TopolearnSets remove duplicates and answer 'is x in here?' in constant time.
A set holds unique, hashable items with no order: tags = {"python", "web"}. Like dict keys, membership tests x in s are O(1) on average, which makes sets the right choice whenever you check "have I seen this?" inside a loop. Note that {} is an empty dict; an empty set is written set().
set(iterable) deduplicates anything: set([1, 2, 2, 3]) == {1, 2, 3}. If you need to keep the original order while removing duplicates, use list(dict.fromkeys(items)) instead, since dicts are ordered.
Modify a set with add(x), discard(x) (no error if absent) and remove(x) (raises KeyError if absent).
Set algebra reads like maths and has operator and method forms:
a | bunion,a & bintersection,a - bdifference,a ^ bsymmetric differencea <= bsubset,a >= bsuperset,a.isdisjoint(b)
The method forms (a.union(b), a.intersection(b)) accept any iterable, not just sets.
A frozenset is the immutable version. Because it's hashable, it can be a dict key or live inside another set, which a regular set can't.
Iteration order of a set is arbitrary and can differ between runs for strings (hash randomisation), so sort the set if you need stable output.
Example
seen = set()
for user in ["ana", "ben", "ana", "cy", "ben"]:
if user in seen:
print("duplicate:", user)
seen.add(user)
print(sorted(seen)) # ['ana', 'ben', 'cy']
backend = {"python", "sql", "docker"}
frontend = {"js", "css", "docker"}
print(backend & frontend) # {'docker'}
print(sorted(backend | frontend))
print(backend - frontend) # {'python', 'sql'} (order may vary)
print(list(dict.fromkeys([3, 1, 3, 2, 1]))) # [3, 1, 2]
print(type({}), type(set())) # <class 'dict'> <class 'set'>Remember
- Sets store unique hashable items with O(1) membership tests
{}is a dict; useset()for an empty set- Use
| & - ^for union, intersection, difference, symmetric difference
Watch out for
- Relying on set iteration order.