Regular expressions in R in R Programming - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if the string contains the word 'cat'.

R Programming

grepl([1], "The cat is sleeping")

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

A"cat"

B"dog"

C"bird"

D"fish"

Attempts:

3 left

💡 Hint

Common Mistakes

Forgetting to put quotes around the pattern.

Using a word not present in the string.

The grepl function checks if the pattern "cat" is found in the string.

2fill in blank

medium

Complete the code to extract all words starting with 'b' from the string.

R Programming

regmatches(text, gregexpr([1], text))

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

A"b"

B"\w+b"

C"b+"

D"\bb\w+"

Attempts:

3 left

💡 Hint

Common Mistakes

Not using word boundary \b.

Using incorrect quantifiers.

The pattern "\bb\w+" matches words starting with 'b'.

3fill in blank

hard

Fix the error in the pattern to match email addresses.

R Programming

grepl([1], email)

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A"[a-zA-Z0-9._%+-]+@[a-zA-Z0-9.-]+\\.[a-zA-Z]{2,}"

B"[a-zA-Z0-9._%+-]+@[a-zA-Z0-9.-]+.[a-zA-Z]{2,}"

C"[a-zA-Z0-9._%+-]+@[a-zA-Z0-9.-]+\.[a-zA-Z]{2,}"

D"[a-zA-Z0-9._%+-]+@[a-zA-Z0-9.-]+\\.[a-zA-Z]{2}"

Attempts:

3 left

💡 Hint

Common Mistakes

Not escaping the dot correctly.

Using single backslash instead of double.

In R strings, a single backslash must be escaped with another backslash. So to match a literal dot, use \\. in the regex, which becomes \. in the string. Option C correctly escapes it.

4fill in blank

hard

Fill both blanks to create a named vector with word lengths for words longer than 3 letters.

R Programming

word_lengths <- c([1] = nchar([2]))

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Awords

Bword

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping the names and values.

Using the wrong variable names.

The c() function creates a named vector. The name is each word and the value is the length of each words element.

5fill in blank

hard

Fill all three blanks to create a list of words longer than 4 letters with their uppercase versions.

R Programming

result <- list([1] = [2], [3] = toupper([2]))

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

Along_words

Bwords

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names inconsistently.

Mixing up the names of list elements.

The list has two elements: long_words holding the words, and long_words holding their uppercase forms. The variable holding the words is words.

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