apply() with lambda functions in Pandas - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of apply() with lambda on DataFrame column

What is the output of the following code snippet?

Pandas

import pandas as pd

df = pd.DataFrame({'A': [1, 2, 3, 4]})
result = df['A'].apply(lambda x: x**2 if x % 2 == 0 else x)
print(result.tolist())

A [1, 2, 3, 4]

B [2, 4, 6, 8]

C [1, 4, 9, 16]

D [1, 4, 3, 16]

💡 Hint

Remember the lambda squares only even numbers.

The lambda function squares the number only if it is even (2 and 4). Odd numbers remain unchanged.

❓ data_output

intermediate

Resulting DataFrame after apply() with lambda on multiple columns

Given the DataFrame and the apply() operation below, what is the resulting DataFrame?

Pandas

import pandas as pd

df = pd.DataFrame({'X': [1, 2], 'Y': [3, 4]})
result = df.apply(lambda row: row['X'] + row['Y'], axis=1)
print(result)

A

X    4
Y    6
dtype: int64

B

0    4
1    6
dtype: int64

C

0    3
1    4
dtype: int64

D

0    1
1    2
dtype: int64

💡 Hint

axis=1 means apply function to each row.

The lambda adds values in columns 'X' and 'Y' for each row: 1+3=4 and 2+4=6.

🔧 Debug

advanced

Identify the error in apply() with lambda on DataFrame

What error does the following code raise?

Pandas

import pandas as pd

df = pd.DataFrame({'A': [1, 2, 3]})
result = df.apply(lambda x: x + 1)
print(result)

A No error, output is DataFrame with each element incremented by 1

B ValueError: The truth value of a Series is ambiguous

C TypeError: unsupported operand type(s) for +: 'str' and 'int'

D AttributeError: 'int' object has no attribute 'apply'

💡 Hint

Check what apply does by default on DataFrame without axis specified.

By default, apply applies the function to each column (Series). Adding 1 to a Series adds 1 to each element, so no error occurs.

🚀 Application

advanced

Using apply() with lambda to categorize data

You have a DataFrame with a column 'score'. You want to create a new column 'grade' where scores >= 90 get 'A', scores >= 80 get 'B', else 'C'. Which code correctly does this?

Pandas

import pandas as pd

df = pd.DataFrame({'score': [95, 82, 74, 88]})

A df['grade'] = df['score'].apply(lambda x: 'A' if x >= 90 else ('B' if x >= 80 else 'C'))

B df['grade'] = df['score'].apply(lambda x: 'A' if x > 90 else 'B' if x > 80 else 'C')

C df['grade'] = df['score'].map(lambda x: 'A' if x >= 90 else 'B' if x >= 80 else 'C')

D df['grade'] = df.apply(lambda x: 'A' if x['score'] >= 90 else 'B' if x['score'] >= 80 else 'C')

💡 Hint

Check the comparison operators and method used.

Option A uses apply on the 'score' column with correct conditions and inclusive >= operators.

🧠 Conceptual

expert

Understanding apply() behavior with lambda on DataFrame axis

Consider a DataFrame df with shape (3, 2). What is the shape of the output when running df.apply(lambda x: x.sum(), axis=1)?

A (3, 2)

B (2,)

C (3,)

D (1,)

💡 Hint

axis=1 applies function to each row, returning one value per row.

Applying sum row-wise returns a Series with one value per row, so shape is (3,).