apply() with lambda functions in Pandas - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to apply a lambda function that doubles each value in the 'score' column.

Pandas

df['double_score'] = df['score'].[1](lambda x: x * 2)

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

Afilter

Bmap

Capply

Dreduce

Attempts:

3 left

💡 Hint

Common Mistakes

Using map instead of apply, which works differently on Series.

Using filter or reduce, which are not pandas Series methods.

The apply() method lets you run a function on each element of a pandas Series. Here, it doubles each score.

2fill in blank

medium

Complete the code to create a new column 'passed' that is True if 'score' is 50 or more, else False.

Pandas

df['passed'] = df['score'].apply(lambda x: x [1] 50)

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

A>=

B<=

C==

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' which would mark scores less than or equal to 50 as passing.

Using '==' which only marks scores exactly 50 as passing.

The lambda checks if each score is greater than or equal to 50 to mark passing.

3fill in blank

hard

Fix the error in the lambda function to correctly add 10 to each value in 'score'.

Pandas

df['score_plus_10'] = df['score'].apply(lambda x: x [1] 10)

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

A-

B/

C*

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using '-' which subtracts instead of adding.

Using '*' or '/' which multiply or divide.

The lambda should add 10 to each score, so the plus sign '+' is correct.

4fill in blank

hard

Fill both blanks to create a dictionary of word lengths for words longer than 3 characters.

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

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

The dictionary comprehension maps each word to its length if the word length is greater than 3.

5fill in blank

hard

Fill all three blanks to create a dictionary with uppercase keys and values greater than 0.

Pandas

result = [1]: [2] for k, v in data.items() if v [3] 0}

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

Ak.upper()

Bv

C>

Dk.lower()

Attempts:

3 left

💡 Hint

Common Mistakes

Using k.lower() instead of k.upper() for keys.

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

The dictionary comprehension uses uppercase keys, keeps values, and filters for values greater than zero.

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