Regular expressions in R in R Programming - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of grepl with pattern and ignore.case

What is the output of this R code snippet?

R Programming

text <- c("Apple", "banana", "Cherry", "date")
result <- grepl("a", text, ignore.case = TRUE)
print(result)

A [1] FALSE TRUE TRUE TRUE

B [1] TRUE FALSE TRUE FALSE

C [1] TRUE TRUE FALSE TRUE

D [1] TRUE TRUE TRUE TRUE

💡 Hint

grepl returns TRUE if the pattern is found in each element. ignore.case = TRUE makes it case insensitive.

The pattern 'a' appears in 'Apple' (A), 'banana' (a), and 'date' (a) ignoring case, but not in 'Cherry'. So the result is [1] TRUE TRUE FALSE TRUE, which is option C.

❓ Predict Output

intermediate

Result of sub function with regex

What is the output of this R code?

R Programming

text <- "The rain in Spain"
result <- sub("ain", "XXX", text)
print(result)

A "The rXXX in Spain"

B "The rain in SpXXX"

C "The rain in Spain"

D "The rXXX in SpXXX"

💡 Hint

sub replaces only the first match of the pattern.

The pattern 'ain' appears twice: in 'rain' and 'Spain'. sub replaces only the first occurrence, so only 'rain' changes to 'rXXX'.

🔧 Debug

advanced

Identify the error in regex pattern

What error does this R code produce?

R Programming

text <- "abc123"
result <- grepl("[a-z{3}", text)
print(result)

A Error: invalid regular expression '[a-z{3}'

B SyntaxError: invalid regular expression

C [1] TRUE

D [1] FALSE

💡 Hint

Check if the regex pattern has balanced brackets.

The pattern '[a-z{3}' is missing a closing bracket ']', so R throws an error about invalid regex.

❓ Predict Output

advanced

Output of regmatches with gregexpr

What is the output of this R code?

R Programming

text <- "one1 two22 three333"
matches <- regmatches(text, gregexpr("\\d+", text))
print(matches)

A [[1]] "one" "two" "three"

B [[1]] "one1" "two22" "three333"

C [[1]] "1" "2" "3"

D [[1]] "1" "22" "333"

💡 Hint

gregexpr finds all matches of one or more digits, regmatches extracts them.

The pattern '\\d+' matches sequences of digits. The text has '1', '22', and '333' as digit sequences.

🧠 Conceptual

expert

Number of matches with lookahead in R regex

How many matches does this R code find?

R Programming

text <- "aaaa"
matches <- gregexpr("a(?=a)", text, perl=TRUE)
length(regmatches(text, matches)[[1]])

A 2

B 3

C 4

D 1

💡 Hint

Lookahead '(?=a)' matches 'a' only if followed by another 'a'.

In 'aaaa', the pattern matches the first three 'a's because each is followed by another 'a'. The last 'a' is not followed by 'a', so total matches are 3.

Practice

1. What does the grepl() function do in R when used with regular expressions?

easy

A. It checks if a pattern exists in a string and returns TRUE or FALSE.

B. It replaces all occurrences of a pattern in a string.

C. It extracts all matches of a pattern from a string.

D. It splits a string into parts based on a pattern.

Solution

1. Step 1: Understand the purpose of grepl()

   grepl() returns a logical value indicating if the pattern is found in the string.

2. Step 2: Compare with other functions

   Functions like gsub() replace text, and strsplit() splits strings, so they do not match grepl()'s behavior.

3. Final Answer:

   It checks if a pattern exists in a string and returns TRUE or FALSE. -> Option A

4. Quick Check:

   grepl() = logical check [OK]

Hint: Remember: grepl() returns TRUE/FALSE if pattern found [OK]

Common Mistakes:

• Confusing grepl() with gsub() which replaces text
• Thinking grepl() extracts matched text
• Assuming grepl() splits strings

2. Which of the following is the correct syntax to replace all occurrences of "cat" with "dog" in the string "cat and cat" using regular expressions in R?

easy

A. sub("cat", "dog", "cat and cat")

B. str_replace_all("cat and cat", "cat", "dog")

C. gsub("cat", "dog", "cat and cat")

D. replace("cat", "dog", "cat and cat")

Solution

1. Step 1: Identify function for replacing all occurrences

   gsub() replaces all matches, while sub() replaces only the first.

