Mark Janssen wrote:
I just thought of something: isn't the obvious solution is for doctest to test the type of an expression's output and if it's in the set of unordered types {set, dict} to run sort() on it? Then the docstring author can just put the (sorted) output of what's expected....
{set, dict} is not the set of unordered types. The set of unordered types is without bound: anyone can create their own unordered types. Even if you limit yourself to the builtins, you forgot frozenset. And then there are non-builtins in the standard library, like OrderedDict, other types like dict_proxy. Sorting the output of an OrderedDict is the wrong thing to do, because the order is significant. So doctest would need to not just recognise mappings and sets, and sort them, but *also* recognise mappings and sets which should *not* be sorted. Remember too, that by the time doctest.OutputChecker sees the output, it only sees it as a string. I don't know how much work it would take to introduce actual type-checks into doctest, but I expect it would be a lot. And one last problem for you to consider: what happens if the output is unsortable? Try this dict for size: { 2+1j: None, 2-1j: None, float('NAN'): None}
Perhaps I'm hashing a dead horse, but I really would like to see this added to the issue's tracker as a requested feature. I may code up the patch myself, but it helps my brain to have it "on the dev stack".
Feel free to add it. For what it's worth, I am a very strong -1 on any suggestion to give doctest "Do What I Mean" powers when it comes to unordered objects. But I would support a "Do What I Say" doctest directive, like NORMALIZE_WHITESPACE, ELLIPSIS, IGNORE_EXCEPTION_DETAIL, e.g. a directive that tells doctest to split both the expected and actual output strings on whitespace, then lexicographically sort them before comparing. This approach doesn't try to be too clever: it's a dumb, understandable test which should fit in nicely with the other tests in doctest.OutputChecker.check_output, perhaps something like this: if optionflags & IGNORE_WORD_ORDER: if sorted(got.split()) == sorted(want.split()): return True It won't solve every doctest ordering problem, but doctest has other heuristics which can be fooled too. It is nice and simple, it solves the first 90% of the problem, and it is under the control of the coder. If you feel like submitting a patch, feel free to use my idea. -- Steven