[Python-ideas] Make range objects orderable

[Joao S. O. Bueno]

Sorry - it looks like Guido's time machine have been there.

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)]
>>>



On 23 April 2015 at 22:11, Terry Reedy <tjreedy at udel.edu> wrote:
> On 4/23/2015 8:31 PM, MRAB wrote:
>>
>> On 2015-04-24 01:15, Ethan Furman wrote:
>>>
>>> On 04/23, Riley Banks wrote:
>>>
>>>> I propose adding the ability to compare range objects using methods
>>>> (e.g.
>>>> range.issubrange) and/or regular operators. Example:
>>>>
>>>>     In [56]: range(0, 10, 3).issubrange(range(10))
>>>>     Out[56]: True
>>>>
>>>>     In [57]: range(0, 10, 3) <= range(10)
>>>>     Out[57]: True
>
>
> This was not obvious to me.  Ranges are not sets.
>
>>>>     In [58]: range(10) <= range(0, 10, 3)
>>>>     Out[58]: False
>>>
>>>
>>> I seem to recall orderable ranges being rejected before.  For example, is
>>> less-than, or subrange, dependent on the values themselves, or on the
>>> lowest and highest values, or on the start and end value, or ...
>
>
> I remember the same.  There is no intuitive answer.  Let us not rehash this
> again.
>
>>> In other words, given:
>>>
>>>    a = range(11)
>>>    b = range(12, step=3)
>>>
>>> is a < b, or b < a?  Why?
>>>
>> It could be functionally equivalent to list(a) < list(b).
>
>
> For range(0, 10, 3) <= range(10), this would give the opposite answer as the
> set interpretation.
>
>
> --
> Terry Jan Reedy
>
>
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