apply() on rows (axis=1) in Pandas - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to apply a function to each row of the DataFrame.

Pandas

import pandas as pd

df = pd.DataFrame({'A': [1, 2], 'B': [3, 4]})
result = df.apply(lambda row: row['A'] + row['B'], axis=[1])
print(result)

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

A1

B2

CNone

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using axis=0 applies the function to columns, not rows.

Forgetting to specify axis parameter.

Setting axis=1 applies the function to each row.

2fill in blank

medium

Complete the code to create a new column 'Sum' by applying a function on each row.

Pandas

import pandas as pd

df = pd.DataFrame({'X': [5, 6], 'Y': [7, 8]})
df['Sum'] = df.apply(lambda row: row['X'] + row['Y'], axis=[1])
print(df)

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

ANone

B0

C2

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using axis=0 applies function to columns, causing errors or wrong results.

Not assigning the result back to a new column.

Use axis=1 to apply the function row-wise and create the new column.

3fill in blank

hard

Fix the error in the code to correctly apply a function on rows to get the max value per row.

Pandas

import pandas as pd

df = pd.DataFrame({'P': [10, 20], 'Q': [30, 15]})
max_values = df.apply(lambda row: max(row), axis=[1])
print(max_values)

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

A2

BNone

C1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using axis=0 applies function column-wise, not row-wise.

Forgetting to specify axis parameter.

Using axis=1 applies the function to each row, allowing max(row) to find the max per row.

4fill in blank

hard

Fill both blanks to create a new column 'Difference' that subtracts column 'B' from 'A' for each row.

Pandas

import pandas as pd

df = pd.DataFrame({'A': [9, 4], 'B': [3, 7]})
df['Difference'] = df.apply(lambda [1]: [2]['A'] - [2]['B'], axis=1)
print(df)

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

Arow

Bdf

Cdata

Dx

Attempts:

3 left

💡 Hint

Common Mistakes

Using a variable name that is not the lambda argument to access row data.

Not matching the lambda argument name with its usage inside the function.

The lambda function takes one argument (commonly named row) representing each row. Use this argument to access columns.

5fill in blank

hard

Fill all three blanks to create a dictionary with keys as row indices and values as the row values from column 'b' where the value in 'a' is greater than 2.

Pandas

import pandas as pd

df = pd.DataFrame({'a': [1, 3], 'b': [4, 5]})
result = { [1]: row[[2]] for index, row in df.iterrows() if row[[3]] > 2 }
print(result)

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

A'A'

B'b'

C'a'

Dindex

Attempts:

3 left

💡 Hint

Common Mistakes

Using column names incorrectly as keys.

Mixing up the filter condition column and value column.

The dictionary uses the row index as key, the value in column 'b' as value, and filters rows where column 'a' is greater than 2.

Practice

1.

What does apply() with axis=1 do in a pandas DataFrame?

easy

A. It applies a function to each column of the DataFrame.

B. It applies a function to each row of the DataFrame.

C. It applies a function to the entire DataFrame at once.

D. It applies a function only to the first row.

Solution

1. Step 1: Understand the axis parameter

   In pandas, axis=1 means the function is applied across columns for each row.

2. Step 2: Apply function on rows

   Using apply() with axis=1 runs the function on each row individually.

3. Final Answer:

   It applies a function to each row of the DataFrame. -> Option B

4. Quick Check:

   apply(axis=1) = row-wise function [OK]

Hint: Remember axis=1 means row-wise operation [OK]

Common Mistakes:

• Confusing axis=1 with column-wise operation
• Thinking apply(axis=1) applies to the whole DataFrame at once
• Assuming it only applies to the first row

2.

Which of the following is the correct syntax to apply a function f to each row of a DataFrame df?

def f(row):
    return row['A'] + row['B']

easy

A. df.apply(f, axis=1)

B. df.apply(f, axis=0)

C. df.apply(f)

D. df.apply(f, axis='rows')

Solution

1. Step 1: Identify axis for row-wise apply

   Row-wise apply requires axis=1.

2. Step 2: Check syntax correctness

   df.apply(f, axis=1) uses axis=1, which is correct. The other options use incorrect axis parameters.

3. Final Answer:

   df.apply(f, axis=1) -> Option A

4. Quick Check:

   Use axis=1 for rows [OK]

Hint: Use axis=1 to apply function row-wise [OK]

Common Mistakes:

• Using axis=0 applies function column-wise
• Omitting axis defaults to axis=0
• Using incorrect axis string values

3.

