Python Set
Syntax[edit  edit source]
 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[edit  edit source]
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
.
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.
Operations on sets[edit  edit source]
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 inplace versions:
method  inplace operation  inplace method 

union  = t  update 
intersection  s &= t  intersection_update 
difference  s = t  difference_update 
symmetric_difference  s ^= t  symmetric_difference_update 
For example:
s = {1, 2}
s.update({3, 4}) # s == {1, 2, 3, 4}
Get the unique elements of a list[edit  edit source]
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 dict
s.
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.
Set of Sets[edit  edit source]
{{1,2}, {3,4}}
leads to:
TypeError: unhashable type: 'set'
Instead, use frozenset
:
{frozenset({1, 2}), frozenset({3, 4})}
Set Operations using Methods and Builtins[edit  edit source]
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 emptydict
. The correct way to create an empty set isset()
.
Intersection[edit  edit source]
a.intersection(b)
returns a new set with elements present in both a
and b
>>> a.intersection(b)
{3, 4}
Union[edit  edit source]
a.union(b)
returns a new set with elements present in either a
and b
>>> a.union(b)
{1, 2, 3, 4, 5}
Difference[edit  edit source]
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[edit  edit source]
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[edit  edit source]
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:
Method  Operator 

a.intersection(b)

a & b

a.union(b)

b

a.difference(b)

a  b

a.symmetric_difference(b)

a ^ b

a.issubset(b)

a <= b

a.issuperset(b)

a >= b

Disjoint sets[edit  edit source]
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[edit  edit source]
The builtin in
keyword searches for occurances
>>> 1 in a
True
>>> 6 in a
False
Length[edit  edit source]
The builtin len()
function returns the number of elements in the set
>>> len(a)
4
>>> len(b)
3
Sets versus multisets[edit  edit source]
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):
Python 2.x2.7
>>> 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.