My understanding is that NumPy does NOT currently support a direct FMA
operation "natively." However, higher-level routines like
`numpy.linalg.solve` that are linked to MKL or BLAS DO take advantage of
FMA within the underlying libraries.
On Mon, Jan 16, 2017 at 10:06 AM, Guido van Rossum <gvanros...@gmail.com>
wrote:
> Does numpy support this?
>
> --Guido (mobile)
>
> On Jan 16, 2017 7:27 AM, "Stephan Houben" <stephan...@gmail.com> wrote:
>
>> Hi Steve,
>>
>> Very good!
>> Here is a version which also handles the nan's, infinities,
>> negative zeros properly.
>>
>> ===============
>> import math
>> from fractions import Fraction
>>
>> def fma2(x, y, z):
>> if math.isfinite(x) and math.isfinite(y) and math.isfinite(z):
>> result = float(Fraction(x)*Fraction(y) + Fraction(z))
>> if not result and not z:
>> result = math.copysign(result, x*y+z)
>> else:
>> result = x * y + z
>> assert not math.isfinite(result)
>> return result
>> ===========================
>>
>> Stephan
>>
>>
>> 2017-01-16 12:04 GMT+01:00 Steven D'Aprano <st...@pearwood.info>:
>>
>>> On Mon, Jan 16, 2017 at 11:01:23AM +0100, Stephan Houben wrote:
>>>
>>> [...]
>>> > So the following would not be a valid FMA fallback
>>> >
>>> > double bad_fma(double x, double y, double z) {
>>> > return x*y + z;
>>> > }
>>> [...]
>>> > Upshot: if we want to provide a software fallback in the Python code,
>>> we
>>> > need to do something slow and complicated like musl does.
>>>
>>> I don't know about complicated. I think this is pretty simple:
>>>
>>> from fractions import Fraction
>>>
>>> def fma(x, y, z):
>>> # Return x*y + z with only a single rounding.
>>> return float(Fraction(x)*Fraction(y) + Fraction(z))
>>>
>>>
>>> When speed is not the number one priority and accuracy is important,
>>> its hard to beat the fractions module.
>>>
>>>
>>> --
>>> Steve
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>>
>>
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