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Dictionaries – Advanced (comprehensions, merging)

A dict comprehension builds a dictionary in one expression; the | operator merges two dicts

What it does: .items() yields (key, value) pairs for each entry in the dict. The pipe | merges dicts - if both have the same key, the right side value wins. zip pairs the two lists, dict() converts those pairs into a dictionary. Add 'if condition' before the closing brace to filter - only pairs where condition is True get included. Unpacking k, v then using v: k reverses which side is the key and which is the value. The right dict overwrites matching keys - think of it as d1 updated with d2. zip(['a','b'], [1,2]) gives ('a',1) then ('b',2) - like a zipper joining two lists.

Common mistakes: Dict comprehensions use curly braces {}, not square brackets []. Dicts don't support +. Use | or {**d1, **d2} to merge.

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Learning

Duo

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

{0:0, 1:1, 2:4, 3:9}[0, 1, 4, 9]

Matches

❌ d1 + d2

↓

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

Applying

Meanings

Term

Dict comprehension

Meaning

{key: value for item in iterable} builds a dictionary concisely.

Practice

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

Testing

Gaps

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

Check Answer

Spots

1# Build a squares dictionary

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

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Quick Questions

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

A. {0:0, 1:1, 2:4, 3:9} ✓

B. [0, 1, 4, 9]

Dict comprehensions use key: value pairs - x maps to x squared for each x

2. d1 | d2 does what?

A. Checks if equal

B. Merges both dicts (d2 wins conflicts) ✓

The pipe | merges dicts - if both have the same key, the right side value wins

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

A. Unpacks dict entries into a new dict ✓

B. Squares the keys

** spreads all key-value pairs from a dict into the surrounding context

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

A. [('a',1), ('b',2)]

B. {'a': 1, 'b': 2} ✓

zip pairs the two lists, dict() converts those pairs into a dictionary

5. Can dict comprehensions have conditions?

A. Yes add if at the end ✓

B. No

Add 'if condition' before the closing brace to filter - only pairs where condition is True get included

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}

Key Term

Dict comprehension

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

Fill in the Blanks

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

2. result = dict(zip(names, scores))

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

4. merged = {**d1, **d2}

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

Spot the 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')

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

Question 2

A d1 = {'a': 1}

B d2 = {'b': 2}

C merged = d1 + d2

D print('Merged:', merged)

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

Question 3

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

B scores = [90, 85]

C result = zip(names, scores)

D print(result)

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

Question 4

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

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

C print('New:', new)

D print('Done')

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

Question 5

A d1 = {'a': 1}

B d2 = {'a': 2}

C merged = d1 | d2

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

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

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