[Python-Dev] PyObject_RichCompareBool identity shortcut

[Guido van Rossum]

On Wed, Apr 27, 2011 at 9:25 PM, Robert Kern <robert.kern at gmail.com> wrote:
> On 2011-04-27 23:01 , Guido van Rossum wrote:
>> And I wouldn't want to change that. It sounds like NumPy wouldn't be
>> much affected if we were to change this (which I'm not saying we
>> would).
>
> Well, I didn't say that. If Python changed its behavior for (float('nan') ==
> float('nan')), we'd have to seriously consider some changes.

Ah, but I'm not proposing anything of the sort! float('nan') returns a
new object each time and two NaNs that are not the same *object* will
still follow the IEEE std. It's just when comparing a NaN-valued
*object* to *itself* (i.e. the *same* object) that I would consider
following the lead of Python's collections.

> We do like to
> keep *some* amount of correspondence with Python semantics. In particular,
> we like our scalar types that match Python types to work as close to the
> Python type as possible. We have the np.float64 type, which represents a C
> double scalar and corresponds to a Python float. It is used when a single
> item is indexed out of a float64 array. We even subclass from the Python
> float type to help working with libraries that may not know about numpy:
>
> [~]
> |5> import numpy as np
>
> [~]
> |6> nan = np.array([1.0, 2.0, float('nan')])[2]
>
> [~]
> |7> nan == nan
> False

Yeah, this is where things might change, because it is the same
*object* left and right.

> [~]
> |8> type(nan)
> numpy.float64
>
> [~]
> |9> type(nan).mro()
> [numpy.float64,
>  numpy.floating,
>  numpy.inexact,
>  numpy.number,
>  numpy.generic,
>  float,
>  object]
>
>
> If the Python float type changes behavior, we'd have to consider whether to
> keep that for np.float64 or change it to match the usual C semantics used
> elsewhere. So there *would* be a dilemma. Not necessarily the most
> nerve-wracking one, but a dilemma nonetheless.

Given what I just said, would it still be a dilemma? Maybe a smaller one?

-- 
--Guido van Rossum (python.org/~guido)