Python Sets

Last Updated : 28 Mar, 2026

Python set is an unordered collection of multiple items having different datatypes. Sets are mutable, unindexed and do not contain duplicates. The order of elements in a set is not preserved and can change.

  • Can store None values.
  • Implemented using hash tables internally.
  • Do not implement interfaces like Serializable or Cloneable.
  • Sets are not inherently thread-safe; synchronization is needed if used across threads.

Creating a Set

Set is created using curly braces {} with comma-separated values. It automatically stores only unique elements and does not maintain any specific order.

Python 
s = {1, 2, 3, 4}
print(s)

Output
{1, 2, 3, 4}

set() function

Sets can also be created using built-in set() function with an iterable object or a sequence by placing the sequence inside curly braces, separated by a 'comma'.

Note: A Python set cannot contain mutable objects like lists or dictionaries as elements because they are unhashable. However, mutable objects like lists can be used as input to create a set using the set() function.

Python 
s = set()
print(s)

s = set("GeeksForGeeks")
print(s)

# Creating a Set with the use of a List
s = set(["GFG", "For", "Geeks"])
print(s)

# Creating a Set with the use of a tuple
t = ("GFG", "for", "Geeks")
print(set(t))

# Creating a Set with the use of a dictionary
d = {"GFG": 1, "for": 2, "Geeks": 3}
print(set(d))

Output
set()
{'e', 'o', 'r', 'F', 'G', 'k', 's'}
{'Geeks', 'GFG', 'For'}
{'for', 'GFG', 'Geeks'}
{'for', 'GFG', 'Geeks'}

Unordered, Unindexed and Mutability

In set, the order of elements is not guaranteed to be the same as the order in which they were added. The output could vary each time we run the program. Also the duplicate items entered are removed by itself.

Sets do not support indexing. Trying to access an element by index (set[0]) raises a TypeError.

Python 
s = {3, 1, 4, 1, 5, 9, 2}

print(s) 
try:
    print(s[0])
except TypeError as e:
    print(e)

Output
{1, 2, 3, 4, 5, 9}
'set' object is not subscriptable

Adding Elements to a Set

We can add items to a set using add() method and update() method. add() method can be used to add only a single item. To add multiple items we use update() method.

Python 
s = {1, 2, 3}
s.add(4)
s.update([5, 6])
print(s)

Output
{1, 2, 3, 4, 5, 6}

Accessing a Set

Because sets are unordered and unindexed, you cannot access elements using a specific index like set[0]. Instead, you must use a loop to iterate through the items or the in keyword to check for an item's existence.

Python 
s = {"Geeks", "For", "Geeks"}

for i in s:
    print(i, end=" ")

print("\n", "Geeks" in s)

Output
For Geeks 
 True

Explanation:

  • for loop accesses each item in the set. Note that because sets are unordered, the items may appear in a different sequence every time the code is executed.
  • Notice that although "Geeks" was added twice in the code, it only appears once in the output because sets automatically remove duplicates.
  • "in" keyword efficiently checks if the string "Geeks" is present in s, returning True if found and False otherwise.

Removing Elements from the Set

1. Using remove() Method or discard() Method: remove() method removes a specified element from the set. If the element is not present in the set, it raises a KeyError.

discard() method also removes a specified element from the set. Unlike remove(), if the element is not found, it does not raise an error.

Python 
s = {1, 2, 3, 4, 5}
s.remove(3)
print(s)  

try:
    s.remove(10)
except KeyError as e:
    print("Error:", e)  

s.discard(4)
print(s)  

s.discard(10)  
print(s)

Output
{1, 2, 4, 5}
Error: 10
{1, 2, 5}
{1, 2, 5}

2. Using pop() Method: pop() method removes and returns an arbitrary element from the set. This means we don't know which element will be removed. If the set is empty, it raises a KeyError.

Note: If the set is unordered then there's no such way to determine which element is popped by using the pop() function. 

Python 
s = {1, 2, 3, 4, 5}
val = s.pop()
print(val)
print(s)

s.clear()  
try:
    s.pop()
except KeyError as e:
    print("Error:", e)

Output
1
{2, 3, 4, 5}
Error: 'pop from an empty set'

3. Using clear() Method: clear() method removes all elements from the set, leaving it empty.

Python 
s = {1, 2, 3, 4, 5}
s.clear()
print(s)

Output
set()

Frozen Sets

A frozenset in Python is a built-in data type that is similar to a set but with one key difference that is immutability. This means that once a frozenset is created, we cannot modify its elements that is we cannot add, remove or change any items in it.

Like regular sets, a frozenset cannot contain duplicate elements. If no parameters are passed, it returns an empty frozenset.  

Python 
fs = frozenset([1, 2, 3, 4, 5])
print(fs)  

s = {3, 1, 4, 1, 5}
fs = frozenset(s)
print(fs)

Output
frozenset({1, 2, 3, 4, 5})
frozenset({1, 3, 4, 5})

Typecasting Objects into Sets

Typecasting objects refers to converting various data types into a set. Python provides the set() constructor to perform this typecasting, allowing us to convert lists, tuples and strings into sets.

Python 
li = [1, 2, 3, 3, 4, 5, 5, 6, 2]
s = set(li)
print(s)

s = "GeeksforGeeks"
s = set(s)
print(s)

d = {1: "One", 2: "Two", 3: "Three"}
s = set(d)
print(s)

Output
{1, 2, 3, 4, 5, 6}
{'G', 'k', 'e', 's', 'o', 'f', 'r'}
{1, 2, 3}

Recommended Problems: Set Operations, Set II

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