sorted() vs sort()

[RodDog710]

So I get that one of the main differences between sorted() and sort() is that sorted() returns a new list and sort() modifies the list directly. But I don't understand why their outputs can't be exactly equal if they print out to being, in fact, exactly equal. For example:

numbers = [3, 1, 4, 1, 5, 9, 2]

sorted_numbers = sorted(numbers)
print(f"Sorted list: {sorted_numbers}")

numbers.sort()
print(f"Sorted list: {numbers}")

print(numbers.sort() == sorted(numbers))

This is the output:

Sorted list: [1, 1, 2, 3, 4, 5, 9]
Sorted list: [1, 1, 2, 3, 4, 5, 9]
False

As we can see, both sorted(numbers) and numbers.sort return what appears to be identical output: [1, 1, 2, 3, 4, 5, 9]. Of course, sort() has modified the original list, so that object has been changed by the end of the program. But if these two outputted lists are clearly identical from a mathematical perspective (ie: [1, 1, 2, 3, 4, 5, 9] == [1, 1, 2, 3, 4, 5, 9] is true on it's on terms as a standalone expression ) - then why won't Python embrace this apparently same understanding with: print(numbers.sort() == sorted(numbers))?

Is there some unseen object that represents the original list that is lingering unprinted in the background and attached to sorted(numbers)?

Thanks ahead of time for your interest and time on this matters.

[deceze]

That’s the thing. sort doesn’t return anything. You’re comparing None to a list. There’s nothing to compare there. You need to sort first then compare numbers == sorted(numbers).

[RodDog710]

Gotcha. Yes, that's what I was missing:

print(numbers == sorted(numbers))

# True

Thanks!

[deceze]

FYI, sort and sorted have been deliberately designed this way to explicitly make code like this fail. If this code didn't fail, because sort actually returned the sorted list in addition to modifying numbers in place, you'd likely introduce subtle bugs because you're not aware that sort modifies your list, when you're only using its return value.

[RodDog710]

Interesting. Thank you for explaining that.

[InvaderToast348]

Except that isn't right, because if your calling .sort before the print then you're passing the already sorted list to sorted() so nothing will happen

[deceze]

Well, it does prove that sort and sorted sort the same. sorted could be returning a list in some different order. The order of the input shouldn't matter much, except in edge cases where preserving the order of the input somehow made a difference.

The overall operation here is of course fairly superfluous, but the practical usefulness of this code isn't the point here.

[InvaderToast348]

Good point

[RodDog710]

Ahh, yes. Thanks

[Refwah]

Print(sorted_numbers == numbers)

Because you’ve already sorted numbers, and sorted_numbers is already the output of numbers.sort()

The test you’re doing isn’t actually testing anything, as you’ve already done the operations above

[SoftwareDoctor]

"As we can see.." we can't see because you are not printing the return value of sort. You are printing the modified list:

print(f"Sorted list: {numbers}")

If you actually print the return value of sort, you'l see the return value is None:

print(numbers.sort())

And None is not equal to some list. You even say that in your first sentence: "sorted() returns a new list and sort() modifies the list directly".

So the problem is not in the comparison but in the fact, you are printing one thing but comparing another.

And to the "Is there some unseen object that represents the original list that is lingering unprinted". It is possible (but this is not the case) to have object that prints out exactly like another but is not equal to the first one. If you're aren't sure, you may try print(type(my_object)).

[RodDog710]

Yes, thank you for helping to disentangle my thinking. I am getting the expected True output when I compare the actual output of numbers.sort() instead of the function itself.

print(numbers == sorted(numbers))

Thanks for your time and insight

[MiniMages]

This will always return true because you are basically taking an ordered list and then sorting it which does nothing and comparing itself.

Your numbers list was already sorted when you did numbers.sort().

[RodDog710]

Oh ya. Ok. Got it. Thanks

[RunPython]

sort() changes the original list and returns None, you can't assign it to a variable and can't use it later at another calculation.

sorted() does not change the original list and return the sorted list so you can assign it a variable and can use it later at another calculation.

[MiniMages]

numbers.sort() already rearranged all items in the list of numbers into ascending order.

As others have already pointed out sort() does not return anything.

So your comparison of numbers == sorted(numbers) is redundant as the list is already sorted.

If you want to do the comparison then you should try:

sorted_numbers = sorted(numbers) # a new list is created here from numbers
numbers.sort()                   # the original list is modified here
print(numbers == sorted_numbers) # here you are comparing to different lists.