apply() on columns in Pandas - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of apply() with a lambda on DataFrame columns

What is the output of this code when applying a lambda function to each column of the DataFrame?

Pandas

import pandas as pd

df = pd.DataFrame({
    'A': [1, 2, 3],
    'B': [4, 5, 6]
})

result = df.apply(lambda x: x.max() - x.min())
print(result)

A

A    2
B    2
dtype: int64

B

A    3
B    6
dtype: int64

C

A    1
B    4
dtype: int64

D

A    6
B    9
dtype: int64

💡 Hint

Think about what max() and min() do for each column.

The lambda function calculates the difference between the maximum and minimum values in each column. For column A, max is 3 and min is 1, difference is 2. For column B, max is 6 and min is 4, difference is 2.

❓ data_output

intermediate

Result of apply() with a custom function on columns

Given this DataFrame, what is the result of applying the custom function to each column?

Pandas

import pandas as pd

def custom_func(col):
    return col.sum() / len(col)

df = pd.DataFrame({
    'X': [10, 20, 30],
    'Y': [5, 15, 25]
})

result = df.apply(custom_func)
print(result)

A

X    30
Y    25
dtype: int64

B

X    60
Y    45
dtype: int64

C

X    10
Y    5
dtype: int64

D

X    20.0
Y    15.0
dtype: float64

💡 Hint

The function calculates the average of each column.

The function sums the values in each column and divides by the number of rows, which is the mean. For X: (10+20+30)/3 = 20. For Y: (5+15+25)/3 = 15.

🔧 Debug

advanced

Identify the error in apply() usage on columns

What error does this code raise when applying the function to each column?

Pandas

import pandas as pd

df = pd.DataFrame({
    'A': [1, 2, 3],
    'B': [4, 5, 6]
})

result = df.apply(lambda x: x["nonexistent"])
print(result)

A KeyError: 'nonexistent'

B AttributeError: 'Series' object has no attribute 'nonexistent'

C TypeError: 'int' object is not subscriptable

D ValueError: cannot index with vector

💡 Hint

Check if the column has the key you are trying to access.

The lambda tries to access a key 'nonexistent' in each column (which is a Series). Since this key does not exist, a KeyError is raised.

❓ visualization

advanced

Visualize the effect of apply() on columns with a plotting function

Which option correctly applies a function to plot the sum of each column as a bar chart?

Pandas

import pandas as pd
import matplotlib.pyplot as plt

df = pd.DataFrame({
    'A': [1, 2, 3],
    'B': [4, 5, 6]
})

# Which code snippet below correctly plots the sum of each column as a bar chart?

A

df.sum().apply(lambda x: x).plot(kind='bar')
plt.show()

B

df.plot(kind='bar').apply(sum)
plt.show()

C

df.apply(sum).plot(kind='bar')
plt.show()

D

df.apply(lambda x: x.plot(kind='bar'))
plt.show()

💡 Hint

Sum the columns first, then plot the result.

Option C sums each column using apply(sum), resulting in a Series with sums, then plots it as a bar chart. Other options either apply plot incorrectly or apply sum in wrong order.

🚀 Application

expert

Use apply() on columns to normalize data

You want to normalize each column in a DataFrame by subtracting the mean and dividing by the standard deviation. Which code snippet correctly does this using apply() on columns?

Pandas

import pandas as pd

df = pd.DataFrame({
    'X': [10, 20, 30],
    'Y': [5, 15, 25]
})

# Choose the correct code to normalize columns

A df.apply(lambda x: (x - df.mean()) / df.std())

B df.apply(lambda x: (x - x.mean()) / x.std())

C df.apply(lambda x: (x - x.median()) / x.std())

D df.apply(lambda x: (x - x.min()) / (x.max() - x.min()))

💡 Hint

Normalization uses mean and standard deviation of each column separately.

Option B correctly normalizes each column by subtracting its own mean and dividing by its own standard deviation. Option B uses overall DataFrame mean/std which is incorrect. Option B uses median instead of mean. Option B does min-max scaling, not normalization.

Practice

1. What does the apply() function do when used on a pandas DataFrame column?

easy

A. It applies a function to each value in the column.

B. It deletes the column from the DataFrame.

C. It sorts the column values in ascending order.

D. It merges the column with another DataFrame.

Solution

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

   The apply() method runs a function on each element of the column, transforming or analyzing the data.

2. Step 2: Compare with other options

   Deleting, sorting, or merging are not actions performed by apply() on a column.

