Set membership testing in Python - Mini Project: Build & Apply

Set Membership Testing

📖 Scenario: You are organizing a small event and have a list of invited guests. You want to check if certain people are on the guest list quickly.

🎯 Goal: Build a program that uses a set to store invited guests and tests if specific people are invited using set membership testing.

📋 What You'll Learn

Create a set with exact invited guest names

Create a variable with a guest name to check

Use set membership testing with the in keyword

Print the result as a boolean value

💡 Why This Matters

🌍 Real World

Checking if a person is on a guest list is a common task in event planning and security.

💼 Career

Set membership testing is useful in many programming jobs for fast lookups, filtering data, and validating inputs.

1

Create the invited guests set

Create a set called invited_guests with these exact names: 'Alice', 'Bob', 'Charlie', 'Diana', and 'Ethan'.

Python

# Create the set invited_guests with the exact names
# Your code here

Hint

Use curly braces {} to create a set with the names separated by commas.

2

Create a guest name to check

Create a variable called guest_to_check and set it to the string 'Charlie'.

Python

invited_guests = {'Alice', 'Bob', 'Charlie', 'Diana', 'Ethan'}
# Create the variable guest_to_check with the value 'Charlie'
# Your code here

Hint

Use an equals sign = to assign the string 'Charlie' to guest_to_check.

3

Check if the guest is invited

Create a variable called is_invited and set it to the result of testing if guest_to_check is in invited_guests using the in keyword.

Python

invited_guests = {'Alice', 'Bob', 'Charlie', 'Diana', 'Ethan'}
guest_to_check = 'Charlie'
# Create is_invited by checking if guest_to_check is in invited_guests
# Your code here

Hint

Use variable = value in set to test membership and store the boolean result.

4

Print the membership test result

Print the value of the variable is_invited to show if the guest is invited.

Python

invited_guests = {'Alice', 'Bob', 'Charlie', 'Diana', 'Ethan'}
guest_to_check = 'Charlie'
is_invited = guest_to_check in invited_guests
# Print the value of is_invited
# Your code here

Hint

Use print(is_invited) to display the result.

Practice

1.

Which operator is used to check if an element exists inside a set in Python?

easy

A. contains

B. has

C. exists

D. in

Solution

1. Step 1: Understand set membership

   In Python, the in keyword checks if an element is present in a set.

2. Step 2: Identify correct operator

   Other options like has, exists, and contains are not valid Python operators for membership testing.

3. Final Answer:

   in -> Option D

4. Quick Check:

   Use in to test membership [OK]

Hint: Remember: use 'in' to check if item is inside a set [OK]

Common Mistakes:

• Using 'has' instead of 'in'
• Trying 'exists' keyword
• Using 'contains' which is not a Python operator

2.

Which of the following is the correct syntax to check if 5 is NOT in the set {1, 2, 3, 4}?

easy

A. 5 not in {1, 2, 3, 4}

B. not 5 in {1, 2, 3, 4}

C. 5 !in {1, 2, 3, 4}

D. 5 notinside {1, 2, 3, 4}

Solution

1. Step 1: Recall correct syntax for 'not in'

   The correct syntax to check if an element is not in a set is: element not in set.

2. Step 2: Evaluate each option

   5 not in {1, 2, 3, 4} uses correct syntax. not 5 in {1, 2, 3, 4} is incorrect because 'not' must come after the element. 5 !in {1, 2, 3, 4} uses invalid operator '!in'. 5 notinside {1, 2, 3, 4} uses a non-existent keyword 'notinside'.

3. Final Answer:

   5 not in {1, 2, 3, 4} -> Option A

4. Quick Check:

   Use 'not in' with element first [OK]

Hint: Write 'element not in set' to check absence [OK]

Common Mistakes:

• Using '!in' instead of 'not in'
• Placing 'not' before element
• Using invalid keywords like 'notinside'

3.

What will be the output of the following code?

my_set = {10, 20, 30}
print(25 in my_set)

medium

A. True

B. False

C. 25

D. Error

Solution

1. Step 1: Understand the set contents

   The set my_set contains 10, 20, and 30. It does not contain 25.

2. Step 2: Evaluate membership test

   The expression 25 in my_set checks if 25 is in the set. Since it is not, the result is False.

3. Final Answer:

   False -> Option B

4. Quick Check:

   25 not in set means False [OK]

Hint: If element missing in set, 'in' returns False [OK]

Common Mistakes:

• Assuming output is the element itself
• Confusing True/False for membership
• Expecting an error for missing element

4.

Find the error in this code snippet:

my_set = {1, 2, 3}
if 4 notin my_set:
    print("4 is not in the set")

medium

A. The keyword 'notin' is invalid

B. The set declaration is wrong

C. Missing colon after if statement

D. Print statement syntax error

Solution

1. Step 1: Check syntax of membership test

   The correct keyword to check absence is not in with a space, not notin.

2. Step 2: Verify other parts of code

   The set declaration is correct, colon after if is present, and print syntax is valid.

3. Final Answer:

   The keyword 'notin' is invalid -> Option A

4. Quick Check:

   Use 'not in' with space, not 'notin' [OK]

Hint: Remember: 'not in' has a space between words [OK]

Common Mistakes:

• Writing 'notin' as one word
• Forgetting colon after if
• Misreading print syntax

5.

Given the list nums = [1, 2, 2, 3, 4, 4, 5], which code snippet correctly prints Found only if 3 is in the unique set of numbers?

A) if 3 in nums:
       print("Found")

B) if 3 in set(nums):
       print("Found")

C) if 3 not in set(nums):
       print("Found")

D) if 3 not in nums:
       print("Found")

hard

A. if 3 in nums:
print("Found")

B. if 3 not in nums:
print("Found")

C. if 3 in set(nums):
print("Found")

D. if 3 not in set(nums):
print("Found")

Solution

1. Step 1: Understand the requirement

   We want to check if 3 is in the unique set of numbers from the list, so duplicates are ignored.

2. Step 2: Analyze each option

   if 3 in nums: print("Found") checks 3 in the list (with duplicates). if 3 in set(nums): print("Found") converts list to set and checks membership correctly. if 3 not in set(nums): print("Found") prints 'Found' if 3 is NOT in the set, which is wrong. if 3 not in nums: print("Found") checks 3 not in list, also wrong.

3. Final Answer:

   if 3 in set(nums): print("Found") -> Option C

4. Quick Check:

   Convert list to set before membership test [OK]

Hint: Convert list to set before membership test for uniqueness [OK]

Common Mistakes:

• Checking membership directly in list with duplicates
• Using 'not in' instead of 'in'
• Confusing list and set membership