Is python's sorted() function guaranteed to be stable?

Question

The documentation doesn't guarantee that. Is there any other place that it is documented?

I'm guessing it might be stable since the sort method on lists is guaranteed to be stable (Notes 9th point: "Starting with Python 2.3, the sort() method is guaranteed to be stable"), and sorted is functionally similar. However, I'm not able to find any definitive source that says so.

Purpose: I need to sort based on a primary key and also a secondary key in cases where the primary key is equal in both records. If sorted() is guaranteed to be stable, I can sort on the secondary key, then sort on the primary key and get the result I need.

PS: To avoid any confusion, I'm using stable in the sense of "a sort is stable if it guarantees not to change the relative order of elements that compare equal".

Answer 1

Yes, the intention of the manual is indeed to guarantee that sorted is stable and indeed that it uses exactly the same algorithm as the sort method. I do realize that the docs aren't 100% clear about this identity; doc patches are always happily accepted!

Answer 2

They are stable.

By the way: you sometimes can ignore knowing whether sort and sorted are stable, by combining a multi-pass sort in a single-pass one.

For example, if you want to sort objects based on their last_name, first_name attributes, you can do it in one pass:

sorted_list= sorted(     your_sequence_of_items,     key= lambda item: (item.last_name, item.first_name)) 

taking advantage of tuple comparison.

This answer, as-is, covers the original question. For further sorting-related questions, there is the Python Sorting How-To.