Set membership testing in Python - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the in keyword do when used with a set in Python?

It checks if a value is present in the set and returns True if it is, otherwise False.

beginner

How do you check if an element is NOT in a set?

Use the not in keyword. For example, if element not in my_set: checks if element is not a member of my_set.

intermediate

Why is set membership testing faster than list membership testing?

Sets use a hash table internally, so checking membership is on average very fast (constant time), while lists check each item one by one (linear time).

beginner

What will be the output of this code?

my_set = {1, 2, 3}
print(2 in my_set)
print(5 in my_set)

True
False

Because 2 is in the set but 5 is not.

beginner

Can you use the in keyword to check membership in other data types besides sets?

Yes! You can use in with lists, tuples, strings, dictionaries (checks keys), and more.

What does element in my_set return if element is not in my_set?

A Raises an error

B False

C None

D True

The expression returns False if the element is not found in the set.

Which data structure generally provides the fastest membership testing?

A Set

B Tuple

C List

D String

Sets use hash tables internally, making membership testing very fast.

What keyword checks if an item is NOT in a set?

A not in

B not

C out

D no in

The 'not in' keyword checks if an item is absent from a set.

What will print('a' in {'a', 'b', 'c'}) output?

A Error

B False

C True

D None

The letter 'a' is in the set, so it prints True.

Which of these is NOT a valid use of in for membership testing?

A Checking if a key is in a dictionary

B Checking if a value is in a list

C Checking if a character is in a string

D Checking if a number is in an integer

You cannot check membership inside an integer because it is not a collection.

Explain how to check if an item is in a set and why this is efficient.

Think about how sets store items differently than lists.

Describe the difference between using 'in' and 'not in' with sets.

One checks if something is there, the other checks if it is missing.

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