apply() on columns in Pandas - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to apply a function to each element in the 'age' column.

Pandas

df['age'] = df['age'].[1](lambda x: x + 1)

Drag options to blanks, or click blank then click option'

Areduce

Bmap

Cfilter

Dapply

Attempts:

3 left

💡 Hint

Common Mistakes

Using map instead of apply, which works differently on Series.

Using filter, which is for filtering elements, not transforming.

The apply method lets you run a function on each element of a column.

2fill in blank

medium

Complete the code to apply a function to each column in the DataFrame.

Pandas

result = df.[1](sum)

Drag options to blanks, or click blank then click option'

Aapplymap

Breduce

Capply

Dmap

Attempts:

3 left

💡 Hint

Common Mistakes

Using applymap, which applies element-wise, not column-wise.

Using map, which is not a DataFrame method.

The apply method on a DataFrame applies a function to each column by default.

3fill in blank

hard

Fix the error in applying a function to each column to get the max value.

Pandas

max_values = df.[1](max)

Drag options to blanks, or click blank then click option'

Afilter

Bapply

Cmap

Dapplymap

Attempts:

3 left

💡 Hint

Common Mistakes

Using applymap which causes an error here.

Using map which is not a DataFrame method.

To get max values per column, use apply with max function.

4fill in blank

hard

Fill both blanks to create a dictionary of word lengths for words longer than 3 characters.

Pandas

lengths = {word: [1] for word in words if len(word) [2] 3}

Drag options to blanks, or click blank then click option'

Alen(word)

B>

C<

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using the word itself as value instead of its length.

Using less than instead of greater than in the condition.

The dictionary comprehension uses len(word) as value and filters words with length greater than 3.

5fill in blank

hard

Fill all three blanks to create a dictionary of uppercase words and their lengths for words longer than 4 characters.

Pandas

result = {{ [1]: [2] for word in words if len(word) [3] 4 }}

Drag options to blanks, or click blank then click option'

Aword.upper()

Blen(word)

C>

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using the original word as key instead of uppercase.

Using less than or equal instead of greater than.

The dictionary comprehension keys are uppercase words, values are lengths, filtered by length greater than 4.

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