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Sets – Advanced (frozen sets, set comprehensions)

A set comprehension builds a set in one expression; frozenset() creates an immutable version of a set

Understanding Sets – Advanced (frozen sets, set comprehensions) in Python

What it does: Curly braces with a single expression (no :) creates a set comprehension. frozensets are hashable and can be used as dictionary keys. frozenset() creates a set that cannot be changed after creation - like a tuple vs list. Frozen means locked - no add(), remove(), or any mutation methods. Mutable objects can't be hashed - use a frozenset if you need a set as a key. x % 2 == 0 is True for even numbers - 0, 2, 4, 6, 8 from range(10). Same curly brace syntax but dict uses key: value pairs while set uses single values. frozensets support all read operations like union and intersection - just no mutations.

Common mistakes: Square brackets make a list comprehension. Use curly braces {} for a set comprehension. Frozen sets are immutable — they have no add() method.

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This module's key terms are also in a printable crossword (Modules 51–63) and word search (Modules 41–63).

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Learning

Duo

{x**2 for x in range(4)} creates?

{0, 1, 4, 9}[0, 1, 4, 9]

Matches

❌ frozenset({1,2}).add(3)

↓

Error: frozenset is immutable — no add method

Applying

Meanings

Term

Set comprehension

Meaning

{expression for item in iterable} builds a set concisely.

Practice

squares = {x**2 for x in range(6)}

Testing

Gaps

squares = {x**2 x in range(5)}

Check Answer

Spots

1numbers = [1, 2, 3, 4, 5]

2squares = [x**2 for x in numbers]

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Quick Questions

1. {x**2 for x in range(4)} creates?

A. [0, 1, 4, 9]

B. {0, 1, 4, 9} ✓

Set comprehensions use curly braces with a single expression - the result is a set, not a list

2. What is a frozenset?

A. A sorted set

B. An immutable set ✓

frozenset() creates a set that cannot be changed after creation - like a tuple vs list

3. Can you add to a frozenset?

A. No it's immutable ✓

B. Yes

Frozen means locked - no add(), remove(), or any mutation methods

4. Can a frozenset be a dict key?

A. Yes it's hashable ✓

B. No

Dict keys must be hashable - frozensets qualify because they can't change

5. Can a regular set be a dict key?

A. No sets are mutable ✓

B. Yes

Mutable objects can't be hashed - use a frozenset if you need a set as a key

Match the Pairs

frozenset({1,2}).add(3) → Error: frozenset is immutable — no add method

[x for x in range(5)] → Error: This is a list, not a set comprehension

{x**2 for x in range(5)} → Valid: Set comprehension

frozenset({1, 2, 3}) → Ok: Create an immutable frozenset

len({1, 2, 2, 3}) → Output: 3

{x for x in range(6) if x%2==0} → Output: {0, 2, 4}

Key Term

Set comprehension

{expression for item in iterable} builds a set concisely.~~frozenset

Fill in the Blanks

1. squares = {x**2 for x in range(5)}

2. evens = {x for x in range(10) if x % 2 == 0}

3. fs = frozenset( {1, 2, 3} )

4. unique = set(['a', 'b', 'a'])

5. d = {frozenset({1, 2}): 'pair'}

Spot the Error

Question 1

A numbers = [1, 2, 3, 4, 5]

B squares = [x**2 for x in numbers]

C print('Squares:', squares)

D print('Done')

Line B — Square brackets make a list comprehension. Use curly braces {} for a set comprehension.

Question 2

A fs = frozenset({1, 2, 3})

B fs.add(4)

C print('Frozen:', fs)

D print('Done')

Line B — Frozen sets are immutable — they have no add() method.

Question 3

A key = frozenset({1, 2})

B d = {{1, 2}: 'pair'}

C print('Dict:', d)

D print('Done')

Line B — Regular sets are mutable and can't be dict keys. Use frozenset({1, 2}) as the key instead.

Question 4

A numbers = range(10)

B evens = {x for x in numbers if x % 2 = 0}

C print('Evens:', evens)

D print('Done')

Line B — Use == for comparison in conditions, not = (which is assignment).

Question 5

A unique = set('hello')

B word = 'hello'

C print(unique) # expects {'hello'}

D print('Done')

Line A — set('hello') splits into individual characters. Use {'hello'} for a set containing the whole word.

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