[Python-Dev] Re: Rich comparisons (original) (raw)

Michael Hudson mwh at python.net
Tue Mar 30 09:58:36 EST 2004

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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