[Python-Dev] Re: Rich comparisons

[Michael Hudson]

Edward Loper <edloper at gradient.cis.upenn.edu> writes:

>> Edward Loper <edloper at gradient.cis.upenn.edu> writes:
>> 
>>>This means that 'nan' is no longer a well-behaved dictionary key:
>>>
>>>     >>> x = {float('nan'):0}
>>>     >>> x[float('nan')] = 1
>>>     >>> print x
>>>     {nan: 0, nan: 1}
>>>
>>>Even worse, we get different behavior if we use the "same copy" of nan:
>>>
>>>     >>> nan = float('nan')
>>>     >>> x = {nan:0}
>>>     >>> x[nan] = 1
>>>     >>> print x
>>>     {nan: 1}
>> Gah.  Still, I think this is more a theoretical objection than
>> anything else? [...]
>> I have at no point claimed that I have given Python 2.4 a coherent
>> ieee 754 floating point story.  All I have tried to do is /reduce/ the
>> level of surprise knocking around, and I think I've succeeded.  If
>> someone (not me!) has the time and energy to do a 'proper job' (and
>> I'd expect working out what that means to be the hard part), then you
>> have my support and pre-emptive thanks.
>
> Tim Delaney's suggestion [1] seemed pretty reasonable.  In particular:
>
>    - Ensure that NaN is a singleton (like True and False).  I.e., the
>      float constructor checks if the float is NaN, and if so returns a
>      singleton.

And we do that... how, exactly?

> Advantages:
>    - We have an easy way to test if a variable is nan: "x is NaN"

There's (potentially, at least) more than one NaN (QNaN & SNaN, f'ex).

>    - nan will work "properly" as a dictionary key

BFD.

> The important question is whether this would slow down other
> operations. As for cmp(), my understanding was that it should just
> return -1 if two unordered objects are not equal.

Then I believe your understanding is flawed.

>>> cmp(2,1)
1

Without wanting to sound patronizing or rude -- but risking it anyway
-- you're not giving the impression of understanding these issues any
more than I do, and I *know* I don't understand them well enough to
get them straight.

Cheers,
mwh

-- 
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