Why custom functions matter in Pandas - Challenge Your Understanding

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of applying a custom function with pandas apply

What is the output of the following code that uses a custom function with pandas apply on a DataFrame column?

Pandas

import pandas as pd

def add_five(x):
    return x + 5

df = pd.DataFrame({'A': [1, 2, 3]})
df['B'] = df['A'].apply(add_five)
print(df)

A

   A  B
0  1  6
1  2  7
2  3  8

B

   A  B
0  6  1
1  7  2
2  8  3

C

   A  B
0  1  5
1  2  6
2  3  7

D

   A  B
0  1  1
1  2  2
2  3  3

💡 Hint

Remember that apply passes each element of the column to the function.

The custom function add_five adds 5 to each element in column 'A'. Applying it creates column 'B' with values 6, 7, 8.

❓ data_output

intermediate

Result of using a lambda function with apply on a DataFrame

What is the resulting DataFrame after applying this lambda function to column 'score'?

Pandas

import pandas as pd

df = pd.DataFrame({'score': [50, 80, 90]})
df['grade'] = df['score'].apply(lambda x: 'Pass' if x >= 60 else 'Fail')
print(df)

A

   score grade
0     50  Fail
1     80  Pass
2     90  Pass

B

   score grade
0     50  Pass
1     80  Pass
2     90  Pass

C

   score grade
0     50  Fail
1     80  Fail
2     90  Pass

D

   score grade
0     50  Fail
1     80  Pass
2     90  Fail

💡 Hint

Check the condition inside the lambda function carefully.

The lambda function assigns 'Pass' if score is 60 or more, otherwise 'Fail'. So 50 is 'Fail', 80 and 90 are 'Pass'.

🔧 Debug

advanced

Identify the error in this custom function used with apply

What error will this code raise when applying the custom function to the DataFrame column?

Pandas

import pandas as pd

def divide_by_zero(x):
    return x / 0

df = pd.DataFrame({'num': [1, 2, 3]})
df['result'] = df['num'].apply(divide_by_zero)

A TypeError

B ZeroDivisionError

C KeyError

D No error, outputs NaN

💡 Hint

Think about what happens when dividing by zero in Python.

Dividing any number by zero raises a ZeroDivisionError in Python, so the apply call will fail with this error.

🚀 Application

advanced

Using a custom function to categorize data in pandas

You want to categorize ages into 'Child', 'Adult', and 'Senior' using a custom function with apply. Which code correctly creates a new column 'category' with these labels?

Pandas

import pandas as pd

df = pd.DataFrame({'age': [5, 25, 70]})

A

def categorize(age):
    if age < 18:
        return 'Child'
    elif age < 65:
        return 'Adult'
    else:
        return 'Senior'

df['category'] = df['age'].apply(categorize)

B

def categorize(age):
    if age <= 18:
        return 'Child'
    elif age <= 65:
        return 'Adult'
    else:
        return 'Senior'

df['category'] = df['age'].apply(categorize)

C

def categorize(age):
    if age <= 18:
        return 'Child'
    elif age < 65:
        return 'Adult'
    else:
        return 'Senior'

df['category'] = df['age'].apply(categorize)

D

def categorize(age):
    if age < 18:
        return 'Child'
    elif age <= 65:
        return 'Adult'
    else:
        return 'Senior'

df['category'] = df['age'].apply(categorize)

💡 Hint

Check the boundary conditions carefully for age groups.

Option D correctly categorizes ages: under 18 is 'Child', 18 to 65 inclusive is 'Adult', above 65 is 'Senior'.

🧠 Conceptual

expert

Why use custom functions with pandas apply instead of vectorized operations?

Which statement best explains why custom functions are important when working with pandas apply?

A Custom functions automatically optimize memory usage in pandas DataFrames.

B Custom functions always run faster than vectorized operations in pandas.

C Custom functions allow applying complex logic that cannot be done with built-in vectorized operations.

D Custom functions prevent the need to import pandas for data manipulation.

💡 Hint

Think about the flexibility custom functions provide.

Custom functions let you apply any logic you want, even if pandas does not have a built-in vectorized method for it.