[Python-ideas] Make range objects orderable

[Guido van Rossum]

On Fri, Apr 24, 2015 at 9:18 PM, Nick Coghlan <ncoghlan at gmail.com> wrote:

> On 25 April 2015 at 00:50, Joao S. O. Bueno <jsbueno at python.org.br> wrote:
> > Since it does not mean anything (as in meaningless) to compare ranges
> without
> > specifying a "key", and since ordering functions in Python do allow a
> > "key" - it looks
> > like this problem is already resolved, whatever the need:
> >
> >>>> sorted([range(10), range(1,14, 3)], key=len)
> > [range(1, 14, 3), range(0, 10)]
> >>>> sorted([range(10), range(1,14, 3)], key=max)
> > [range(0, 10), range(1, 14, 3)]
> >>>> sorted([range(10), range(1,14, 3)], key=min)
> > [range(0, 10), range(1, 14, 3)]
>
> ranges() are actually defined as memory-efficient tuples these days
> (https://docs.python.org/3/library/stdtypes.html#ranges), and have
> supported "==" and "!=" using sequence comparison semantics since 3.3,
> so there's a defined natural order for them these days (the ordering
> of the equivalent tuples).
>
> That said, I'm still not sure it makes sense to allow ordering them.
> The cases I can think of where it might make sense (such as ordering
> graph nodes) would all use an actual tuple instead.
>

Seems everybody has their own view on how range() objects should be
ordered. That's one good reason against a builtin ordering.

-- 
--Guido van Rossum (python.org/~guido)
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