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).