2. Step 2: Check syntax correctness

   gsub(pattern, replacement, string) is the correct order and usage in R.

3. Final Answer:

   gsub("cat", "dog", "cat and cat") -> Option C

4. Quick Check:

   Replace all = gsub() [OK]

Hint: Use gsub() to replace all matches, sub() only first [OK]

Common Mistakes:

• Using sub() when all occurrences need replacement
• Using non-existent replace() function in base R
• Confusing argument order in functions

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

text <- c("apple", "banana", "apricot", "berry")
matches <- grepl("^a", text)
matches

medium

A. [FALSE, TRUE, FALSE, TRUE]

B. [FALSE, FALSE, FALSE, FALSE]

C. [TRUE, TRUE, TRUE, TRUE]

D. [TRUE, FALSE, TRUE, FALSE]

Solution

1. Step 1: Understand the pattern "^a"

   The caret (^) means the string must start with 'a'. So strings starting with 'a' match.

2. Step 2: Check each string in text

   "apple" and "apricot" start with 'a' (TRUE), "banana" and "berry" do not (FALSE).

3. Final Answer:

   [TRUE, FALSE, TRUE, FALSE] -> Option D

4. Quick Check:

   ^a matches start with 'a' = [TRUE, FALSE, TRUE, FALSE] [OK]

Hint: ^ means start of string; check which words start with 'a' [OK]

Common Mistakes:

• Ignoring ^ and matching 'a' anywhere
• Confusing TRUE/FALSE order in output
• Assuming grepl returns matched strings instead of logical

4. The following R code is intended to replace the first digit in a string with "#" but does not work as expected:

text <- "abc123"
result <- sub("[0-9]", "#", text)
print(result)

What is the problem and how to fix it?

medium

A. Pattern should be "\\d" instead of "[0-9]" for digits in R regex.

B. No problem; code works correctly and replaces first digit.

C. Use gsub() instead of sub() to replace all digits, not just first.

D. The replacement string "#" must be escaped as "\\#".

Solution

1. Step 1: Check the pattern and function

   [0-9] matches any digit; sub() replaces the first match only.

2. Step 2: Verify the output

   Replacing first digit '1' with '#' in "abc123" results in "abc#23", which is expected.

3. Final Answer:

   No problem; code works correctly and replaces first digit. -> Option B

4. Quick Check:

   sub() replaces first digit correctly [OK]

Hint: sub() replaces first match; [0-9] matches digits [OK]

Common Mistakes:

• Thinking [0-9] is invalid in R regex
• Confusing sub() with gsub() for all replacements
• Believing replacement '#' needs escaping

5. You have a vector of email addresses:

emails <- c("[e-mail]", "admin@sitecom", "[e-mail]")

Which regular expression pattern would correctly identify emails ending with ".com" using grepl()?

hard

A. "\\.com$"

B. ".com"

C. "com$"

D. "\\.com"

Solution

1. Step 1: Understand the pattern requirements

   We want to match emails ending exactly with ".com". The dot must be escaped as "\\." to match a literal dot, and "$" means end of string.

2. Step 2: Analyze each option

   "\\.com$" uses "\\.com$" which matches '.com' at the end. ".com" matches any character before 'com', not necessarily end. "com$" misses the dot. "\\.com" matches '.com' anywhere, not necessarily at end.

3. Final Answer:

   "\\.com$" -> Option A

4. Quick Check:

   Escape dot and use $ for end match [OK]

Hint: Escape dot with \\. and use $ to match string end [OK]

Common Mistakes:

• Not escaping dot, so matches any character
• Missing $ causing partial matches
• Confusing string end anchor with start anchor