On Oct 12, 2011 9:37 PM, "Sven Marnach" sven@marnach.net wrote:
Steven D'Aprano schrieb am Do, 13. Okt 2011, um 04:33:49 +1100:
When implementing '==' and '!=' for range objects, it would be natural to implement the other comparison operators, too (lexicographically, as for all other sequence types).
I don't agree. Equality makes sense for ranges: two ranges are equal if they have the same start, stop and step values.
No, two ranges should be equal if they represent the same sequence, i.e. if they compare equal when converted to a list:
range(0) == range(4, 4, 4) range(5, 10, 3) == range(5, 11, 3) range(3, 6, 3) == range(3, 4)
But order comparisons don't have any sensible meaning: range objects are numeric ranges, integer-valued intervals, not generic lists, and it is meaningless to say that one range is less than or greater than another.
Well, it's meaningless unless you define what it means. Range objects are equal if they compare equal after converting to a list. You could define '<' or '>' the same way. All built-in sequence types support lexicographical comparison, so I thought it would be natural to bring the only one that behaves differently in line. (Special cases aren't special enough...)
This is just to explain my thoughts, I don't have a strong opinion on this one.
I'll try and prepare a patch for '==' and '!=' and add it to the issue tracker.
Cheers, Sven _______________________________________________ Python-ideas mailing list Python-ideas@python.org http://mail.python.org/mailman/listinfo/python-ideas
If you consider a range to represent a special type of set, which it is since it always contains unique values, then comparison operators do make sense. E.g. range(4,8) < range(2,9) is a subset comparison.
+1 for equality checks
David