Set membership testing in Python - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if 5 is in the set.

Python

numbers = {1, 3, 5, 7}
if 5 [1] numbers:
    print("Found 5")

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

Ain

Bnot in

Ccontains

Dinside

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'not in' instead of 'in'.

Using 'contains' which is not a Python keyword.

Using 'inside' which is not valid syntax.

The keyword in checks if an item is in a set.

2fill in blank

medium

Complete the code to check if 'apple' is NOT in the set.

Python

fruits = {'banana', 'orange', 'grape'}
if 'apple' [1] fruits:
    print("Apple is missing")

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

Ahas

Bin

Ccontains

Dnot in

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'in' instead of 'not in'.

Using 'contains' or 'has' which are not Python keywords.

The keyword not in checks if an item is NOT in a set.

3fill in blank

hard

Fix the error in the code to correctly test membership.

Python

colors = {'red', 'blue', 'green'}
if 'yellow' [1] colors:
    print("Yellow found")
else:
    print("Yellow not found")

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

Ain

Bnot in

Cinside

Dcontains

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'inside' or 'contains' which cause syntax errors.

Using 'not in' when checking for presence.

Use in to check if 'yellow' is in the set.

4fill in blank

hard

Fill both blanks to create a set of squares for numbers greater than 3.

Python

squares = {x[1]2 for x in range(1, 7) if x [2] 3}

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

A**

B+

C>

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' instead of '**' for squaring.

Using '<' instead of '>' for filtering.

Use ** to square numbers and > to filter numbers greater than 3.

5fill in blank

hard

Fill all three blanks to create a set of uppercase words longer than 4 letters.

Python

result = {word[1] for word in words if len(word) [2] 4 and word.isalpha() [3] True}

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

A.upper()

B>

C==

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' or other operators instead of '.upper()'.

Using '<' instead of '>' for length check.

Using '=' instead of '==' for comparison.

Use .upper() to convert words to uppercase, > to check length greater than 4, and == to check if isalpha() returns True.

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