Regex operations in Pandas - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to select rows where the 'Name' column contains the letter 'a'.

Pandas

filtered = df[df['Name'].str.[1]('a')]

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

Areplace

Bmatch

Cfindall

Dcontains

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'match' instead of 'contains' which only checks the start of the string.

Using 'replace' which changes strings instead of filtering.

The contains method checks if the pattern exists anywhere in the string.

2fill in blank

medium

Complete the code to replace all digits in the 'Code' column with '#' symbol.

Pandas

df['Code'] = df['Code'].str.[1]('\d', '#')

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

Areplace

Bcontains

Cmatch

Dfindall

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'contains' which only checks for presence but does not replace.

Using 'match' which only checks the start of the string.

The replace method replaces all occurrences of the regex pattern with the given string.

3fill in blank

hard

Fix the error in the code to find all words starting with 'a' in the 'Text' column.

Pandas

df['Words'] = df['Text'].str.[1]('\ba\w*')

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

Afindall

Breplace

Ccontains

Dmatch

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'match' which only checks the start of the string and returns a boolean.

Using 'contains' which returns a boolean, not the matched words.

The findall method returns all matches of the regex pattern in each string.

4fill in blank

hard

Fill both blanks to create a dictionary with words as keys and their lengths as values, only for words longer than 3 letters.

Pandas

lengths = {word: [1] for word in words if len(word) [2] 3}

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

Alen(word)

B>

C<

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using the word itself as the value instead of its length.

Using '<' instead of '>' which selects shorter words.

We want the length of each word as the value, and only include words longer than 3 letters.

5fill in blank

hard

Fill all three blanks to create a dictionary with uppercase words as keys, their counts as values, only for counts greater than 1.

Pandas

result = [1]: [2] for word, count in counts.items() if count [3] 1}

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

Aword.upper()

Bcount

C>

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using the original word instead of uppercase for keys.

Using '<' instead of '>' which filters wrong counts.

The keys are uppercase words, values are counts, and we filter counts greater than 1.

Practice

1. What does the str.contains method do in Pandas when used with a regex pattern?

easy

A. It extracts the first match of the regex pattern from each string.

B. It replaces all matches of the regex pattern with a new string.

C. It sorts the DataFrame rows based on regex matches.

D. It checks if each string in a column contains the regex pattern and returns true or false.

Solution

1. Step 1: Understand str.contains usage

   str.contains checks if the regex pattern appears anywhere in each string of the column and returns a boolean Series.

2. Step 2: Differentiate from other methods

   str.replace changes text, str.extract pulls out matches, and sorting is unrelated.

3. Final Answer:

   It checks if each string in a column contains the regex pattern and returns true or false. -> Option D

4. Quick Check:

   str.contains = Boolean check [OK]

Hint: Contains returns true/false if pattern found in each string [OK]

Common Mistakes:

• Confusing contains with replace or extract
• Thinking contains modifies the data
• Assuming contains returns matched text

2. Which of the following is the correct syntax to replace every sequence of consecutive digits in a Pandas column df['col'] with the character 'X' using regex?

easy

A. df['col'].str.replace('\d+', 'X')

B. df['col'].str.replace('[0-9]', 'X')

C. df['col'].replace('[0-9]', 'X')

D. df['col'].str.extract('[0-9]', 'X')

Solution

1. Step 1: Identify correct method for regex replacement

   Use str.replace on the Series to replace regex matches. df['col'].replace('[0-9]', 'X') is wrong because replace without str is for exact values, not regex.

2. Step 2: Check regex pattern and syntax

   df['col'].str.replace('[0-9]', 'X') matches single digits only ('123' -> 'XXX'). df['col'].str.replace('\d+', 'X') matches sequences of one or more digits ('123' -> 'X').

