Python sorted() Function (original) (raw)

Last Updated : 08 Mar, 2025

sorted() function returns a new sorted list from the elements of any iterable like (e.g., list, tuples, strings ). It creates and returns a new sorted list and leaves the original iterable unchanged.

Let’s start with a basic example of sorting a list of numbers using the sorted() function.

Python `

a = [4, 1, 3, 2]

Using sorted() to create a new sorted list without modifying the original list

b = sorted(a) print(b)

`

Table of Content

Syntax of sorted() function

sorted(iterable, key=None, reverse=False)

**Parameters:

Return Type:

Sorting in ascending order

When no additional parameters are provided then It arranges the element in increasing order.

Python `

a = [5, 2, 9, 1, 3]

#Sorted() arranges the list in ascending order b = sorted(a) print(b)

`

Sorting in descending order

To sort an iterable in descending order, set the **reverse argument to **True.

Python `

a = [5, 2, 9, 1, 5, 6]

"reverse= True" helps sorted() to arrange the element

#from largest to smallest elements res = sorted(a, reverse=True) print(res)

`

Sorting using _key parameter

The **key parameter is an optional argument that allows us to customize the sort order.

Sorting Based on String Length:

Python `

a = ["apple", "banana", "cherry", "date"] res = sorted(a, key=len) print(res)

`

Output

['date', 'apple', 'banana', 'cherry']

**Explanation: The **key parameter is set to **len, which sorts the words by their length in ascending order.

Sorting a List of Dictionaries:

Python `

a = [ {"name": "Alice", "score": 85}, {"name": "Bob", "score": 91}, {"name": "Eve", "score": 78} ]

Use sorted() to sort the list 'a' based on the 'score' key

sorted() returns a new list with dictionaries sorted by the 'score'

b = sorted(a, key=lambda x: x['score']) print(b)

`

Output

[{'name': 'Eve', 'score': 78}, {'name': 'Alice', 'score': 85}, {'name': 'Bob', 'score': 91}]

**Explanation: key=lambda x: x[‘score’] specifies that the sorting should be done using the ‘score’ value from each dictionary

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