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Python Sets: Unique Values, Membership and Set Operations

Learn Python sets: store unique values, remove duplicates from a list, add and remove items, fast membership tests, and set operations (union, intersection, difference, symmetric difference).

Beginner5 min readUpdated
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A set is an unordered collection of unique values. Adding something that is already there does nothing. Sets are the fastest way to remove duplicates and to check whether a value is present, and they support the union and intersection operations you may remember from maths.

Creating a set#

Use curly braces, or set() to convert another collection:

Python
tags = {"python", "sql", "python", "web"}
print(len(tags), "sql" in tags)
Output
3 True

The duplicate "python" was stored only once.

Removing duplicates#

The most common use of sets:

Python
emails = ["a@x.com", "b@x.com", "a@x.com", "c@x.com", "b@x.com"]
unique = set(emails)
print(len(emails), len(unique))
print(sorted(unique))
Output
5 3
['a@x.com', 'b@x.com', 'c@x.com']

Sets have no order, so printing one directly can show items in any order. sorted() returns them as an ordered list.

If you need to keep the original order while removing duplicates, use a dictionary instead:

Python
nums = [3, 1, 3, 2, 1]
print(list(dict.fromkeys(nums)))
Output
[3, 1, 2]

Adding and removing#

Python
s = set()
s.add("apple")
s.add("banana")
s.add("apple")       # already there: no change
s.discard("cherry")  # not there: no error
s.remove("banana")
print(s)
Output
{'apple'}

remove() raises an error if the item is missing; discard() does not:

Python
s = {"apple"}
s.remove("cherry")
Output
KeyError: 'cherry'

Sets have no indexes, so s[0] is an error. Loop over a set, or convert it to a list.

Set operations#

Who knows Python, SQL, or both?

Python
python_devs = {"Asha", "Ravi", "Meera"}
sql_devs = {"Ravi", "Meera", "John"}

print(sorted(python_devs | sql_devs))   # union: in either
print(sorted(python_devs & sql_devs))   # intersection: in both
print(sorted(python_devs - sql_devs))   # difference: only in the first
print(sorted(python_devs ^ sql_devs))   # symmetric difference: in exactly one
Output
['Asha', 'John', 'Meera', 'Ravi']
['Meera', 'Ravi']
['Asha']
['Asha', 'John']
OperatorMethodResult
a | ba.union(b)everything in a or b
a & ba.intersection(b)only what is in both
a - ba.difference(b)in a but not b
a ^ ba.symmetric_difference(b)in one but not both

Subset and overlap checks:

Python
print({1, 2} <= {1, 2, 3}, {1, 2, 3} >= {3}, {1, 2}.isdisjoint({3, 4}))
Output
True True True

Fast membership tests#

Checking x in some_set is very fast, even with millions of items, because sets use hashing. Checking x in some_list has to look through the list item by item. If you test membership many times, convert to a set first.

Only immutable items#

Set items must be hashable (immutable): numbers, strings and tuples work; lists and dictionaries do not:

Python
s = {[1, 2]}
Output
TypeError: unhashable type: 'list'

Common mistakes#

Practice#

Practice exercise

Given words = ["apple", "banana", "apple", "cherry", "banana", "apple"], find the words that appear more than once, printed as a sorted list.

Show hint

Keep two sets: seen and dupes. If a word is already in seen, add it to dupes.

Show solution
Python
words = ["apple", "banana", "apple", "cherry", "banana", "apple"]
seen = set()
dupes = set()

for w in words:
    if w in seen:
        dupes.add(w)
    seen.add(w)

print(sorted(dupes))
Output
['apple', 'banana']

Using a set for dupes means "apple" is recorded once even though it repeats twice.

Key takeaway#

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