Python set MCQ questions with answers

Answer: A

A set keeps one copy of each value, so the six items shrink to three. sorted() returns those values as a sorted list.

Answer: B

set() takes each character of the string, and the two 'l' characters count once, leaving h, e, l and o. For a set holding the whole word, write {'hello'}.

Answer: C

| is union: every value that is in either set, each once. & is intersection: only the values that are in both sets, here 3 and 4.

Answer: D

Adding a value that is already in a set changes nothing, so add(20) has no effect. Only 30 is new, which gives three items.

Answer: B

discard() removes a value if it is present and does nothing otherwise, so discard(5) raises no error and discard(2) removes 2. remove(5) would have raised KeyError.

Answer: A, KeyError: 4

remove() on a set raises KeyError when the value is missing. A list raises ValueError in the same situation, which makes it the tempting wrong answer. Use discard() to remove without an error.

Answer: D, It stores each value once and has no index positions

A set holds unique values with no defined order and no indexing. It is mutable (add(), remove()), but its items must be hashable, so lists cannot go in. frozenset is the immutable version.

Answer: A

set(nums) keeps one copy of each value and len() counts them, giving 3. {nums} tries to put the whole list inside a set and raises TypeError, because lists are not hashable.

Answer: A

a - b keeps the values of a that are not in b. a ^ b is the symmetric difference: the values that are in exactly one of the two sets, so the shared 3 and 4 drop out.

Answer: D

add() inserts its argument as one item, the string 'ab'. update() treats its argument as an iterable and adds each character, 'a' and 'b'. The set ends with three strings, which sorted() lists alphabetically.

Answer: B

<= tests for a subset and >= for a superset, so the first and last are True. < tests for a proper subset, which also requires the sets to differ, so a < a is False.

Answer: C

Methods whose names end in _update change the set in place and return None, like list.sort(). a keeps only the values that are also in the other set. a.intersection(...) would return a new set instead.

Answer: C

A frozenset is an immutable set, so it is hashable and can be an item of another set. Sets ignore order, so frozenset({1, 2}) and frozenset({2, 1}) are equal and count once.

Answer: B, TypeError: 'set' object is not subscriptable

Sets have no positions, so they do not support indexing at all and raise TypeError. Convert first, for example sorted(colors)[0], when you need an item by position.

Answer: A, A frozenset cannot be changed, so it is hashable and can be a dictionary key

A frozenset supports union, intersection and subset tests like a set but has no methods that change it. Because it is immutable it is hashable, so it can be a dictionary key or an item of another set.

Answer: B, An empty dictionary

Empty braces mean an empty dictionary, so an empty set must be written set(). Braces with items and no colons, such as {1, 2}, do make a set.

Answer: D

1, 1.0 and True are equal and have the same hash, so the set stores only one of them. The string 'True' is a different value, so the set has two items.

Answer: D

a |= {3} updates the set in place, so b, which names the same set, gains 3. a = a | {4} builds a new set and rebinds only a; b still holds {1, 2, 3}.

Answer: C, TypeError: unsupported operand type(s) for |: 'set' and 'list'

Set methods such as union() accept any iterable, so the second line works. The | operator requires a set on both sides, so evens | [8] raises TypeError. Write evens | {8} instead.

Answer: C

Dictionary keys are unique and keep insertion order, so dict.fromkeys() drops repeats in first-seen order. A set also drops repeats but has no defined order, sorted() gives alphabetical order, and the count() filter removes repeated names entirely.