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 <gvanrossum@gmail.com> wrote:
Does numpy support this?
--Guido (mobile)
On Jan 16, 2017 7:27 AM, "Stephan Houben" <stephanh42@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 <steve@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.
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