Lambda with filter() in Python - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this code using lambda and filter?

Consider the following Python code that uses lambda with filter(). What will be printed?

Python

numbers = [1, 2, 3, 4, 5, 6]
result = list(filter(lambda x: x % 2 == 0, numbers))
print(result)

A [2, 4, 6]

B [1, 2, 3, 4, 5, 6]

C []

D [1, 3, 5]

💡 Hint

Remember, filter keeps items where the lambda returns True.

The lambda function checks if a number is even (x % 2 == 0). filter() keeps only those numbers. So the output is the list of even numbers: [2, 4, 6].

❓ Predict Output

intermediate

What does this filter with lambda return?

What will be the output of this code snippet?

Python

words = ['apple', 'banana', 'cherry', 'date']
result = list(filter(lambda w: len(w) > 5, words))
print(result)

A ['banana', 'cherry']

B ['apple', 'banana', 'cherry']

C ['date']

D []

💡 Hint

Check the length of each word carefully.

The lambda keeps words with length greater than 5. 'banana' and 'cherry' have length 6, so they are kept.

❓ Predict Output

advanced

Output of filter with lambda checking nested list elements

What is the output of this code?

Python

data = [[1, 2], [3, 4], [5], []]
result = list(filter(lambda x: len(x) == 2, data))
print(result)

A []

B [[1, 2], [3, 4], [5]]

C [[5], []]

D [[1, 2], [3, 4]]

💡 Hint

Filter keeps lists with exactly two elements.

Only the first two lists have length 2, so they are kept.

❓ Predict Output

advanced

What error does this code raise?

What error will this code produce when run?

Python

numbers = [1, 2, 3]
result = list(filter(lambda x: x > 1, numbers))
print(result[3])

A TypeError

B SyntaxError

C IndexError

D No error, prints 3

💡 Hint

Check the length of the filtered list and the index accessed.

The filtered list is [2, 3], which has indices 0 and 1. Accessing index 3 causes IndexError.

🧠 Conceptual

expert

Which option produces the filtered list of odd squares?

Given nums = range(1, 6), which option produces a list of squares of odd numbers only using filter() and lambda?

A list(filter(lambda x: (x**0.5) % 2 != 0, [n**2 for n in nums]))

B list(filter(lambda x: x % 2 != 0, [n**2 for n in nums]))

C list(filter(lambda x: x % 2 == 1, nums))

D list(filter(lambda n: n % 2 != 0, nums))

💡 Hint

Filter after squaring, check odd squares.

Option B filters the squared numbers to keep only odd squares. Option B filters by sqrt parity using floating-point modulo, which is unreliable due to precision issues. Option B filters nums, not squares. Option B filters nums, not squares.

Practice

1. What does the following code do?
filter(lambda x: x > 5, [2, 7, 4, 10])

easy

A. Selects numbers greater than 5 from the list

B. Selects numbers less than 5 from the list

C. Returns the sum of numbers greater than 5

D. Sorts the list in ascending order

Solution

1. Step 1: Understand the lambda condition

   The lambda function checks if each number is greater than 5.

2. Step 2: Apply filter with the lambda

   filter() keeps only numbers where the lambda returns True, so numbers > 5.

3. Final Answer:

   Selects numbers greater than 5 from the list -> Option A

4. Quick Check:

   filter with lambda x > 5 = select numbers > 5 [OK]

Hint: filter + lambda picks items where condition is True [OK]

Common Mistakes:

• Thinking filter sums or sorts the list
• Confusing greater than with less than
• Assuming filter changes the original list

2. Which of these is the correct syntax to filter even numbers from a list nums using lambda and filter?

easy

A. filter(lambda x: x % 2 == 0, nums)

B. filter(lambda x: x // 2 == 0, nums)

C. filter(lambda x: x % 2, nums)

D. filter(lambda x: x == 2, nums)

Solution

1. Step 1: Identify condition for even numbers

   Even numbers have zero remainder when divided by 2, so x % 2 == 0.

2. Step 2: Check filter syntax

   filter(lambda x: x % 2 == 0, nums) correctly filters even numbers.

3. Final Answer:

   filter(lambda x: x % 2 == 0, nums) -> Option A

4. Quick Check:

   Even check = x % 2 == 0 [OK]

Hint: Use x % 2 == 0 to check even numbers [OK]

Common Mistakes:

• Using floor division (//) instead of modulo (%)
• Using x % 2 without comparison (filters odd numbers)
• Checking equality to 2 instead of remainder zero

3. What is the output of this code?

nums = [1, 3, 6, 8, 11]
result = list(filter(lambda x: x < 7, nums))
print(result)

medium

A. [7, 8, 11]

B. [6, 8, 11]

C. [1, 3, 6, 8]

D. [1, 3, 6]

Solution

1. Step 1: Understand the filter condition

   The lambda keeps numbers less than 7.

2. Step 2: Apply condition to each list item

   1 < 7 (keep), 3 < 7 (keep), 6 < 7 (keep), 8 < 7 (no), 11 < 7 (no).

3. Final Answer:

   [1, 3, 6] -> Option D

4. Quick Check:

   Filter x < 7 = [1, 3, 6] [OK]

Hint: Filter keeps items where lambda returns True [OK]

Common Mistakes:

• Including numbers not meeting condition
• Confusing less than with greater than
• Not converting filter object to list before printing

4. Find the error in this code snippet:

nums = [10, 15, 20]
filtered = filter(lambda x: x % 2 = 0, nums)
print(list(filtered))

medium

A. Incorrect list conversion syntax

B. Missing parentheses after filter

C. Using '=' instead of '==' in lambda condition

D. Lambda function missing argument

Solution

1. Step 1: Check lambda condition syntax

   The condition uses '=' which is assignment, not comparison.

2. Step 2: Correct operator usage

   It should be '==' to compare if x % 2 equals 0.

3. Final Answer:

   Using '=' instead of '==' in lambda condition -> Option C

4. Quick Check:

   Comparison needs '==' not '=' [OK]

Hint: Use '==' for comparison inside lambda [OK]

Common Mistakes:

• Using '=' instead of '==' in conditions
• Forgetting to convert filter result to list
• Missing lambda argument

5. You have a list of words: words = ['apple', '', 'banana', ' ', 'cherry', None]. Which code correctly filters out empty strings, strings with only spaces, and None values using lambda and filter?

hard

A. list(filter(lambda x: x == '', words))

B. list(filter(lambda x: x and x.strip(), words))

C. list(filter(lambda x: x is None, words))

D. list(filter(lambda x: x.strip() == '', words))

Solution

1. Step 1: Understand filtering criteria

   We want to remove empty strings (''), strings with only spaces (' '), and None values.

2. Step 2: Analyze each option

   list(filter(lambda x: x and x.strip(), words)) checks x is truthy (not None or empty) and x.strip() removes spaces; if non-empty after strip, it keeps the word.

3. Final Answer:

   list(filter(lambda x: x and x.strip(), words)) -> Option B

4. Quick Check:

   Filter removes empty, spaces, None with x and x.strip() [OK]

Hint: Use x and x.strip() to remove empty/space/None [OK]

Common Mistakes:

• Filtering only empty strings but not spaces or None
• Checking equality to None incorrectly
• Calling strip() on None causing error