3. Final Answer:

   It applies a function to each value in the column. -> Option A

4. Quick Check:

   apply() runs function on each column value [OK]

Hint: Remember: apply() runs a function on each column value [OK]

Common Mistakes:

• Thinking apply() deletes or sorts data
• Confusing apply() with merge or drop functions
• Assuming apply() works on entire DataFrame without axis

2. Which of the following is the correct syntax to apply a lambda function that doubles each value in the 'age' column of a DataFrame df?

easy

A. df['age'].apply = lambda x: x * 2

B. df.apply['age'](lambda x: x * 2)

C. df['age'].apply(lambda x: x * 2)

D. apply(df['age'], lambda x: x * 2)

Solution

1. Step 1: Identify correct method call syntax

   The correct way to use apply on a column is df['column'].apply(function). df['age'].apply(lambda x: x * 2) matches this.

2. Step 2: Check lambda function usage

   df['age'].apply(lambda x: x * 2) correctly passes a lambda function that doubles the value. Other options misuse syntax or assignment.

3. Final Answer:

   df['age'].apply(lambda x: x * 2) -> Option C

4. Quick Check:

   Use df['col'].apply(lambda x: ...) [OK]

Hint: Use df['col'].apply(lambda x: ...) for column functions [OK]

Common Mistakes:

• Using square brackets with apply like a list
• Assigning apply instead of calling it
• Calling apply as a standalone function

3. Given the DataFrame df below:

   score
0     10
1     20
2     30
3     40

What is the output of df['score'].apply(lambda x: x // 15)?

medium

A. 0 10
1 20
2 30
3 40
Name: score, dtype: int64

B. 0 0
1 1
2 2
3 3
Name: score, dtype: int64

C. 0 0
1 0
2 1
3 2
Name: score, dtype: int64

D. 0 0
1 1
2 2
3 2
Name: score, dtype: int64

Solution

1. Step 1: Apply the lambda function to each value

   Divide each score by 15 using floor division:
   10 // 15 = 0
   20 // 15 = 1
   30 // 15 = 2
   40 // 15 = 2

2. Step 2: Write the resulting Series

   The output Series is:
   0 0
   1 1
   2 2
   3 2

3. Final Answer:

   0 0 1 1 2 2 3 2 Name: score, dtype: int64 -> Option D

4. Quick Check:

   Floor division on each value = output [OK]

Hint: Apply lambda to each value and calculate carefully [OK]

Common Mistakes:

• Using normal division instead of floor division
• Misreading floor division results
• Confusing output format with DataFrame instead of Series

4. What is wrong with this code snippet?

df['new_col'] = df['value'].apply(x: x + 5)

How can it be fixed?

medium

A. Missing lambda keyword; fix: df['value'].apply(lambda x: x + 5)

B. apply() cannot add numbers; use map() instead

C. Should use square brackets around x; fix: apply([x]: x + 5)

D. Function must be defined outside apply; fix: def f(x): return x+5

Solution

1. Step 1: Identify syntax error in lambda function

   The code uses x: x + 5 without the lambda keyword, which is required to define an anonymous function.

2. Step 2: Correct the syntax

   Add lambda before the function: df['value'].apply(lambda x: x + 5).

3. Final Answer:

   Missing lambda keyword; fix: df['value'].apply(lambda x: x + 5) -> Option A

4. Quick Check:

   Lambda keyword needed for inline functions [OK]

Hint: Always use 'lambda' before inline functions in apply() [OK]

Common Mistakes:

• Omitting the lambda keyword
• Using incorrect brackets or syntax for lambda
• Thinking apply() can't use inline functions

5. You have a DataFrame df with a column 'temperature' in Celsius. You want to create a new column 'temp_f' with Fahrenheit values using apply(). Which code correctly does this?

hard

A. df['temp_f'] = df.apply(lambda c: (c * 9/5) + 32)

B. df['temp_f'] = df['temperature'].apply(lambda c: (c * 9/5) + 32)

C. df['temp_f'] = df['temperature'].apply(lambda c: c + 32 * 9/5)

D. df['temp_f'] = df['temperature'].apply(lambda c: c * (9/5 + 32))

Solution

1. Step 1: Recall the formula for Celsius to Fahrenheit

   The correct formula is Fahrenheit = (Celsius * 9/5) + 32.

2. Step 2: Check each option's lambda function

   df['temp_f'] = df['temperature'].apply(lambda c: (c * 9/5) + 32) correctly applies the formula to each value in 'temperature'. df['temp_f'] = df.apply(lambda c: (c * 9/5) + 32) applies to whole DataFrame, C and D have incorrect formulas.

3. Final Answer:

   df['temp_f'] = df['temperature'].apply(lambda c: (c * 9/5) + 32) -> Option B

4. Quick Check:

   Use correct formula inside apply() on the column [OK]

Hint: Use correct formula inside lambda for each column value [OK]

Common Mistakes:

• Applying function to whole DataFrame instead of column
• Mixing order of operations in formula
• Forgetting parentheses in formula