applymap() for DataFrame-wide operations in Pandas - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of applymap() with a simple function

What is the output of this code when applying applymap() to double each element in the DataFrame?

Pandas

import pandas as pd

df = pd.DataFrame({'A': [1, 2], 'B': [3, 4]})
result = df.applymap(lambda x: x * 2)
print(result)

A

   A  B
0  2  6
1  4  8

B

   A  B
0  1  3
1  2  4

C

   A  B
0  1  6
1  4  8

D

   A  B
0  2  3
1  4  8

💡 Hint

applymap applies the function to every element individually.

The lambda function doubles each element. So 1 becomes 2, 2 becomes 4, 3 becomes 6, and 4 becomes 8.

❓ data_output

intermediate

Result of applymap() with a conditional function

Given this DataFrame, what is the result of applying applymap() with a function that replaces odd numbers with 0 and keeps even numbers unchanged?

Pandas

import pandas as pd

df = pd.DataFrame({'X': [1, 4], 'Y': [3, 6]})
result = df.applymap(lambda x: 0 if x % 2 != 0 else x)
print(result)

A

   X  Y
0  0  0
1  4  6

B

   X  Y
0  1  3
1  4  6

C

   X  Y
0  0  3
1  4  0

D

   X  Y
0  1  0
1  0  6

💡 Hint

Check which numbers are odd and replace them with zero.

1 and 3 are odd, so they become 0. 4 and 6 are even, so they stay the same.

🔧 Debug

advanced

Identify the error in applymap() usage

What error will this code raise when trying to use applymap() with a function that returns a list?

Pandas

import pandas as pd

df = pd.DataFrame({'A': [1, 2], 'B': [3, 4]})
result = df.applymap(lambda x: [x, x*2])
print(result)

A ValueError: Must produce scalar value

B No error, prints DataFrame with lists as elements

C AttributeError: 'list' object has no attribute 'applymap'

D TypeError: unhashable type: 'list'

💡 Hint

applymap applies the function to each element and can return any object.

applymap applies the function to each element and stores the returned object. Returning a list is allowed, so no error occurs.

🚀 Application

advanced

Using applymap() to format DataFrame values

You want to format all float numbers in a DataFrame to show only two decimal places as strings. Which applymap() function achieves this?

Pandas

import pandas as pd

df = pd.DataFrame({'A': [1.2345, 2.3456], 'B': [3.4567, 4.5678]})
result = df.applymap(??? )
print(result)

A lambda x: str(round(x, 2))

B lambda x: f"{x:.2f}"

C lambda x: format(x, '.2f') if isinstance(x, float) else x

D lambda x: round(x, 2)

💡 Hint

Only floats should be formatted as strings, other types remain unchanged.

Option C formats only floats to strings with two decimals, leaving other types untouched.

🧠 Conceptual

expert

Understanding applymap() vs apply() on DataFrames

Which statement correctly describes the difference between applymap() and apply() when used on a pandas DataFrame?

A applymap() applies a function along rows; apply() applies a function element-wise to each cell.

B applymap() and apply() are interchangeable and produce the same results.

C applymap() works only on Series; apply() works only on DataFrames.

D applymap() applies a function element-wise to each cell; apply() applies a function along rows or columns.

💡 Hint

Think about whether the function is applied to each cell or to rows/columns.

applymap() works element-wise on each cell of a DataFrame. apply() works on rows or columns (Series) and can aggregate or transform.

Practice

1. What does the applymap() function do in a pandas DataFrame?

easy

A. Applies a function to each column separately.

B. Applies a function to every single cell in the DataFrame.

C. Applies a function to each row separately.

D. Applies a function only to numeric columns.

Solution

1. Step 1: Understand the purpose of applymap()

   applymap() is designed to apply a function to every individual cell in a DataFrame, not just rows or columns.

2. Step 2: Compare with other pandas functions

   Functions like apply() work on rows or columns, but applymap() works cell-by-cell.

3. Final Answer:

   Applies a function to every single cell in the DataFrame. -> Option B

4. Quick Check:

   applymap() = cell-wise function application [OK]

Hint: Remember: applymap() works on each cell, not rows or columns [OK]

Common Mistakes:

• Confusing applymap() with apply()
• Thinking applymap() works on Series
• Assuming applymap() only works on numeric data

2. Which of the following is the correct syntax to use applymap() on a DataFrame df with a function func?

easy

A. df.applymap(func)

B. df.applymap = func()

C. df.applymap[func]

D. applymap(df, func)

Solution

1. Step 1: Recall the method call format

   In pandas, applymap() is a method of DataFrame objects and is called as df.applymap(func).

