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

[Charles R Harris]

On Sun, Feb 14, 2016 at 1:11 PM, Charles R Harris <charlesr.harris at gmail.com
> wrote:

>
>
> 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
>
>
However, I do note that some languages offer two versions of modulus, one
floor based and the other trunc based (effectively fmod). What I wanted is
to keep the remainder consistent with the floor function in the C library.

Chuck
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