str.contains() for pattern matching in Pandas - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to filter rows where the 'Name' column contains the substring 'John'.

Pandas

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

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

Asearch

Bmatch

Cfind

Dcontains

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'match' instead of 'contains' which looks for exact matches.

Using 'find' or 'search' which are not pandas string methods.

The str.contains() method checks if the string contains the given pattern.

2fill in blank

medium

Complete the code to filter rows where the 'Email' column contains the pattern '@gmail.com'.

Pandas

gmail_users = df[df['Email'].str.[1]('@gmail.com')]

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

Acontains

Bendswith

Cstartswith

Dmatch

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'match' which checks the whole string.

Using 'startswith' or 'endswith' which check only the start or end.

str.contains() checks if the string contains the specified substring anywhere.

3fill in blank

hard

Fix the error in the code to filter rows where 'City' contains 'York' ignoring case.

Pandas

nyc = df[df['City'].str.contains('York', [1]=False)]

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

Acase_sensitive

Bignore_case

Ccase

Dcase=False

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'case_sensitive' or 'ignore_case' which are not valid parameters.

Passing True instead of False to the 'case' parameter.

The correct parameter to ignore case is case=False.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that maps words to their lengths only if the word length is greater than 3.

Pandas

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

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

Alen(word)

Bword

C>

D<

Attempts:

3 left

💡 Hint

Common Mistakes

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

Using '<' instead of '>' in the condition.

The dictionary maps each word to its length, filtering words longer than 3 characters.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps uppercase words to their counts only if the count is greater than 1.

Pandas

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

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

Aword.upper()

Bcount

Cword

Dcount.upper()

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'count.upper()' which is invalid since count is a number.

Using 'word' as key without uppercasing.

The dictionary keys are uppercase words, values are counts, iterating over word_counts items.