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,).

Practice

1. What does the apply() function with a lambda do in pandas?

easy

A. It sorts the DataFrame by column values.

B. It applies a small function to each element in a DataFrame column or row.

C. It merges two DataFrames based on a key.

D. It creates a new DataFrame from scratch.

Solution

1. Step 1: Understand the purpose of apply() with lambda

   The apply() method lets you run a function on each element or row/column of a DataFrame or Series.

2. Step 2: Recognize what lambda does

   A lambda is a small anonymous function used inside apply() to specify what to do with each element.

3. Final Answer:

   It applies a small function to each element in a DataFrame column or row. -> Option B

4. Quick Check:

   apply() + lambda = element-wise function [OK]

Hint: apply() runs a function on each item quickly [OK]

Common Mistakes:

• Thinking apply() sorts data
• Confusing apply() with merge or join
• Believing apply() creates new DataFrames

2. Which of the following is the correct syntax to use apply() with a lambda on a pandas Series named data to double each value?

easy

A. data.apply(lambda x: x * 2)

B. data.apply(x => x * 2)

C. apply(data, lambda x: x * 2)

D. data.apply(lambda x x * 2)

Solution

1. Step 1: Recall correct lambda syntax in Python

   Python lambda functions use the format lambda x: expression, so lambda x: x * 2 is correct.

2. Step 2: Check apply() usage on Series

   The method call is data.apply(lambda x: x * 2). Other options have syntax errors or wrong order.

3. Final Answer:

   data.apply(lambda x: x * 2) -> Option A

4. Quick Check:

   Correct lambda syntax = data.apply(lambda x: x * 2) [OK]

Hint: Use colon after lambda parameter: lambda x: ... [OK]

Common Mistakes:

• Using arrow syntax (=>) from other languages
• Missing colon after lambda parameter
• Calling apply() as a standalone function

3. Given the DataFrame:

import pandas as pd
df = pd.DataFrame({'A': [1, 2, 3]})
df['B'] = df['A'].apply(lambda x: x + 5)
print(df)

What is the output?

medium

A. Error: lambda function syntax

B. A B
0 1 5
1 2 6
2 3 7

C. A B
0 1 6
1 2 7
2 3 8

D. A B
0 6 1
1 7 2
2 8 3

Solution

1. Step 1: Understand the lambda function applied

   The lambda adds 5 to each value in column 'A'. So 1+5=6, 2+5=7, 3+5=8.

2. Step 2: Check DataFrame after assignment

   Column 'B' gets these new values, while 'A' remains unchanged.

3. Final Answer:

   A B 0 1 6 1 2 7 2 3 8 -> Option C

4. Quick Check:

   Original + 5 = New column [OK]

Hint: apply() changes each value as per lambda [OK]

Common Mistakes:

• Mixing up columns A and B
• Forgetting to add 5 inside lambda
• Expecting error from correct syntax

4. What is wrong with this code snippet?

df['C'] = df['A'].apply(lambda x x * 3)

medium

A. Missing colon after lambda parameter causes syntax error.

B. apply() cannot be used on DataFrame columns.

C. lambda functions cannot multiply values.

D. Column 'C' does not exist yet.

Solution

1. Step 1: Check lambda function syntax

   Python lambda requires a colon after parameters: lambda x: x * 3. The code misses the colon.

2. Step 2: Understand error cause

   Missing colon causes a syntax error, so code won't run.

3. Final Answer:

   Missing colon after lambda parameter causes syntax error. -> Option A

4. Quick Check:

   lambda syntax error = missing colon [OK]

Hint: Always put colon after lambda parameters [OK]

Common Mistakes:

• Forgetting colon after lambda parameter
• Thinking apply() can't be used on columns
• Assuming new columns must exist before assignment

5. You have a DataFrame df with a column 'score' containing numbers. You want to create a new column 'grade' that says 'Pass' if the score is 50 or more, and 'Fail' otherwise. Which code correctly uses apply() with a lambda to do this?

hard

A. df['grade'] = df.apply(lambda x: 'Pass' if x['score'] >= 50 else 'Fail')

B. df['grade'] = df['score'].apply(lambda x: if x >= 50 then 'Pass' else 'Fail')

C. df['grade'] = df['score'].apply(lambda x: x >= 50 ? 'Pass' : 'Fail')

D. df['grade'] = df['score'].apply(lambda x: 'Pass' if x >= 50 else 'Fail')

Solution

1. Step 1: Understand conditional expression in lambda

   Python uses value_if_true if condition else value_if_false syntax inside lambda.

2. Step 2: Check application target and syntax

   Applying to the Series df['score'] with lambda x: 'Pass' if x >= 50 else 'Fail' is correct, as x is each scalar score. Applying to the whole df (without axis=1) passes column Series to lambda, where x['score'] causes KeyError. Syntaxes like ? : ternary or if-then are invalid in Python.

3. Final Answer:

   df['grade'] = df['score'].apply(lambda x: 'Pass' if x >= 50 else 'Fail') -> Option D

4. Quick Check:

   Correct if-else in lambda = df['grade'] = df['score'].apply(lambda x: 'Pass' if x >= 50 else 'Fail') [OK]

Hint: Use Python's if-else inside lambda: val_if_true if cond else val_if_false [OK]

Common Mistakes:

• Using ?: ternary operator from other languages
• Applying lambda on whole DataFrame without axis
• Writing if-else without proper Python syntax