TRUE

💡 Hint

grepl returns TRUE for elements containing the pattern, FALSE otherwise.

Only "dog" and "cow" contain the letter 'o', so their positions are TRUE, others FALSE.

🧠 Conceptual

advanced

Difference between grep and grepl

Which statement correctly describes the difference between grep() and grepl() in R?

A grep() returns the count of matches; grepl() returns the matching strings.

B Both grep() and grepl() return matching values but in different orders.

C grep() returns indices or matching values; grepl() returns a logical vector indicating matches.

D grep() returns a logical vector; grepl() returns indices of matches.

💡 Hint

Think about what each function returns: positions vs logical flags.

grep() returns the indices or matching values (if value=TRUE), while grepl() returns a logical vector indicating which elements matched.

❓ Predict Output

advanced

Output of grep with fixed = TRUE

What is the output of this R code?

vec <- c("a.c", "abc", "a-c", "a.c")
grep("a.c", vec, fixed = TRUE)

R Programming

vec <- c("a.c", "abc", "a-c", "a.c")
grep("a.c", vec, fixed = TRUE)

A [1] 1 4

B [1] 1 2 4

C [1] 1 3 4

D [1] 2 3

💡 Hint

fixed = TRUE treats the pattern as a plain string, not a regular expression.

With fixed = TRUE, the pattern "a.c" matches exactly "a.c" strings at positions 1 and 4.

❓ Predict Output

expert

Count of matches using grepl and sum

What is the value of count after running this R code?

vec <- c("red", "green", "blue", "yellow", "redgreen")
count <- sum(grepl("red", vec))
count

R Programming

vec <- c("red", "green", "blue", "yellow", "redgreen")
count <- sum(grepl("red", vec))
count

A 4

B 2

C 1

D 3

💡 Hint

grepl returns TRUE for each element containing "red"; sum counts TRUE as 1.

The elements "red" and "redgreen" contain "red", so count is 2.

Practice

1. What does the grepl function in R return when searching for a pattern in a character vector?

easy

A. The matching elements themselves

B. The positions of elements matching the pattern

C. A logical vector indicating TRUE or FALSE for each element

D. The count of matches found

Solution

1. Step 1: Understand the purpose of grepl

   grepl checks each element of a vector and returns TRUE if the pattern is found, otherwise FALSE.

2. Step 2: Compare with other functions

   grep returns positions or values, but grepl returns logical TRUE/FALSE values for each element.

3. Final Answer:

   A logical vector indicating TRUE or FALSE for each element -> Option C

4. Quick Check:

   grepl returns logical vector [OK]

Hint: Remember: grepl gives TRUE/FALSE, grep gives positions [OK]

Common Mistakes:

• Confusing grepl with grep output
• Expecting grepl to return matching values
• Thinking grepl returns counts

2. Which of the following is the correct syntax to find the positions of elements containing the word "apple" in vector fruits using grep ignoring case?

easy

A. grep("apple", fruits, ignore.case = TRUE)

B. grepl("apple", fruits, ignore.case = TRUE)

C. grep(fruits, "apple", ignore.case = TRUE)

D. grepl(fruits, "apple", ignore.case = TRUE)

Solution

1. Step 1: Identify correct argument order for grep

   The first argument is the pattern, second is the vector to search.

2. Step 2: Check function name and parameters

   grep("apple", fruits, ignore.case = TRUE) correctly searches for "apple" ignoring case and returns positions.

3. Final Answer:

   grep("apple", fruits, ignore.case = TRUE) -> Option A

4. Quick Check:

   Pattern first, vector second in grep [OK]

Hint: Pattern goes first, vector second in grep [OK]

Common Mistakes:

• Swapping pattern and vector arguments
• Using grepl instead of grep for positions
• Omitting ignore.case parameter

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

words <- c("Cat", "dog", "caterpillar", "Dogma")
grep("dog", words, ignore.case = TRUE)

medium

A. [1] TRUE TRUE FALSE FALSE

B. [2] 2 4

C. [1] "dog" "Dogma"

D. [1] 2 4

Solution

1. Step 1: Understand grep output

   grep returns positions of elements matching the pattern.

2. Step 2: Identify matching elements ignoring case

   "dog" matches "dog" (position 2) and "Dogma" (position 4) ignoring case.

3. Final Answer:

   [1] 2 4 -> Option D

4. Quick Check:

   grep returns positions [1] 2 4 [OK]

Hint: grep returns positions, not values or logicals [OK]

Common Mistakes:

• Expecting matching values instead of positions
• Confusing grep with grepl output
• Misreading output format with index labels

4. The following code is intended to return a logical vector indicating which elements contain "blue" ignoring case, but it gives an error. What is the error?

colors <- c("Blue", "red", "green", "BLUE")
grepl(colors, "blue", ignore.case = TRUE)

medium

A. ignore.case is not a valid argument for grepl

B. The pattern and vector arguments are swapped in grepl

C. colors vector contains uppercase letters causing error

D. grepl cannot be used with character vectors

Solution

1. Step 1: Check argument order for grepl

   grepl expects pattern first, then the vector to search.

2. Step 2: Identify swapped arguments in code

   The code uses grepl(colors, "blue", ignore.case = TRUE) which swaps pattern and vector.

3. Final Answer:

   The pattern and vector arguments are swapped in grepl -> Option B

4. Quick Check:

   Pattern first, vector second in grepl [OK]

Hint: Pattern always comes first in grepl [OK]

Common Mistakes:

• Swapping pattern and vector arguments
• Thinking ignore.case is invalid
• Assuming case in vector causes error

5. You have a vector names <- c("Anna", "Annie", "Annette", "Bob", "An"). You want to find all names that start with "Ann" ignoring case. Which code correctly returns the matching names?

hard

A. names[grepl("^Ann", names, ignore.case = TRUE)]

B. grep("Ann", names, ignore.case = TRUE)

C. names[grep("Ann", names)]

D. grepl(names, "^Ann", ignore.case = TRUE)

Solution

1. Step 1: Use grepl with pattern and vector correctly

   To filter names starting with "Ann", use grepl("^Ann", names, ignore.case = TRUE) which returns TRUE for matching elements.

2. Step 2: Subset names vector using logical vector

   Use names[logical_vector] to get matching names.

3. Final Answer:

   names[grepl("^Ann", names, ignore.case = TRUE)] -> Option A

4. Quick Check:

   grepl + logical indexing filters names [OK]

Hint: Use grepl with ^ for start and subset vector [OK]

Common Mistakes:

• Using grep without subsetting vector
• Swapping pattern and vector in grepl
• Not using ^ to match start of string