apply() function for custom logic in Data Analysis Python - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of apply() with a custom function on DataFrame column

What is the output of the following code snippet?

Data Analysis Python

import pandas as pd

df = pd.DataFrame({'score': [45, 82, 77, 90, 56]})

def grade(x):
    if x >= 80:
        return 'A'
    elif x >= 60:
        return 'B'
    else:
        return 'C'

df['grade'] = df['score'].apply(grade)
print(df['grade'].tolist())

A ['B', 'A', 'B', 'A', 'C']

B ['C', 'A', 'B', 'A', 'C']

C ['C', 'A', 'A', 'A', 'C']

D ['C', 'B', 'B', 'A', 'C']

💡 Hint

Check the conditions in the grade function carefully for each score.

Scores 45 and 56 are below 60, so grade 'C'. Scores 82 and 90 are 80 or above, so grade 'A'. Score 77 is between 60 and 79, so grade 'B'.

❓ data_output

intermediate

Resulting DataFrame after apply() with lambda function

What is the resulting DataFrame after applying the lambda function below?

Data Analysis Python

import pandas as pd

df = pd.DataFrame({'value': [10, 15, 20, 25]})
df['double'] = df['value'].apply(lambda x: x * 2)
print(df)

A

   value  double
0     10     100
1     15     225
2     20     400
3     25     625

B

   value  double
0     10      10
1     15      15
2     20      20
3     25      25

C

   value  double
0     10      20
1     15      30
2     20      40
3     25      50

D

   value  double
0     10      5
1     15      7
2     20     10
3     25     12

💡 Hint

The lambda function multiplies each value by 2.

Each 'value' is doubled in the 'double' column.

🔧 Debug

advanced

Identify the error in apply() usage with a custom function

What error does the following code raise?

Data Analysis Python

import pandas as pd

df = pd.DataFrame({'temp': [30, 22, 25]})

def convert(x):
    return x * 9/5 + 32

df['temp_f'] = df.apply(convert)
print(df)

A TypeError: unsupported operand type(s) for *: 'DataFrame' and 'float'

B ValueError: The truth value of a DataFrame is ambiguous

C KeyError: 'temp_f'

D No error, prints DataFrame with 'temp_f' column

💡 Hint

Check how apply() is used on the DataFrame without specifying axis.

apply() without axis defaults to applying function on columns (Series), so 'x' is a Series, and multiplying Series by float works, but the function expects a scalar. Actually, the function works on Series but returns a Series, so the assignment fails. But in this case, the error is TypeError because the function tries to multiply DataFrame by float if axis is not specified. Actually, the error is because apply() applies function to each column (Series), so x is a Series, and the function returns a Series, so the assignment fails. The actual error is TypeError due to unsupported operand types when trying to multiply DataFrame by float. The correct usage is to specify axis=1 or apply on a Series.

🚀 Application

advanced

Using apply() to create a new column based on multiple columns

Given the DataFrame below, which option correctly creates a new column 'status' that is 'Pass' if score >= 50 and attendance >= 75, else 'Fail'?

Data Analysis Python

import pandas as pd

df = pd.DataFrame({'score': [55, 48, 70, 40], 'attendance': [80, 70, 90, 85]})

A df['status'] = df.apply(lambda row: 'Pass' if row.score >= 50 or row.attendance >= 75 else 'Fail', axis=0)

B df['status'] = df.apply(lambda x: 'Pass' if x['score'] >= 50 and x['attendance'] >= 75 else 'Fail')

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

D df['status'] = df.apply(lambda row: 'Pass' if row['score'] >= 50 and row['attendance'] >= 75 else 'Fail', axis=1)

💡 Hint

Remember to use axis=1 to apply function row-wise.

Option D correctly applies the function row-wise (axis=1) and checks both conditions with 'and'. Option D misses axis=1, so applies column-wise causing error. Option D only checks score, ignoring attendance. Option D uses axis=0 and 'or' instead of 'and'.

🧠 Conceptual

expert

Understanding apply() behavior with axis parameter

Which statement best describes the behavior of the apply() function when used on a DataFrame with axis=0 and axis=1?

A axis=0 applies the function to each column (Series), axis=1 applies the function to each row (Series)

B axis=0 applies the function to each row (Series), axis=1 applies the function to each column (Series)

C axis=0 applies the function element-wise, axis=1 applies the function to the entire DataFrame

D axis=0 and axis=1 both apply the function to the entire DataFrame but in different orders

💡 Hint

Think about rows and columns as Series objects.

In pandas, axis=0 means apply function to each column (each column is a Series). axis=1 means apply function to each row (each row is a Series).

Practice

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

easy

A. It merges two DataFrames together.

B. It sorts the DataFrame by a column.

C. It filters rows based on a condition.

D. It applies a custom function to each row or column.

Solution

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

   The apply() function is used to run a custom function on each row or column of a DataFrame.

2. Step 2: Compare with other DataFrame methods

   Sorting, filtering, and merging are different operations and not done by apply().

3. Final Answer:

   It applies a custom function to each row or column. -> Option D

4. Quick Check:

   apply() = custom function application [OK]

Hint: Remember: apply() runs your function on rows or columns [OK]

Common Mistakes:

• Confusing apply() with sort or filter methods
• Thinking apply() merges DataFrames
• Assuming apply() only works on columns

2. Which of the following is the correct syntax to apply a function f row-wise on a DataFrame df?

easy

A. df.apply(f, axis=0)

B. df.apply(f)

C. df.apply(f, axis=1)

D. df.apply(f, axis=2)

Solution

1. Step 1: Recall axis parameter meaning

   In pandas, axis=1 means apply function row-wise, while axis=0 means column-wise.

