Why doesn’t list.sort() return the sorted list in Python?

Example

In this example, let’s first see the usage of list.sort() before moving further. Here, we have created a List and sorted in Ascending order using the sort() method −

# Creating a List
myList = ["Jacob", "Harry", "Mark", "Anthony"]

# Displaying the List
print("List = ",myList)

# Sort the Lists in Ascending Order
myList.sort()

# Display the sorted List
print("Sort (Ascending Order) = ",myList)

Output

List =  ['Jacob', 'Harry', 'Mark', 'Anthony']
Sort (Ascending Order) =  ['Anthony', 'Harry', 'Jacob', 'Mark']

Where performance matters more, making a copy of the list just to sort it wouldn’t be considered good and is wasteful. Therefore, list.sort() sorts the list in place. This method does not return the sorted list. This way, you won’t be fooled into accidentally overwriting a list when you need a sorted copy but also need to keep the unsorted version around.

Return a new list using the built-in sorted() function instead. This function creates a new list from a provided iterable, sorts it and returns it.

Using sorted() to Sort a List of Dictionaries using values

Example

We have now used the sorted() method to sort a list of dictionaries.

# List of dictionaries
d = [
   {"name" : "Sam", "marks" : 98},
   {"name" : "Tom", "marks" : 93},
   {"name" : "Jacob", "marks" : 97}
]

# Display the Dictionary
print("Dictionary = \n",d)

# Sorting using values with the lambda function
print("Sorted = \n",sorted(d, key = lambda item: item['marks']))

Output

('Dictionary = \n', [{'name': 'Sam', 'marks': 98}, {'name': 'Tom', 'marks': 93}, {'name': 'Jacob', 'marks': 97}])
('Sorted = \n', [{'name': 'Tom', 'marks': 93}, {'name': 'Jacob', 'marks': 97}, {'name': 'Sam', 'marks': 98}])