What is the difference between sorted(list) vs list.sort()?

list.sort() sorts the list and replaces the original list, whereas sorted(list) returns a sorted copy of the list, without changing the original list.

• When is one preferred over the other?
• Which is more efficient? By how much?
• Can a list be reverted to the unsorted state after list.sort() has been performed?

_{Please use Why do these list operations (methods) return None, rather than the resulting list? to close questions where OP has inadvertently assigned the result of .sort(), rather than using sorted or a separate statement. Proper debugging would reveal that .sort() had returned None, at which point "why?" is the remaining question.}

sorted() returns a new sorted list, leaving the original list unaffected. list.sort() sorts the list in-place, mutating the list indices, and returns None (like all in-place operations).

sorted() works on any iterable, not just lists. Strings, tuples, dictionaries (you'll get the keys), generators, etc., returning a list containing all elements, sorted.

• Use list.sort() when you want to mutate the list, sorted() when you want a new sorted object back. Use sorted() when you want to sort something that is an iterable, not a list yet.

• For lists, list.sort() is faster than sorted() because it doesn't have to create a copy. For any other iterable, you have no choice.

• No, you cannot retrieve the original positions. Once you called list.sort() the original order is gone.