2. Step 2: Check other options for syntax errors

   Options B, C, and D do not follow valid Python or pandas syntax for method calls.

3. Final Answer:

   df.applymap(func) -> Option A

4. Quick Check:

   Method call syntax = df.applymap(func) [OK]

Hint: Use dot notation and parentheses: df.applymap(func) [OK]

Common Mistakes:

• Using assignment instead of method call
• Using square brackets instead of parentheses
• Calling applymap as a standalone function

3. Given the DataFrame df below:

   A  B
0  1  4
1  2  5
2  3  6

What is the output of df.applymap(lambda x: x * 2)?

medium

A. A B
0 1 4
1 2 5
2 3 6

B. A B
0 1 8
1 4 10
2 6 12

C. A B
0 2 8
1 4 10
2 6 12

D. Error: lambda function not supported

Solution

1. Step 1: Understand the lambda function

   The lambda function doubles each value: lambda x: x * 2.

2. Step 2: Apply the function to each cell

   Each cell in the DataFrame is multiplied by 2, so 1 becomes 2, 4 becomes 8, etc.

3. Final Answer:

   A B 0 2 8 1 4 10 2 6 12 -> Option C

4. Quick Check:

   Each cell doubled = output B [OK]

Hint: applymap applies function cell-wise; multiply each cell by 2 [OK]

Common Mistakes:

• Thinking applymap works column-wise
• Expecting original DataFrame unchanged
• Assuming lambda functions cause errors

4. What is wrong with this code snippet?

import pandas as pd
df = pd.DataFrame({'A':[1,2], 'B':[3,4]})
result = df.applymap(lambda x: x + 1)

Why might result be the same as df?

medium

A. The DataFrame contains non-numeric data, so applymap fails silently.

B. The lambda function is not applied because applymap was not called correctly.

C. Nothing is wrong; result will be different from df.

D. The code is correct, but the result is not assigned back to df.

Solution

1. Step 1: Check the code correctness

   The code correctly creates a DataFrame and applies applymap() with a lambda that adds 1.

2. Step 2: Understand the result assignment

   The result is stored in result, which will be different from df because each cell is incremented by 1.

3. Final Answer:

   The code is correct, but the result is not assigned back to df. -> Option D

4. Quick Check:

   applymap() works and returns new DataFrame [OK]

Hint: applymap returns new DataFrame; original stays unchanged [OK]

Common Mistakes:

• Expecting applymap to modify df in place
• Thinking applymap fails on numeric data
• Confusing assignment of result with in-place change

5. You have a DataFrame df with mixed data types:

   A     B
0  1   foo
1  2   bar
2  3   baz

You want to convert all numeric cells to their squares but leave non-numeric cells unchanged. Which applymap() function achieves this?

hard

A. df.applymap(lambda x: x**2 if isinstance(x, (int, float)) else x)

B. df.applymap(lambda x: x**2)

C. df.applymap(lambda x: str(x))

D. df.applymap(lambda x: x**2 if isinstance(x, str) else x)

Solution

1. Step 1: Identify the requirement

   We want to square numeric values only and keep strings unchanged.

2. Step 2: Analyze each option

   df.applymap(lambda x: x**2 if isinstance(x, (int, float)) else x) checks if the cell is int or float, then squares it; else returns original. df.applymap(lambda x: x**2) squares all cells, causing error on strings. df.applymap(lambda x: str(x)) converts all to strings, losing numeric info. df.applymap(lambda x: x**2 if isinstance(x, str) else x) squares strings (causing TypeError) and leaves numerics unchanged.

3. Final Answer:

   df.applymap(lambda x: x**2 if isinstance(x, (int, float)) else x) -> Option A

4. Quick Check:

   Use type check inside applymap to selectively transform [OK]

Hint: Use isinstance check inside applymap to handle mixed types [OK]

Common Mistakes:

• Applying math operations to strings causing errors
• Not checking data type before transformation
• Converting all data to string losing numeric info