Sets in Python
A Set is an unordered collection data type that is iterable, mutable and has no duplicate elements. Python’s set class represents the mathematical notion of a set. The major advantage of using a set, as opposed to a list, is that it has a highly optimized method for checking whether a specific element is contained in the set. This is based on a data structure known as a hash table. Since sets are unordered, we cannot access items using indexes like we do in lists.
Python3
# Python program to# demonstrate sets# Same as {"a", "b", "c"}myset = set(["a", "b", "c"])print(myset)# Adding element to the setmyset.add("d")print(myset) |
{'c', 'b', 'a'}
{'d', 'c', 'b', 'a'}
Frozen Sets
Frozen sets in Python are immutable objects that only support methods and operators that produce a result without affecting the frozen set or sets to which they are applied. While elements of a set can be modified at any time, elements of the frozen set remain the same after creation.
If no parameters are passed, it returns an empty frozenset.
Python
# Python program to demonstrate differences# between normal and frozen set# Same as {"a", "b","c"}normal_set = set(["a", "b","c"])print("Normal Set")print(normal_set)# A frozen setfrozen_set = frozenset(["e", "f", "g"])print("\nFrozen Set")print(frozen_set)# Uncommenting below line would cause error as# we are trying to add element to a frozen set# frozen_set.add("h") |
Normal Set set(['a', 'c', 'b']) Frozen Set frozenset(['e', 'g', 'f'])
Internal working of Set
This is based on a data structure known as a hash table.
If Multiple values are present at the same index position, then the value is appended to that index position, to form a Linked List. In, Python Sets are implemented using dictionary with dummy variables, where key beings the members set with greater optimizations to the time complexity.
Set Implementation:-

Sets with Numerous operations on a single HashTable:-

Methods for Sets
Adding elements
Insertion in set is done through set.add() function, where an appropriate record value is created to store in the hash table. Same as checking for an item, i.e., O(1) on average. However, in worst case it can become O(n).
Python3
# A Python program to# demonstrate adding elements# in a set# Creating a Setpeople = {"Jay", "Idrish", "Archi"}print("People:", end = " ")print(people)# This will add Daxit# in the setpeople.add("Daxit")# Adding elements to the# set using iteratorfor i in range(1, 6): people.add(i)print("\nSet after adding element:", end = " ")print(people) |
People: {'Idrish', 'Archi', 'Jay'}
Set after adding element: {1, 2, 3, 4, 5, 'Idrish', 'Archi', 'Jay', 'Daxit'}
Union
Two sets can be merged using union() function or | operator. Both Hash Table values are accessed and traversed with merge operation perform on them to combine the elements, at the same time duplicates are removed. Time Complexity of this is O(len(s1) + len(s2)) where s1 and s2 are two sets whose union needs to be done.
Python3
# Python Program to# demonstrate union of# two setspeople = {"Jay", "Idrish", "Archil"}vampires = {"Karan", "Arjun"}dracula = {"Deepanshu", "Raju"}# Union using union()# functionpopulation = people.union(vampires)print("Union using union() function")print(population)# Union using "|"# operatorpopulation = people|draculaprint("\nUnion using '|' operator")print(population) |
Union using union() function
{'Karan', 'Idrish', 'Jay', 'Arjun', 'Archil'}
Union using '|' operator
{'Deepanshu', 'Idrish', 'Jay', 'Raju', 'Archil'}
Intersection
This can be done through intersection() or & operator. Common Elements are selected. They are similar to iteration over the Hash lists and combining the same values on both the Table. Time Complexity of this is O(min(len(s1), len(s2)) where s1 and s2 are two sets whose union needs to be done.
