No, the python ** gets translated to a pow statement by cython. I think the issue is that for some reason, i'm getting stuck in the gcc slow pow function....
if i let e2 and e1 be 1 and replace f**2 (which would call pow) with f*f, my execution time drops to this: 10000 loops, best of 3: 108 µs per loop over 6x improvement just by avoid a few measly pow statements... anyone know why i'm stuck in slowpow? On Tue, Sep 29, 2009 at 6:15 PM, Sturla Molden <[email protected]> wrote: > Sturla Molden skrev: >> Chris Colbert skrev: >> >>> and within that loop it is these statements that take the bulk of the time: >>> >>> F = ((f1**2)**(1/e2) + (f2**2)**(1/e2))**(e2/e1) + (f3**2)**(1/e1) >>> >>> temperr = (C4 * (F**(e1) - 1))**2 >>> >>> and replacing the powers with serial multiplications don't really help >>> any... >>> >>> >> >> Does this help? >> >> cdef extern from "math.h": >> double pow(double, double) >> >> F = pow(pow((f1*f1),(1/e2)) + pow((f2*f2),(1/e2)),(e2/e1)) \ >> + pow((f3*f3),(1/e1)) >> >> > cdef double tmp > > tmp = C4 * (pow(F,e1) - 1) > > temperr = tmp*tmp > > > > > _______________________________________________ > Cython-dev mailing list > [email protected] > http://codespeak.net/mailman/listinfo/cython-dev > _______________________________________________ Cython-dev mailing list [email protected] http://codespeak.net/mailman/listinfo/cython-dev
