RE: Set and {} comparison confusion
From: Roman Yakovenko (roman.yakovenko_at_actimize.com)
Date: 09/09/04
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Date: Thu, 9 Sep 2004 14:14:08 +0300 To: <python-list@python.org>
> -----Original Message-----
> From: Roman Yakovenko
> Sent: Thursday, September 09, 2004 2:11 PM
> To: python-list@python.org
> Subject: RE: Set and {} comparison confusion
>
>
>
> > -----Original Message-----
> > From: Alex Martelli [mailto:aleaxit@yahoo.com]
> >
> >
> > Roman Yakovenko <roman.yakovenko@actimize.com> wrote:
> > ...
> > > classes have __eq__, __ne__. Classes are mutable - I can't define
> > > __hash__ function. __lt__ - I can implement but it will be
> > meaningless.
> >
> > As long as it respects the fundamental semantics
> constraints such as:
> > a < b and b < c imply a < c
> > a < b implies not (a == b)
> > a < b implies (a != b)
> > not (a < a) for any a
> > and so on, your __lt__will not be 'meaningless' but very useful.
>
> Well I am not talking about algebra. Those rules are clear to me.
>
> > Basically, __lt__ is meaningful if it's transitive,
> > non-reflective, and
> > compatible with your == and != (which I assume are compatible
> > with each
> > other); a transitive non-reflective < defines implicitly an
> > equivalence
> > relation, a eqv b <--> not (a < b or b < a), and you need your == to
> > express exactly that equivalence relation... so if your == is
> > meaningful, your < can't really be 'meaningless'!-)
>
>
> I don't agree with you. I can compare properties of some object.
> For example I can compare cows:
^^^^^ pythons :-)
> one python is longer then other ( length )
> one python is heavier then other ( weight )
> one python is older then other ( age )
>
> But I can't define meaningful operator "<" on pythons.
> I can compare pythons only property by property. I don't thing
> that next code is meaningful
>
> def __lt__( self, other_python ):
> return self.length < other_python.length \
> and self.weight < other_python.weight \
> and self.age < other_python.age
>
> I see this code as meaningless on python's.
>
> > > Thank you for help. I think I have a dicision:
> > > 1. I will implement meaningless __lt__
> > > 2. I will sort ( I don't have duplicated items ) every time
> > I need to compare
> > > 2.1 Because sort is happen in place next time it will take
> > less time to sort.
> >
> > Yes, that does seem to make sense to me. Once two lists without
> > duplicates are sorted, they're equal as sets iff they're ==
> as lists;
> > and yes, maintaining sorted order is typically quite cheap in
> > Python due
> > to Tim Peters' powerful natural mergesort algorithm as implemented
> > inside list.sort (though it might be worth having a look at
> the bisect
> > module, it's likely going to be faster to just sort the lists each
> > time).
> >
> >
> > > Again - Thanks for help. It was very usefull. It seems that
> > I had wrong
> > expectation
> > > from set - " unordered set collection based only on
> > comparison operators".
> > > My mistake.
> >
> > Ah, isn't set documented to need hashable elements? It
> should be. Of
> > course, _why_ your class appears to be hashable when it
> defines __eq__
> > and not __hash__ I dunno -- Python should diagnose that,
> but it does't
> > appear to...
>
> It is my fault too. All my classes derives from object.
> object do implements __hash__
> function. I think I should raise exception if somebody tries
> to insert object into
> set \ dictionary
>
>
> > Alex
> > --
> > http://mail.python.org/mailman/listinfo/python-list
> >
> --
> http://mail.python.org/mailman/listinfo/python-list
>
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- Maybe in reply to: Roman Yakovenko: "Set and {} comparison confusion"
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