On 9/12/2015 1:04 PM, Alexander Belopolsky wrote:
On Sat, Sep 12, 2015 at 1:20 AM, Terry Reedy <tjre...@udel.edu
<mailto:tjre...@udel.edu>> wrote:
A mathematician has no problem with 'a'+'b' != 'b'+'a'.
I doubt it. A binary operation denoted + (and called addition) is
almost universally a commutative operation. A non-commutative binary
operation is usually denoted * (and called multiplication).
I am aware of the single-operation group theory convention, but the
context was sequence concatenation and scalar multiplication, where '*'
is repeated '+'. But these details are not directly relevant to DateTimes.
After closure,
I perhaps should have said 'completeness'. In any case, I was referring
to 'a op b' existing for all a and b in the set.
Agree, and we have a solution for PEP 495 which preserves == as and
equivalence (symmetric, reflexive and transitive) relationship.
Datetime members, are rather unusual beasts. They are triples
consisting of a member of a discrete sequence (with some odd gaps),
Your correction, summarized, is that there are no gaps, so the set is
simpler than I thought. Skipping on to the heart of the matter.
I am not concerned about '-'. My main concern is about order
operations.
Is not '<' defined, in the obvious way, in terms of '-' and the sign of
the resul?
I am happy with the solution I have for ==, but I am still
struggling with the non-transitivity of <.
I am guessing that the 'struggle' is at least partly this: "Is
non-transitivity of < necessary given other constraints, including
back-compatibility, or is there another solution possible that would
also be <-transitive?"
Comparison operations are special because they are used implicitly in
other operations. The < operator is used implicitly in bisect. If it
does not satisfy the (partial?) order properties, bisect may enter an
infinite loop.
and, if we are stuck with <-intransitivity, what do we do? If
back-compatibility allowed, I might suggest defining 'lt' or 'less'
rather than '__lt__' so that sort and bisect don't work with DateTimes.
Then document that the function is not transitive.
--
Terry Jan Reedy
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