Why apply() on columns in Pandas? - Purpose & Use Cases

The Big Idea

What if you could update every column in your data with just one simple command?

The Scenario

Imagine you have a big table of sales data in a spreadsheet. You want to change every price by adding tax or convert all dates to a new format. Doing this by clicking and typing for each column feels like a never-ending chore.

The Problem

Manually editing each cell or column is slow and tiring. It's easy to make mistakes, like forgetting a cell or typing wrong numbers. When the data grows, this manual work becomes impossible to finish on time.

The Solution

The apply() function lets you quickly run a small task on every column in your table. You write the task once, and apply() does the work for all columns automatically. This saves time and avoids errors.

Before vs After

✗ Before

for col in df.columns:
    df[col] = [x * 1.1 for x in df[col]]

✓ After

df = df.apply(lambda col: col * 1.1)

What It Enables

With apply(), you can transform entire columns of data in one simple step, making your work faster and more reliable.

Real Life Example

A store manager wants to increase all product prices by 10% after tax changes. Instead of changing each price manually, they use apply() to update all price columns instantly.

Key Takeaways

Manual edits on columns are slow and error-prone.

apply() runs a function on every column automatically.

This makes data changes faster, easier, and safer.

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