isin() for value matching in Pandas - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to select rows where the 'color' column values are in the list ['red', 'blue'].

Pandas

filtered = df[df['color'].[1](['red', 'blue'])]

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

Acontains

Bmatch

Cisin

Dequals

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'contains' which is for string matching, not list membership.

Using 'equals' which compares exact values, not membership.

The isin() method checks if each element in the 'color' column is in the given list.

2fill in blank

medium

Complete the code to filter rows where the 'age' column values are in the list [25, 30, 35].

Pandas

result = data[data['age'].[1]([25, 30, 35])]

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

Aisin

Bcontains

Cin_list

Dmatch

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'contains' which is for substring checks.

Using 'in_list' which is not a pandas method.

The isin() method is used to filter rows where 'age' matches any value in the list.

3fill in blank

hard

Fix the error in the code to correctly filter rows where 'status' is either 'active' or 'pending'.

Pandas

filtered_data = df[df['status'].[1](['active', 'pending'])]

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

Aisin

Bmatch

Ccontains

Dequals

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'equals' which only compares single values.

Using 'contains' which is for string matching.

The isin() method correctly checks if 'status' values are in the given list.

4fill in blank

hard

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

Pandas

{word: [1] for word in words if [2]

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

Alen(word)

Bword

Clen(word) > 3

Dword > 3

Attempts:

3 left

💡 Hint

Common Mistakes

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

Checking if the word is greater than 3 instead of its length.

The dictionary comprehension maps each word to its length using len(word), and filters words with length greater than 3 using len(word) > 3.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps uppercase words to their lengths only if the length is greater than 4.

Pandas

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

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

Aword.upper()

Blen(word)

Clen(word) > 4

Dword.lower()

Attempts:

3 left

💡 Hint

Common Mistakes

Using lowercase words as keys instead of uppercase.

Using the word itself as value instead of its length.

Filtering by word length less than or equal to 4.

The dictionary comprehension uses word.upper() as keys, maps to len(word) as values, and filters words with length greater than 4 using len(word) > 4.

Practice

1. What does the isin() function do in pandas?

easy

A. Checks if each value in a Series or DataFrame is in a given list and returns True or False.

B. Sorts the values in a DataFrame or Series.

C. Calculates the sum of values in a DataFrame column.

D. Removes duplicate values from a DataFrame.

Solution

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

   isin() checks if each element in a pandas Series or DataFrame exists in a list or another iterable.

2. Step 2: Identify the output type

   The function returns a boolean Series or DataFrame with True where the value is found and False otherwise.

3. Final Answer:

   Checks if each value in a Series or DataFrame is in a given list and returns True or False. -> Option A

4. Quick Check:

   isin() = membership check [OK]

Hint: Remember: isin() returns True/False for membership [OK]

Common Mistakes:

• Thinking isin() sorts data
• Confusing isin() with aggregation functions
• Assuming it removes duplicates

2. Which of the following is the correct syntax to filter rows in a DataFrame df where column col has values in the list [1, 2, 3]?

easy

A. df.isin(col, [1, 2, 3])

B. df.col.isin([1, 2, 3]) = True

C. df[df.isin(col, [1, 2, 3])]

D. df[df.col.isin([1, 2, 3])]

Solution

1. Step 1: Use isin() on the column

   The correct way is to call isin() on the Series: df.col.isin([1, 2, 3]).

2. Step 2: Use boolean indexing to filter rows

   Wrap the boolean Series inside df[ ... ] to select rows where condition is True.

3. Final Answer:

   df[df.col.isin([1, 2, 3])] -> Option D

4. Quick Check:

   Filter rows with boolean mask = df[df.col.isin([1, 2, 3])] [OK]

Hint: Call isin() on column, then filter with df[...] [OK]

Common Mistakes:

• Calling isin() on DataFrame with wrong arguments
• Using assignment instead of filtering
• Incorrect method call syntax

3. Given the DataFrame:

import pandas as pd
df = pd.DataFrame({'A': [10, 20, 30, 40], 'B': ['x', 'y', 'z', 'x']})

What is the output of df['B'].isin(['x', 'z'])?

medium

A. [True, False, True, True]

B. [False, True, False, False]

C. [True, True, True, True]

D. [False, False, False, False]

Solution

1. Step 1: Check each value in column 'B'

   Values are ['x', 'y', 'z', 'x']. We check if each is in ['x', 'z'].

2. Step 2: Mark True if value is in list, else False

   'x' is in list -> True, 'y' not in list -> False, 'z' in list -> True, 'x' in list -> True.

3. Final Answer:

   [True, False, True, True] -> Option A

4. Quick Check:

   Membership check matches [True, False, True, True] [OK]

Hint: Check each value against list, True if found [OK]

Common Mistakes:

• Mixing True/False order
• Confusing column values
• Assuming all True or all False

4. What is wrong with this code snippet?

df = pd.DataFrame({'num': [1, 2, 3, 4]})
mask = df.isin([1, 3])
filtered = df[mask]

medium

A. The DataFrame df must have a column named 'mask' to filter.

B. The isin() method is called on the whole DataFrame, resulting in a DataFrame of booleans, which cannot be used directly for filtering.

C. The variable mask should be converted to a list before filtering.

D. The list passed to isin() must be a string, not integers.

Solution

1. Step 1: Understand df.isin([1, 3]) output

   This returns a DataFrame of booleans with the same shape as df, not a single boolean Series.

2. Step 2: Why filtering with DataFrame mask fails

   Using a DataFrame of booleans as an index mask causes an error because pandas expects a boolean Series or array matching the index length.

3. Final Answer:

   The isin() method is called on the whole DataFrame, resulting in a DataFrame of booleans, which cannot be used directly for filtering. -> Option B

4. Quick Check:

   Mask must be Series, not DataFrame [OK]

Hint: Use isin() on a column, not whole DataFrame for filtering [OK]

Common Mistakes:

• Using DataFrame mask instead of Series mask for filtering
• Passing wrong data types to isin()
• Trying to filter with a DataFrame of booleans

5. You have a DataFrame df with a column status containing values like 'new', 'pending', 'done', and 'cancelled'. You want to create a new column is_active that is True only if status is 'new' or 'pending'. Which code correctly does this using isin()?

hard

A. df['is_active'] = df['status'] == ['new', 'pending']

B. df['is_active'] = df.isin(['new', 'pending'])

C. df['is_active'] = df['status'].isin(['new', 'pending'])

D. df['is_active'] = df['status'].isin('new', 'pending')

Solution

1. Step 1: Use isin() on the 'status' column with a list of target values

   Correct syntax is df['status'].isin(['new', 'pending']) which returns a boolean Series.

2. Step 2: Assign the boolean Series to new column 'is_active'

   This creates a new column with True where status is 'new' or 'pending', False otherwise.

3. Final Answer:

   df['is_active'] = df['status'].isin(['new', 'pending']) -> Option C

4. Quick Check:

   Use isin() on column with list [OK]

Hint: Assign isin() result directly to new column [OK]

Common Mistakes:

• Calling isin() on whole DataFrame
• Passing multiple arguments instead of a list
• Using equality operator with list