If I had the time machine for comparisons
On Wed, Oct 12, 2011 at 11:18 PM, Guido van Rossum <guido@python.org> wrote:
(I'm also -1 on adding ordering comparisons; there's little disagreement on that issue.)
If I had a time machine, I would allow comparisons to return "unordered" as well. Right now, objects are comparable or not based strictly on the type, even though comparison is inherently about the values. I think that range(3) < range(10) is obviously true, even though it isn't clear whether or not range(3, 15, 2) < range(7, -8, -1) is true. Here we err by not allowing the first comparison; other objects (like dicts) we err by forcing an arbitrary ordering. -jJ
On Thu, Oct 13, 2011 at 7:06 AM, Jim Jewett <jimjjewett@gmail.com> wrote:
Have you used Python3 lately? It doesn't allow dict ordering. In general Python expresses unordered by raising an exception (often TypeError). Even though range(1) is "obviously" < range(2), there are so many unobvious cases that supporting this one special case isn't worth it. -- --Guido van Rossum (python.org/~guido)
On 10/13/2011 10:06 AM, Jim Jewett wrote:
Based on what property? Is that because: a) len(range(3)) < len(range(10)) b) max(range(3)) < min(range(10)) c) ((min(range(3)) < min(range(10))) and (max(range(3)) < max(range(10))) I guess your argument is that in this degenerate case, all of these properties are true so it is "obviously true". Personally, if I can't pin-point the exact reason that "x < y", then it's not obvious even if for every definition of "<" I can come up with it is a true statement, because there is not one obvious definition that is true. In other words, I don't know what "x < y" means for ranges in general so I can't reason about it in general, therefore this special case is not useful or obvious. In the absence of explicit arguments to the "<" operand, I would be baffled as to what to expect as the result. However, I can understand the desire to test for range equality, and the definition for that is significantly more obvious. -- Scott Dial scott@scottdial.com
On Thu, Oct 13, 2011 at 7:06 AM, Jim Jewett <jimjjewett@gmail.com> wrote:
Have you used Python3 lately? It doesn't allow dict ordering. In general Python expresses unordered by raising an exception (often TypeError). Even though range(1) is "obviously" < range(2), there are so many unobvious cases that supporting this one special case isn't worth it. -- --Guido van Rossum (python.org/~guido)
On 10/13/2011 10:06 AM, Jim Jewett wrote:
Based on what property? Is that because: a) len(range(3)) < len(range(10)) b) max(range(3)) < min(range(10)) c) ((min(range(3)) < min(range(10))) and (max(range(3)) < max(range(10))) I guess your argument is that in this degenerate case, all of these properties are true so it is "obviously true". Personally, if I can't pin-point the exact reason that "x < y", then it's not obvious even if for every definition of "<" I can come up with it is a true statement, because there is not one obvious definition that is true. In other words, I don't know what "x < y" means for ranges in general so I can't reason about it in general, therefore this special case is not useful or obvious. In the absence of explicit arguments to the "<" operand, I would be baffled as to what to expect as the result. However, I can understand the desire to test for range equality, and the definition for that is significantly more obvious. -- Scott Dial scott@scottdial.com
participants (3)
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Guido van Rossum -
Jim Jewett -
Scott Dial