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().
a.intersection(b) returns a new set with elements present in both a and b
>>> a.intersection(b)
{3, 4}
a.union(b) returns a new set with elements present in either a and b
>>> a.union(b)
{1, 2, 3, 4, 5}
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}
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)
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) | a|b | 
a.difference(b) | a - b | 
a.symmetric_difference(b) | a ^ b | 
a.issubset(b) | a <= b | 
a.issuperset(b) | a >= b | 
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
The builtin in keyword searches for occurances
>>> 1 in a
True
>>> 6 in a
False
The builtin len() function returns the number of elements in the set
>>> len(a)
4
>>> len(b)
3