On Wed, Apr 27, 2011 at 8:42 PM, Robert Kern <robert.k...@gmail.com> wrote:
> On 2011-04-27 22:16 , Guido van Rossum wrote:
>> So does NumPy also follow Python's behavior about ignoring the NaN
>> special-casing when doing array ops?
>
> By "ignoring the NaN special-casing", do you mean that identity is checked
> first? When we use dtype=object arrays (arrays that contain Python objects
> as their data), yes:
>
> [~]
> |1> nan = float('nan')
>
> [~]
> |2> import numpy as np
>
> [~]
> |3> a = np.array([1, 2, nan], dtype=object)
>
> [~]
> |4> nan in a
> True
>
> [~]
> |5> float('nan') in a
> False
>
>
> Just like lists:
>
> [~]
> |6> nan in [1, 2, nan]
> True
>
> [~]
> |7> float('nan') in [1, 2, nan]
> False
>
>
> Actually, we go a little further by using PyObject_RichCompareBool() rather
> than PyObject_RichCompare() to implement the array-wise comparisons in
> addition to containment:
>
> [~]
> |8> a == nan
> array([False, False, True], dtype=bool)
Hm, this sounds like NumPy always considers a NaN equal to *itself* as
long as objects are concerned.
> [~]
> |9> [x == nan for x in [1, 2, nan]]
> [False, False, False]
>
>
> But for dtype=float arrays (which contain C doubles, not Python objects) we
> use C semantics. Literally, we use whatever C's == operator gives us for the
> two double values. Since there is no concept of identity for this case,
> there is no cognate behavior of Python to match.
>
> [~]
> |10> b = np.array([1.0, 2.0, nan], dtype=float)
>
> [~]
> |11> b == nan
> array([False, False, False], dtype=bool)
>
> [~]
> |12> nan in b
> False
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).
Thanks!
--
--Guido van Rossum (python.org/~guido)
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