Lambda vs regular functions in Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a lambda function that adds 5 to a number.

Python

add_five = lambda x: x [1] 5
print(add_five(10))

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

A-

B+

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using '-' instead of '+' will subtract instead of add.

Using '*' or '/' will multiply or divide, not add.

The lambda function adds 5 to the input number using the + operator.

2fill in blank

medium

Complete the regular function to return the square of a number.

Python

def square(num):
    return num [1] num

print(square(4))

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

A/

B+

C-

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' will add the number to itself, not square it.

Using '-' or '/' will not give the square.

The function returns the number multiplied by itself to get the square.

3fill in blank

hard

Fix the error in the lambda function that should multiply two numbers.

Python

multiply = lambda a, b: a [1] b
print(multiply(3, 7))

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

A*

B+

C-

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' adds instead of multiplies.

Using '-' or '/' will not multiply.

The lambda function multiplies two numbers using the '*' operator.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that maps numbers to their squares using a lambda.

Python

square_func = lambda x: x [1] x
squares = {num: square_func(num) for num in range(1, 6) if num [2] 3}
print(squares)

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

A*

B>

C<

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' in lambda adds instead of squares.

Using '<' includes numbers less than 3, not greater.

The lambda multiplies the number by itself to get the square. The comprehension includes numbers greater than 3.

5fill in blank

hard

Fill all three blanks to create a lambda that filters even numbers and returns their squares in a dictionary.

Python

is_even = lambda n: n [1] 2 == 0
square = lambda x: x [2] x
result = {num: square(num) for num in range(1, 11) if is_even(num) [3] True}
print(result)

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

A%

B*

C==

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '!=' instead of '==' changes the condition.

Using '+' instead of '*' does not square the number.

Using '/' instead of '%' does not check evenness.

The lambda 'is_even' checks if a number divided by 2 has remainder 0 using '%'. The 'square' lambda multiplies the number by itself. The dictionary comprehension includes numbers where 'is_even(num) == True'.