Sort a list of tuples by multiple elements in Python

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Borislav Hadzhiev

Last updated: Jun 29, 2022

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Sort a list of tuples by multiple elements in Python #

To sort a list of tuples by multiple elements in Python:

  1. Pass the list to the sorted() function.
  2. Use the key argument to select the elements at the specific indices in each tuple.
  3. The sorted() function will sort the list of tuples by the specified elements.
main.py
list_of_tuples = [(1, 3, 100), (2, 3, 50), (3, 2, 75)] # ✅ sort list of tuples by second and third elements sorted_list = sorted( list_of_tuples, key=lambda t: (t[1], t[2]) ) print(sorted_list) # 👉️ [(3, 2, 75), (2, 3, 50), (1, 3, 100)]

The sorted function takes an iterable and returns a new sorted list from the items in the iterable.

The function takes an optional key argument that can be used to sort by different criteria.

The key argument can be set to a function that determines the sorting criteria.

The example sorts the list of tuples by the elements at index 1 and 2 in each tuple.

Since the second item in the third tuple is the lowest, it gets moved to the front.

The second items in the first and second tuples are equal (both 3), so the third items get compared and since 50 is less than 100, it takes second place.

If you need to sort a list of tuples by multiple elements in descending order, set the reverse argument to True in the call to the sorted() function.

main.py
list_of_tuples = [(1, 3, 100), (3, 2, 75), (2, 3, 50)] sorted_list = sorted( list_of_tuples, key=lambda t: (t[1], t[2]), reverse=True ) print(sorted_list) # 👉️ [(1, 3, 100), (2, 3, 50), (3, 2, 75)]
If the reverse argument is set to True, then the elements are sorted as if each comparison were reversed.

2 of the tuples have 3 as the second element, and since we set the reverse argument to True, the tuple with the greater 3rd element gets moved to the front.

The last tuple in the list is the one with the lowest second element.

You can also use the operator.itemgetter() method to specify the indexes you want to sort by.

main.py
from operator import itemgetter list_of_tuples = [(1, 3, 100), (2, 3, 50), (3, 2, 75)] # ✅ sort list of tuples by second and third elements sorted_list = sorted( list_of_tuples, key=itemgetter(1, 2), ) print(sorted_list) # 👉️ [(3, 2, 75), (2, 3, 50), (1, 3, 100)]

The operator.itemgetter method returns a callable object that fetches the item at the specified index.

For example, x = itemgetter(1) and then calling x(my_tuple), returns my_tuple[1].

Using the itemgetter method is faster than using a lambda function, but it is also a bit more implicit.

Alternatively, you can use the list.sort() method to sort the list of tuples by multiple elements in place.

main.py
from operator import itemgetter list_of_tuples = [(1, 3, 100), (2, 3, 50), (3, 2, 75)] list_of_tuples.sort(key=itemgetter(1, 2)) print(list_of_tuples) # 👉️ [(3, 2, 75), (2, 3, 50), (1, 3, 100)]

The list.sort method sorts the list in place and it uses only < comparisons between items.

The method takes the following 2 keyword-only arguments:

NameDescription
keya function that takes 1 argument and is used to extract a comparison key from each list element
reversea boolean value indicating whether each comparison should be reversed

Note that the list.sort method mutates the list in place and returns None.

You can use the sorted() function if you need a new list instance rather than an in-place mutation.

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