Regular expressions in R in R Programming - Mini Project: Build & Apply

Regular Expressions in R

📖 Scenario: You work in a small library. You have a list of book titles mixed with some extra text. You want to find which titles mention the word "data" in any form.

🎯 Goal: Build a simple R program that uses regular expressions to find and print book titles containing the word "data".

📋 What You'll Learn

Create a character vector called books with given titles

Create a regular expression pattern variable called pattern

Use grep() with pattern on books to find matching titles

Print the matching titles using print()

💡 Why This Matters

🌍 Real World

Libraries, bookstores, or data analysts often need to search text data for keywords quickly and accurately.

💼 Career

Knowing regular expressions and text search functions in R is useful for data cleaning, text mining, and report generation.

1

Create the book titles vector

Create a character vector called books with these exact entries: "Data Science Basics", "Introduction to R", "Advanced Data Analysis", "Cooking Recipes", "Big Data Trends"

R Programming

# Create the character vector books with the given titles
# Your code here

Hint

Use the c() function to create a vector of strings.

2

Create the regular expression pattern

Create a variable called pattern and set it to the regular expression string "data" to match the word "data" in any case.

R Programming

books <- c("Data Science Basics", "Introduction to R", "Advanced Data Analysis", "Cooking Recipes", "Big Data Trends")
# Create the pattern variable with the string "data"
# Your code here

Hint

Just assign the string "data" to the variable pattern.

3

Find titles matching the pattern

Use grep() with pattern on books to find indices of titles containing "data" ignoring case. Store the result in a variable called matches.

R Programming

books <- c("Data Science Basics", "Introduction to R", "Advanced Data Analysis", "Cooking Recipes", "Big Data Trends")
pattern <- "data"
# Use grep() with ignore.case=TRUE to find matching titles
# Your code here

Hint

Use grep(pattern, books, ignore.case = TRUE) to find matches ignoring case.

4

Print the matching book titles

Use print() to display the titles from books at the positions stored in matches.

R Programming

books <- c("Data Science Basics", "Introduction to R", "Advanced Data Analysis", "Cooking Recipes", "Big Data Trends")
pattern <- "data"
matches <- grep(pattern, books, ignore.case = TRUE)
# Print the matching titles
# Your code here

Hint

Use print(books[matches]) to show the matching titles.

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