3. Final Answer:

   df['col'].str.replace('\d+', 'X') -> Option A

4. Quick Check:

   Use str.replace with regex pattern [OK]

Hint: Use str.replace with regex pattern and replacement string [OK]

Common Mistakes:

• Using DataFrame replace instead of Series.str.replace
• Wrong regex pattern syntax
• Using str.extract instead of replace

3. Given the DataFrame:

import pandas as pd
df = pd.DataFrame({'text': ['apple123', 'banana', 'cherry456', 'date789']})
df['numbers'] = df['text'].str.extract('(\d+)')

What will be the content of the new column numbers?

medium

A. ['123', 'banana', '456', '789']

B. ['apple', NaN, 'cherry', 'date']

C. ['123', NaN, '456', '789']

D. [NaN, NaN, NaN, NaN]

Solution

1. Step 1: Understand str.extract with regex

   The regex (\d+) captures one or more digits. str.extract returns the first match or NaN if none.

2. Step 2: Apply regex to each string

   'apple123' matches '123', 'banana' has no digits so NaN, 'cherry456' matches '456', 'date789' matches '789'.

3. Final Answer:

   ['123', NaN, '456', '789'] -> Option C

4. Quick Check:

   Extract digits or NaN if none [OK]

Hint: Extract returns first regex group or NaN if no match [OK]

Common Mistakes:

• Expecting extract to return full string
• Confusing extract with contains or replace
• Not handling NaN for no matches

4. Consider this code snippet:

import pandas as pd
df = pd.DataFrame({'col': ['abc123', 'def456', 'ghi789']})
df['col'].str.match('[a-z]+\d')

What is the error or issue in this code?

medium

A. The regex pattern misses the last digit, so match returns false for all rows.

B. The code raises a SyntaxError due to incorrect regex syntax.

C. The method str.match does not accept regex patterns.

D. The code runs correctly and returns true for all rows.

Solution

1. Step 1: Analyze the regex pattern

   The pattern [a-z]+\d matches one or more lowercase letters followed by exactly one digit from the start of the string.

2. Step 2: Compare pattern with data strings

   The strings 'abc123', 'def456', 'ghi789' have multiple digits at the end, but the pattern only expects one digit. Since str.match matches from the start and the pattern does not cover the full digit sequence, it returns False for all rows.

3. Final Answer:

   The regex pattern misses the last digit, so match returns false for all rows. -> Option A

4. Quick Check:

   str.match matches from start; pattern must cover full prefix for True [OK]

Hint: str.match anchors at start (prefix match), pattern must cover full prefix [OK]

Common Mistakes:

• Thinking str.match requires full string match
• Confusing str.match with str.fullmatch
• Thinking regex syntax is invalid here

5. You have a DataFrame df with a column 'emails' containing email addresses mixed with invalid entries. How would you create a new column 'valid' that is true only for rows where 'emails' contains a valid email pattern using regex in Pandas?

hard

A. df['valid'] = df['emails'].str.extract(r'^[\w\.-]+@[\w\.-]+\.\w+$')

B. df['valid'] = df['emails'].str.contains(r'^[\w\.-]+@[\w\.-]+\.\w+$')

C. df['valid'] = df['emails'].str.replace(r'^[\w\.-]+@[\w\.-]+\.\w+$', 'true')

D. df['valid'] = df['emails'].str.match(r'\w+@\w+')

Solution

1. Step 1: Choose method to check pattern presence

   str.contains returns true/false if regex matches anywhere in string, suitable for validation.

2. Step 2: Use correct regex for full email validation

   The regex ^[\w\.-]+@[\w\.-]+\.\w+$ matches a full email pattern from start to end. df['valid'] = df['emails'].str.extract(r'^[\w\.-]+@[\w\.-]+\.\w+$') extracts matched text, not boolean. df['valid'] = df['emails'].str.replace(r'^[\w\.-]+@[\w\.-]+\.\w+$', 'true') replaces text, not boolean. df['valid'] = df['emails'].str.match(r'\w+@\w+') uses str.match but regex is incomplete and may match partial emails.

3. Final Answer:

   df['valid'] = df['emails'].str.contains(r'^[\w\.-]+@[\w\.-]+\.\w+$') -> Option B

4. Quick Check:

   Use str.contains with full regex for validation [OK]

Hint: Use str.contains with full regex pattern for true/false validation [OK]

Common Mistakes:

• Using str.extract which returns strings, not booleans
• Using str.replace which changes data instead of checking
• Using incomplete regex that matches partial emails