with other sets

# Intersection    
{1, 2, 3, 4, 5}.intersection({3, 4, 5, 6})  # {3, 4, 5}
{1, 2, 3, 4, 5} & {3, 4, 5, 6}              # {3, 4, 5}

# Union
{1, 2, 3, 4, 5}.union({3, 4, 5, 6})  # {1, 2, 3, 4, 5, 6}
{1, 2, 3, 4, 5} | {3, 4, 5, 6}       # {1, 2, 3, 4, 5, 6}

# Difference
{1, 2, 3, 4}.difference({2, 3, 5})  # {1, 4}
{1, 2, 3, 4} - {2, 3, 5}            # {1, 4}

# Symmetric difference with
{1, 2, 3, 4}.symmetric_difference({2, 3, 5})  # {1, 4, 5}
{1, 2, 3, 4} ^ {2, 3, 5}                      # {1, 4, 5}

# Superset check
{1, 2}.issuperset({1, 2, 3})  # False
{1, 2} >= {1, 2, 3}           # False

# Subset check
{1, 2}.issubset({1, 2, 3})  # True
{1, 2} <= {1, 2, 3}         # True

# Disjoint check
{1, 2}.isdisjoint({3, 4})  # True
{1, 2}.isdisjoint({1, 4})  # False

with single elements

# Existence check
2 in {1,2,3}      # True
4 in {1,2,3}      # False
4 not in {1,2,3}  # True

# Add and Remove
s = {1,2,3}
s.add(4)        # s == {1,2,3,4}

s.discard(3)    # s == {1,2,4}
s.discard(5)    # s == {1,2,4}

s.remove(2)     # s == {1,4}
s.remove(2)     # KeyError!

Set operations return new sets, but have the corresponding in-place versions:

For example:

s = {1, 2}
s.update({3, 4})   # s == {1, 2, 3, 4}

Let's say you've got a list of restaurants -- maybe you read it from a file. You care about the unique restaurants in the list. The best way to get the unique elements from a list is to turn it into a set:

restaurants = ["McDonald's", "Burger King", "McDonald's", "Chicken Chicken"]
unique_restaurants = set(restaurants)
print(unique_restaurants)
# prints {'Chicken Chicken', "McDonald's", 'Burger King'}

Note that the set is not in the same order as the original list; that is because sets are unordered, just like dicts.

This can easily be transformed back into a List with Python's built in list function, giving another list that is the same list as the original but without duplicates:

list(unique_restaurants)
# ['Chicken Chicken', "McDonald's", 'Burger King']

It's also common to see this as one line:

# Removes all duplicates and returns another list
list(set(restaurants))

Now any operations that could be performed on the original list can be done again.

{{1,2}, {3,4}}

leads to:

TypeError: unhashable type: 'set'

Instead, use frozenset:

{frozenset({1, 2}), frozenset({3, 4})}

We define two sets a and b

>>> a = {1, 2, 2, 3, 4}
>>> b = {3, 3, 4, 4, 5}

NOTE: {1} creates a set of one element, but {} creates an empty dict. The correct way to create an empty set is set().

Intersection

a.intersection(b) returns a new set with elements present in both a and b

>>> a.intersection(b)
{3, 4}

Union

a.union(b) returns a new set with elements present in either a and b

>>> a.union(b)
{1, 2, 3, 4, 5}

Difference

a.difference(b) returns a new set with elements present in a but not in b

>>> a.difference(b)
{1, 2}
>>> b.difference(a)
{5}

Symmetric Difference

a.symmetric_difference(b) returns a new set with elements present in either a or b but not in both

>>> a.symmetric_difference(b)
{1, 2, 5}
>>> b.symmetric_difference(a)
{1, 2, 5}

NOTE: a.symmetric_difference(b) == b.symmetric_difference(a)

Subset and superset

c.issubset(a) tests whether each element of c is in a.

a.issuperset(c) tests whether each element of c is in a.

>>> c = {1, 2}
>>> c.issubset(a)
True
>>> a.issuperset(c)
True

The latter operations have equivalent operators as shown below:

Disjoint sets

Sets a and d are disjoint if no element in a is also in d and vice versa.

>>> d = {5, 6}
>>> a.isdisjoint(b) # {2, 3, 4} are in both sets
False
>>> a.isdisjoint(d)
True

# This is an equivalent check, but less efficient
>>> len(a & d) == 0
True

# This is even less efficient
>>> a & d == set()
True

Testing membership

The builtin in keyword searches for occurances

>>> 1 in a
True
>>> 6 in a
False

Length

The builtin len() function returns the number of elements in the set

>>> len(a)
4
>>> len(b)
3

Sets are unordered collections of distinct elements. But sometimes we want to work with unordered collections of elements that are not necessarily distinct and keep track of the elements' multiplicities.

Consider this example:

>>> setA = {'a','b','b','c'}
>>> setA
set(['a', 'c', 'b'])

By saving the strings 'a', 'b', 'b', 'c' into a set data structure we've lost the information on the fact that 'b' occurs twice. Of course saving the elements to a list would retain this information

>>> listA = ['a','b','b','c']
>>> listA
['a', 'b', 'b', 'c']

but a list data structure introduces an extra unneeded ordering that will slow down our computations.

For implementing multisets Python provides the Counter class from the collections module (starting from version 2.7):

>>> from collections import Counter
>>> counterA = Counter(['a','b','b','c'])
>>> counterA
Counter({'b': 2, 'a': 1, 'c': 1})

Counter is a dictionary where where elements are stored as dictionary keys and their counts are stored as dictionary values. And as all dictionaries, it is an unordered collection.

Syntax

• empty_set = set() # initialize an empty set
• literal_set = {'foo', 'bar', 'baz'} # construct a set with 3 strings inside it
• set_from_list = set(['foo', 'bar', 'baz']) # call the set function for a new set
• set_from_iter = set(x for x in range(30)) # use arbitrary iterables to create a set
• set_from_iter = {x for x in [random.randint(0,10) for i in range(10)]} # alternative notation

Remarks

Sets are unordered and have very fast lookup time (amortized O(1) if you want to get technical). It is great to use when you have a collection of things, the order doesn't matter, and you'll be looking up items by name a lot. If it makes more sense to look up items by an index number, consider using a list instead. If order matters, consider a list as well.

Sets are mutable and thus cannot be hashed, so you cannot use them as dictionary keys or put them in other sets, or anywhere else that requires hashable types. In such cases, you can use an immutable frozenset (opens new window).

The elements of a set must be hashable. This means that they have a correct __hash__ method, that is consistent with __eq__. In general, mutable types such as list or set are not hashable and cannot be put in a set. If you encounter this problem, consider using dict and immutable keys.