On Tue, Nov 21, 2017 at 4:47 AM, Josh B. <[email protected]> wrote:
> Now for the question: Is this useful? I ask because this leads to the
> following behavior:
>
>>>> unordered = MyColl([1, 2, 3])
>>>> ordered = MyOrderedColl([3, 2, 1])
>>>> s = {ordered, unordered}
>>>> len(s)
> 1
>>>> s = {ordered}
>>>> unordered in s
> True
>>>> # etc.
>
> In other words, sets and mappings can't tell unordered and ordered apart;
> they're treated like the same values.
>
> However, I'm less confident that this kind of behavior is useful for MyColl
> and MyOrderedColl. Could anyone who feels more certain one way or the other
> please explain the rationale and possibly even give some real-world examples?
>
This isn't a consequence of __hash__, it's a consequence of __eq__.
You have declared that MyColl and MyOrderedColl are equal, therefore
only one of them stays in the set.
But what you have is the strangeness of non-transitive equality, which
is likely to cause problems.
>>> unordered = MyColl([1, 2, 3])
>>> ordered1 = MyColl([3, 2, 1])
>>> ordered2 = MyColl([2, 1, 3])
unordered is equal to each of the others, but they're not equal to
each other. So if you put them into a set, you'll get results that
depend on order. Here's a simpler form of non-transitive equality:
>>> class Oddity(int):
... def __eq__(self, other):
... if other - 5 <= self <= other + 5:
... return True
... return int(self) == other
... def __hash__(self):
... return 1
...
>>> x, y, z = Oddity(5), Oddity(10), Oddity(15)
>>> x == y, y == z, x == z
(True, True, False)
>>> {x, y, z}
{5, 15}
>>> {y, x, z}
{10}
Setting __hash__ to a constant value is safe (but inefficient); it's
all based on how __eq__ works. So the question is: are you willing to
accept the bizarre behaviour of non-transitive equality?
ChrisA
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