[Numpy-discussion] Modulus (remainder) function corner cases

[Charles R Harris]

On Sun, Feb 14, 2016 at 12:54 PM, Charles R Harris <
charlesr.harris at gmail.com> wrote:

>
>
> On Sun, Feb 14, 2016 at 12:35 PM, Nils Becker <nilsc.becker at gmail.com>
> wrote:
>
>> 2016-02-13 17:42 GMT+01:00 Charles R Harris <charlesr.harris at gmail.com>:
>>
>>> The Fortran modulo function, which is the same basic function as in my
>>>> branch, does not specify any bounds on the result for floating
>>>> numbers, but gives only the formula,  modulus(a, b) = a - b*floor(a/b),
>>>> which has the advantage of being simple and well defined ;)
>>>>
>>>
>>>
>> In the light of the libm-discussion I spent some time looking at floating
>> point functions and their accuracy. I would vote in favor of keeping an
>> implementation that uses the fmod-function of the system library and bends
>> it to adhere to the python convention (sign of divisor). There is probably
>> a reason why the fmod-implementation is not as simple as "a - b*floor(a/b)"
>> [1].
>>
>> One obvious problem with the simple expression arises when a/b = 0.0 in
>> floating point. E.g.
>>
>> In [43]: np.__version__
>> Out[43]: '1.10.4'
>> In [44]: x = np.float64(1e-320)
>> In [45]: y = np.float64(-1e10)
>> In [46]: x % y # this uses libm's fmod on my system
>>
>
I'm not too worried about denormals. However, this might be considered a
bug in the floor function

In [16]: floor(-1e-330)
Out[16]: -0.0

Chuck
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