What is the output of the following code?

import pandas as pd
df = pd.DataFrame({'A': [1, 2], 'B': [3, 4]})
def sum_row(row):
    return row['A'] + row['B']
df['Sum'] = df.apply(sum_row, axis=1)
print(df)

medium

A. Error: KeyError

B. A B Sum
0 1 3 3
1 2 4 4

C. A B Sum
0 1 3 1
1 2 4 2

D. A B Sum
0 1 3 4
1 2 4 6

Solution

1. Step 1: Understand the function sum_row

   The function adds values from columns 'A' and 'B' for each row.

2. Step 2: Apply function row-wise and add new column

   Applying with axis=1 sums each row's 'A' and 'B'. For row 0: 1+3=4, row 1: 2+4=6.

3. Final Answer:

   A B Sum 0 1 3 4 1 2 4 6 -> Option D

4. Quick Check:

   Row sums = 4, 6 [OK]

Hint: Sum columns per row with apply(axis=1) [OK]

Common Mistakes:

• Using axis=0 sums columns, not rows
• Forgetting to assign result back to DataFrame
• Mixing up row and column indices

4.

Identify the error in the following code snippet:

import pandas as pd
df = pd.DataFrame({'X': [10, 20], 'Y': [30, 40]})
def multiply(row):
    return row['x'] * row['Y']
df['Product'] = df.apply(multiply, axis=1)
print(df)

medium

A. KeyError because 'x' column does not exist (case sensitive).

B. TypeError due to wrong axis parameter.

C. No error, output is correct.

D. SyntaxError in function definition.

Solution

1. Step 1: Check column names in DataFrame

   Columns are 'X' and 'Y' with uppercase letters.

2. Step 2: Check function usage of column names

   Function uses 'x' (lowercase), which does not match 'X' (uppercase), causing KeyError.

3. Final Answer:

   KeyError because 'x' column does not exist (case sensitive). -> Option A

4. Quick Check:

   Column names are case sensitive [OK]

Hint: Check exact column names and case sensitivity [OK]

Common Mistakes:

• Ignoring case sensitivity in column names
• Using wrong axis parameter
• Assuming no error without testing

5.

You have a DataFrame df with columns 'Name', 'Math', and 'English'. You want to create a new column 'Result' that shows 'Pass' if the average of 'Math' and 'English' is 50 or more, otherwise 'Fail'. Which code correctly uses apply() with axis=1 to do this?

hard

A. df['Result'] = df.apply(lambda row: 'Pass' if row['Math'] + row['English'] >= 50 else 'Fail', axis=1)

B. df['Result'] = df.apply(lambda col: 'Pass' if (col['Math'] + col['English']) / 2 >= 50 else 'Fail', axis=0)

C. df['Result'] = df.apply(lambda row: 'Pass' if (row['Math'] + row['English']) / 2 >= 50 else 'Fail', axis=1)

D. df['Result'] = df.apply(lambda row: 'Pass' if row['Math'] >= 50 and row['English'] >= 50 else 'Fail')

Solution

1. Step 1: Understand the condition for passing

   Pass if average of 'Math' and 'English' >= 50, else Fail.

2. Step 2: Check each option's logic and axis

   df['Result'] = df.apply(lambda row: 'Pass' if (row['Math'] + row['English']) / 2 >= 50 else 'Fail', axis=1) correctly uses axis=1 for row-wise apply and calculates average properly. df['Result'] = df.apply(lambda col: 'Pass' if (col['Math'] + col['English']) / 2 >= 50 else 'Fail', axis=0) uses axis=0 (column-wise), which is wrong. df['Result'] = df.apply(lambda row: 'Pass' if row['Math'] >= 50 and row['English'] >= 50 else 'Fail') misses axis=1 and uses wrong condition. df['Result'] = df.apply(lambda row: 'Pass' if row['Math'] + row['English'] >= 50 else 'Fail', axis=1) checks sum >= 50, which is incorrect for average >= 50.

3. Final Answer:

   df['Result'] = df.apply(lambda row: 'Pass' if (row['Math'] + row['English']) / 2 >= 50 else 'Fail', axis=1) -> Option C

4. Quick Check:

   Use axis=1 and correct average condition [OK]

Hint: Use axis=1 and check average inside lambda [OK]

Common Mistakes:

• Using axis=0 applies function column-wise
• Forgetting to divide sum by 2 for average
• Using wrong logical conditions