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.