[Numpy-discussion] numpy.nansum() behavior in 1.3.0

[Keith Goodman]

On Mon, Jun 1, 2009 at 4:50 PM,  <josef.pktd at gmail.com> wrote:
> On Mon, Jun 1, 2009 at 7:43 PM,  <josef.pktd at gmail.com> wrote:
>> On Mon, Jun 1, 2009 at 7:30 PM, Robert Kern <robert.kern at gmail.com> wrote:
>>> On Mon, Jun 1, 2009 at 15:31,  <josef.pktd at gmail.com> wrote:
>>>> On Mon, Jun 1, 2009 at 4:06 PM, Alan G Isaac <aisaac at american.edu> wrote:
>>>>> On 6/1/2009 3:38 PM josef.pktd at gmail.com apparently wrote:
>>>>>> Here's a good one:
>>>>>>
>>>>>>>>> np.isnan([]).all()
>>>>>> True
>>>>>>>>> np.isnan([]).any()
>>>>>> False
>>>>>
>>>>>
>>>>>  >>> all([])
>>>>> True
>>>>>  >>> any([])
>>>>> False
>>>>
>>>> also:
>>>>
>>>>>>> y
>>>> array([], dtype=float64)
>>>>>>> (y>0).all()
>>>> True
>>>>>>> (y>0).any()
>>>> False
>>>>>>> ((y>0)>0).sum()
>>>> 0
>>>>
>>>> I don't know what's the logic, but it causes the bug in np.nansum.
>>>
>>> You will have to special-case empty arrays, then.
>>>
>>
>> is np.size the right check for non-empty array, including subtypes?
>>
>> i.e.
>>
>> if y.size and mask.all():
>>        return np.nan
>>
>> or more explicit
>> if y.size > 0 and mask.all():
>>        return np.nan
>>
>
> Actually, now I think this is the wrong behavior, nansum should never
> return nan.
>
>>>> np.nansum([np.nan, np.nan])
> 1.#QNAN
>
> shouldn't this be zero

The doc string says it is zero: "Return the sum of array elements over
a given axis treating Not a Numbers (NaNs) as zero." Treating NaNs
differently in different cases is harder to explain.