On 12/16/19 6:30 PM, Tim Peters wrote:

If it's desired that "insertion order" be consistent across runs, platforms, and releases, then what "insertion order" _means_ needs to be rigorously defined & specified for all set operations. This was comparatively trivial for dicts, because there are, e.g., no commutative binary operators defined on dicts.

My intuition is that figuring out sensible semantics here is maybe not trivial, but hardly impossible. If this proposal goes anywhere I'd be willing to contribute to figuring it out.

If you want to insist that `a | b` first list all the elements of a, and then all the elements of b that weren't already in a, regardless of cost, then you create another kind of unintuitive surprise: in general the result of "a | b" will display differently than the result of "b | a" (although the results will compare equal), and despite that the _user_ didn't "insert" anything.

Call me weird--and I won't disagree--but I find nothing unintuitive about that. After all, that's already the world we live in: there are any number of sets that compare as equal but display differently. In current Python:

>>> a = {'a', 'b', 'c'} >>> d = {'d', 'e', 'f'} >>> a | d {'f', 'e', 'd', 'a', 'b', 'c'} >>> d | a {'f', 'b', 'd', 'a', 'e', 'c'} >>> a | d == d | a True

This is also true for dicts, in current Python, which of course do maintain insertion order. Dicts don't have the | operator, so I approximate the operation by duplicating the dict (which AFAIK preserves insertion order) and using update.

>>> aa = {'a': 1, 'b': 1, 'c': 1} >>> dd = {'d': 1, 'e': 1, 'f': 1} >>> x = dict(aa) >>> x.update(dd) >>> y = dict(dd) >>> y.update(aa) >>> x {'a': 1, 'b': 1, 'c': 1, 'd': 1, 'e': 1, 'f': 1} >>> y {'d': 1, 'e': 1, 'f': 1, 'a': 1, 'b': 1, 'c': 1} >>> x == y True

Since dicts already behave in exactly that way, I don't think it would be too surprising if sets behaved that way too. In fact, I think it's a little surprising that they behave differently, which I suppose was my whole thesis from the beginning.

Cheers,

//arry/