2. Step 2: Identify correct axis for row-wise

   Since the question asks for row-wise application, axis=1 is correct.

3. Final Answer:

   df.apply(f, axis=1) -> Option C

4. Quick Check:

   Row-wise apply = axis=1 [OK]

Hint: Use axis=1 for rows, axis=0 for columns [OK]

Common Mistakes:

• Using axis=0 for row-wise apply
• Omitting axis parameter and expecting row-wise
• Using invalid axis values like 2

3. Given the DataFrame:

import pandas as pd
df = pd.DataFrame({'A': [1, 2], 'B': [3, 4]})

What is the output of:

df['Sum'] = df.apply(lambda row: row['A'] + row['B'], axis=1)
print(df)

medium

A. A B Sum
0 1 3 3
1 2 4 4

B. A B Sum
0 1 3 4
1 2 4 6

C. A B Sum
0 1 3 1
1 2 4 2

D. Error: axis parameter missing

Solution

1. Step 1: Understand the lambda function

   The lambda adds values from columns 'A' and 'B' for each row.

2. Step 2: Apply row-wise addition

   Row 0: 1 + 3 = 4; Row 1: 2 + 4 = 6.

3. Final Answer:

   A B Sum 0 1 3 4 1 2 4 6 -> Option B

4. Quick Check:

   Sum column = A + B per row [OK]

Hint: Sum columns row-wise with apply and axis=1 [OK]

Common Mistakes:

• Forgetting axis=1 causes wrong operation
• Adding columns incorrectly (e.g., column-wise)
• Expecting apply to work without lambda

4. What is wrong with this code snippet?

df = pd.DataFrame({'X': [10, 20], 'Y': [30, 40]})
df['Result'] = df.apply(lambda x: x['X'] * 2 if x['Y'] > 35 else x['X'], axis=0)

medium

A. Using axis=0 applies function column-wise, causing a KeyError.

B. Lambda function syntax is incorrect.

C. DataFrame is missing required columns.

D. No error; code runs correctly.

Solution

1. Step 1: Check axis parameter usage

   axis=0 applies function to each column (Series), not rows.

2. Step 2: Understand lambda function context

   Lambda expects row with keys 'X' and 'Y', but column Series only has index labels, causing KeyError.

3. Final Answer:

   Using axis=0 applies function column-wise, causing a KeyError. -> Option A

4. Quick Check:

   axis=0 = columns, lambda expects rows [OK]

Hint: Use axis=1 for row-wise logic to avoid key errors [OK]

Common Mistakes:

• Confusing axis=0 and axis=1
• Expecting row keys when applying column-wise
• Ignoring error messages about missing keys

5. You have a DataFrame df with columns "score" and "passed" (boolean). You want to create a new column "grade" where:
- If passed is True and score >= 90, grade is 'A'
- If passed is True and score >= 75 but < 90, grade is 'B'
- Otherwise, grade is 'F'
Which apply() function correctly implements this logic?

hard

A. df['grade'] = df.apply(lambda r: 'A' if r['passed'] and r['score'] >= 90 else ('B' if r['passed'] and r['score'] >= 75 else 'F'), axis=1)

B. df['grade'] = df.apply(lambda r: 'A' if r['score'] >= 90 else 'B' if r['score'] >= 75 else 'F', axis=0)

C. df['grade'] = df.apply(lambda r: 'A' if r['passed'] or r['score'] >= 90 else 'F', axis=1)

D. df['grade'] = df.apply(lambda r: 'A' if r['passed'] and r['score'] > 90 else 'B', axis=1)

Solution

1. Step 1: Check axis parameter for row-wise logic

   axis=1 is needed to apply function on each row.

2. Step 2: Verify conditional logic matches requirements

   df['grade'] = df.apply(lambda r: 'A' if r['passed'] and r['score'] >= 90 else ('B' if r['passed'] and r['score'] >= 75 else 'F'), axis=1) correctly checks if passed is True and score thresholds for 'A' and 'B', else 'F'.

3. Step 3: Identify errors in other options

   df['grade'] = df.apply(lambda r: 'A' if r['score'] >= 90 else 'B' if r['score'] >= 75 else 'F', axis=0) uses axis=0 (wrong axis). df['grade'] = df.apply(lambda r: 'A' if r['passed'] or r['score'] >= 90 else 'F', axis=1) uses 'or' instead of 'and'. df['grade'] = df.apply(lambda r: 'A' if r['passed'] and r['score'] > 90 else 'B', axis=1) misses 'F' grade and uses wrong comparison for 90.

4. Final Answer:

   df['grade'] = df.apply(lambda r: 'A' if r['passed'] and r['score'] >= 90 else ('B' if r['passed'] and r['score'] >= 75 else 'F'), axis=1) -> Option A

5. Quick Check:

   Correct logic and axis=1 for row-wise apply [OK]

Hint: Use axis=1 and nested if-else for multi-condition grading [OK]

Common Mistakes:

• Using axis=0 instead of axis=1
• Using 'or' instead of 'and' in conditions
• Missing else condition for failing grade