| title | Python Sets - Python Cheatsheet |
|---|---|
| description | Python comes equipped with several built-in data types to help us organize our data. These structures include lists, dictionaries, tuples and sets. |
Python comes equipped with several built-in data types to help us organize our data. These structures include lists, dictionaries, tuples and sets.
From the Python 3 documentation A set is an unordered collection with no duplicate elements. Basic uses include membership testing and eliminating duplicate entries.Read Python Sets: What, Why and How for a more in-deep reference.
There are two ways to create sets: using curly braces {} and the built-in function set()
{} or you will get an empty dictionary instead.
>>> s = {1, 2, 3}
>>> s = set([1, 2, 3])
>>> s = {} # this will create a dictionary instead of a set
>>> type(s)
# <class 'dict'>A set automatically remove all the duplicate values.
>>> s = {1, 2, 3, 2, 3, 4}
>>> s
# {1, 2, 3, 4}And as an unordered data type, they can't be indexed.
>>> s = {1, 2, 3}
>>> s[0]
# Traceback (most recent call last):
# File "<stdin>", line 1, in <module>
# TypeError: 'set' object does not support indexingUsing the add() method we can add a single element to the set.
>>> s = {1, 2, 3}
>>> s.add(4)
>>> s
# {1, 2, 3, 4}And with update(), multiple ones:
>>> s = {1, 2, 3}
>>> s.update([2, 3, 4, 5, 6])
>>> s
# {1, 2, 3, 4, 5, 6}Both methods will remove an element from the set, but remove() will raise a key error if the value doesn't exist.
>>> s = {1, 2, 3}
>>> s.remove(3)
>>> s
# {1, 2}
>>> s.remove(3)
# Traceback (most recent call last):
# File "<stdin>", line 1, in <module>
# KeyError: 3discard() won't raise any errors.
>>> s = {1, 2, 3}
>>> s.discard(3)
>>> s
# {1, 2}
>>> s.discard(3)union() or | will create a new set that with all the elements from the sets provided.
>>> s1 = {1, 2, 3}
>>> s2 = {3, 4, 5}
>>> s1.union(s2) # or 's1 | s2'
# {1, 2, 3, 4, 5}intersection or & will return a set with only the elements that are common to all of them.
>>> s1 = {1, 2, 3}
>>> s2 = {2, 3, 4}
>>> s3 = {3, 4, 5}
>>> s1.intersection(s2, s3) # or 's1 & s2 & s3'
# {3}difference or - will return only the elements that are unique to the first set (invoked set).
>>> s1 = {1, 2, 3}
>>> s2 = {2, 3, 4}
>>> s1.difference(s2) # or 's1 - s2'
# {1}
>>> s2.difference(s1) # or 's2 - s1'
# {4}symetric_difference or ^ will return all the elements that are not common between them.
>>> s1 = {1, 2, 3}
>>> s2 = {2, 3, 4}
>>> s1.symmetric_difference(s2) # or 's1 ^ s2'
# {1, 4}