Dictionaries – Advanced (comprehensions, merging)

Name:

Date:

Class:

Quick Questions

Shade in the A or B circle for each question.

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

A {0:0, 1:1, 2:4, 3:9}

B [0, 1, 4, 9]

2. d1 | d2 does what?

A Checks if equal

B Merges both dicts (d2 wins conflicts)

3. {**d1, **d2} what does ** do?

A Squares the keys

B Unpacks dict entries into a new dict

4. dict(zip(['a','b'], [1,2])) creates?

A {'a': 1, 'b': 2}

B [('a',1), ('b',2)]

5. Can dict comprehensions have conditions?

A Yes add if at the end

B No

Match the Pairs

Draw a line connecting each item on the left to its match on the right.

Left column		Right column
1d1 + d2		AValid: Dictionary comprehension
2{k for k, v in d}		BOk: Merge two dicts using unpacking
3{k: v for k, v in d.items()}		CError: This is a set, not a dict comprehension
4{d1, d2}		DOutput: {'a': 1, 'b': 2}
5{x: x**2 for x in range(4)}		EError: Use d1 \| d2 or {d1, d2} to merge
6dict(zip(['a','b'], [1,2]))		FOutput: {0: 0, 1: 1, 2: 4, 3: 9}

Definitions

Dict comprehension

Fill in the Blanks

Complete each line of code by filling in the missing parts.

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

result = dict(___(names, scores))

pos = {k: v for k, v in d.items() __ v > 0}

merged = {____d1, **d2}

inverted = {v: _ for k, v in d.items()}

Spot the Error

Shade in the letter of the line that contains an error.

Question 1

A # Build a squares dictionary

B squares = [x: x**2 for x in range(5)]

C print('Squares:', squares)

D print('Done')

Question 2

A d1 = {'a': 1}

B d2 = {'b': 2}

C merged = d1 + d2

D print('Merged:', merged)

Question 3

A names = ['Alice', 'Bob']

B scores = [90, 85]

C result = zip(names, scores)

D print(result)

Question 4

A d = {'a': 1, 'b': 2}

B new = {k: v for k in d}

C print('New:', new)

D print('Done')

Question 5

A d1 = {'a': 1}

B d2 = {'a': 2}

C merged = d1 | d2

D print(merged['a']) # expects 1

Answer Key

Dictionaries – Advanced (comprehensions, merging)

Section A: Quick Questions

A — {0:0, 1:1, 2:4, 3:9}

B — Merges both dicts (d2 wins conflicts)

B — Unpacks dict entries into a new dict

A — {'a': 1, 'b': 2}

A — Yes add if at the end

Section B: Match the Pairs

d1 + d2 → Error: Use d1 | d2 or {**d1, **d2} to merge

{k for k, v in d} → Error: This is a set, not a dict comprehension

{k: v for k, v in d.items()} → Valid: Dictionary comprehension

{**d1, **d2} → Ok: Merge two dicts using unpacking

{x: x**2 for x in range(4)} → Output: {0: 0, 1: 1, 2: 4, 3: 9}

dict(zip(['a','b'], [1,2])) → Output: {'a': 1, 'b': 2}

Section C: Definitions

Dict comprehension — {key: value for item in iterable} builds a dictionary concisely.~~Merge operator

Section D: Fill in the Blanks

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

result = dict(zip(names, scores))

pos = {k: v for k, v in d.items() if v > 0}

merged = {**d1, **d2}

inverted = {v: k for k, v in d.items()}

Section E: Spot the Error

B — Dict comprehensions use curly braces {}, not square brackets []

C — Dicts don't support +. Use | or {**d1, **d2} to merge.

D — zip returns a zip object. Wrap in dict() to get a dictionary.

B — Need .items() to get both key and value and unpack as k, v

D — In d1 | d2, d2 wins for shared keys. merged['a'] is 2, not 1.