Python3
# Python program to# demonstrate intersection# of two setsset1 = set()set2 = set()for i in range(5): set1.add(i)for i in range(3,9): set2.add(i)# Intersection using# intersection() functionset3 = set1.intersection(set2)print("Intersection using intersection() function")print(set3)# Intersection using# "&" operatorset3 = set1 & set2print("\nIntersection using '&' operator")print(set3) |
Intersection using intersection() function
{3, 4}
Intersection using '&' operator
{3, 4}
Difference
To find difference in between sets. Similar to find difference in linked list. This is done through difference() or – operator. Time complexity of finding difference s1 – s2 is O(len(s1))
Python3
# Python program to# demonstrate difference# of two setsset1 = set()set2 = set()for i in range(5): set1.add(i)for i in range(3,9): set2.add(i)# Difference of two sets# using difference() functionset3 = set1.difference(set2)print(" Difference of two sets using difference() function")print(set3)# Difference of two sets# using '-' operatorset3 = set1 - set2print("\nDifference of two sets using '-' operator")print(set3) |
Difference of two sets using difference() function
{0, 1, 2}
Difference of two sets using '-' operator
{0, 1, 2}
Clearing sets
Clear() method empties the whole set.
Python3
# Python program to# demonstrate clearing# of setset1 = {1,2,3,4,5,6}print("Initial set")print(set1)# This method will remove# all the elements of the setset1.clear()print("\nSet after using clear() function")print(set1) |
Initial set
{1, 2, 3, 4, 5, 6}
Set after using clear() function
set()
However, there are two major pitfalls in Python sets:
- The set doesn’t maintain elements in any particular order.
- Only instances of immutable types can be added to a Python set.
Time complexity of Sets
Operation Average case Worst Case notes x in s O(1) O(n) Union s|t O(len(s)+len(t)) Intersection s&t; O(min(len(s), len(t)) O(len(s) * len(t)) replace “min” with “max” if t is not a set Multiple intersection s1&s2;&..&sn; (n-1)*O(l) where l is max(len(s1),..,len(sn)) Difference s-t O(len(s))
Operators for Sets
Sets and frozen sets support the following operators:
Operators Notes key in s containment check key not in s non-containment check s1 == s2 s1 is equivalent to s2 s1 != s2 s1 is not equivalent to s2 s1 <= s2 s1 is subset of s2 s1 < s2 s1 is proper subset of s2 s1 >= s2 s1 is superset of s2 s1 > s2 s1 is proper superset of s2 s1 | s2 the union of s1 and s2 s1 & s2 the intersection of s1 and s2 s1 – s2 the set of elements in s1 but not s2 s1 ˆ s2 the set of elements in precisely one of s1 or s2
Code Snippet to illustrate all Set operations in Python
Python
# Python program to demonstrate working# of# Set in Python# Creating two setsset1 = set()set2 = set()# Adding elements to set1for i in range(1, 6): set1.add(i)# Adding elements to set2for i in range(3, 8): set2.add(i)print("Set1 = ", set1)print("Set2 = ", set2)print("\n")# Union of set1 and set2set3 = set1 | set2# set1.union(set2)print("Union of Set1 & Set2: Set3 = ", set3)# Intersection of set1 and set2set4 = set1 & set2# set1.intersection(set2)print("Intersection of Set1 & Set2: Set4 = ", set4)print("\n")# Checking relation between set3 and set4if set3 > set4: # set3.issuperset(set4) print("Set3 is superset of Set4")else if set3 < set4: # set3.issubset(set4) print("Set3 is subset of Set4")else : # set3 == set4 print("Set3 is same as Set4")# displaying relation between set4 and set3if set4 < set3: # set4.issubset(set3) print("Set4 is subset of Set3") print("\n")# difference between set3 and set4set5 = set3 - set4print("Elements in Set3 and not in Set4: Set5 = ", set5)print("\n")# check if set4 and set5 are disjoint setsif set4.isdisjoint(set5): print("Set4 and Set5 have nothing in common\n")# Removing all the values of set5set5.clear()print("After applying clear on sets Set5: ")print("Set5 = ", set5) |
('Set1 = ', set([1, 2, 3, 4, 5]))
('Set2 = ', set([3, 4, 5, 6, 7]))
('Union of Set1 & Set2: Set3 = ', set([1, 2, 3, 4, 5, 6, 7]))
('Intersection of Set1 & Set2: Set4 = ', set([3, 4, 5]))
Set3 is superset of Set4
Set4 is subset of Set3
('Elements in Set3 and not in Set4: Set5 = ', set([1, 2, 6, 7]))
Set4 and Set5 have nothing in common
After applying clear on sets Set5:
('Set5 = ', set([]))
Recent articles on Python Set.
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