I have this C++ function, which I can call from Python with the code below. The performance is only half compared to running pure C++. Is there a way to get their performance at the same level? I compile both codes with -Ofast -march=native
flags. I do not understand where I can lose 50%, because most of the time should be spent in the C++ kernel. Is Cython making a memory copy that I can avoid?
namespace diff
{
void diff_cpp(double* __restrict__ at, const double* __restrict__ a, const double visc,
const double dxidxi, const double dyidyi, const double dzidzi,
const int itot, const int jtot, const int ktot)
{
const int ii = 1;
const int jj = itot;
const int kk = itot*jtot;
for (int k=1; k<ktot-1; k++)
for (int j=1; j<jtot-1; j++)
for (int i=1; i<itot-1; i++)
{
const int ijk = i + j*jj + k*kk;
at[ijk] += visc * (
+ ( (a[ijk+ii] - a[ijk ])
- (a[ijk ] - a[ijk-ii]) ) * dxidxi
+ ( (a[ijk+jj] - a[ijk ])
- (a[ijk ] - a[ijk-jj]) ) * dyidyi
+ ( (a[ijk+kk] - a[ijk ])
- (a[ijk ] - a[ijk-kk]) ) * dzidzi
);
}
}
}
I have this .pyx
file
# import both numpy and the Cython declarations for numpy
import cython
import numpy as np
cimport numpy as np
# declare the interface to the C code
cdef extern from "diff_cpp.cpp" namespace "diff":
void diff_cpp(double* at, double* a, double visc, double dxidxi, double dyidyi, double dzidzi, int itot, int jtot, int ktot)
@cython.boundscheck(False)
@cython.wraparound(False)
def diff(np.ndarray[double, ndim=3, mode="c"] at not None,
np.ndarray[double, ndim=3, mode="c"] a not None,
double visc, double dxidxi, double dyidyi, double dzidzi):
cdef int ktot, jtot, itot
ktot, jtot, itot = at.shape[0], at.shape[1], at.shape[2]
diff_cpp(&at[0,0,0], &a[0,0,0], visc, dxidxi, dyidyi, dzidzi, itot, jtot, ktot)
return None
I call this function in Python
import numpy as np
import diff
import time
nloop = 20;
itot = 256;
jtot = 256;
ktot = 256;
ncells = itot*jtot*ktot;
at = np.zeros((ktot, jtot, itot))
index = np.arange(ncells)
a = (index/(index+1))**2
a.shape = (ktot, jtot, itot)
# Check results
diff.diff(at, a, 0.1, 0.1, 0.1, 0.1)
print("at={0}".format(at.flatten()[itot*jtot+itot+itot//2]))
# Time the loop
start = time.perf_counter()
for i in range(nloop):
diff.diff(at, a, 0.1, 0.1, 0.1, 0.1)
end = time.perf_counter()
print("Time/iter: {0} s ({1} iters)".format((end-start)/nloop, nloop))
This is the setup.py
:
from distutils.core import setup
from distutils.extension import Extension
from Cython.Distutils import build_ext
import numpy
setup(
cmdclass = {'build_ext': build_ext},
ext_modules = [Extension("diff",
sources=["diff.pyx"],
language="c++",
extra_compile_args=["-Ofast -march=native"],
include_dirs=[numpy.get_include()])],
)
And here the C++ reference file that reaches twice the performance:
#include <iostream>
#include <iomanip>
#include <cstdlib>
#include <stdlib.h>
#include <cstdio>
#include <ctime>
#include "math.h"
void init(double* const __restrict__ a, double* const __restrict__ at, const int ncells)
{
for (int i=0; i<ncells; ++i)
{
a[i] = pow(i,2)/pow(i+1,2);
at[i] = 0.;
}
}
void diff(double* const __restrict__ at, const double* const __restrict__ a, const double visc,
const double dxidxi, const double dyidyi, const double dzidzi,
const int itot, const int jtot, const int ktot)
{
const int ii = 1;
const int jj = itot;
const int kk = itot*jtot;
for (int k=1; k<ktot-1; k++)
for (int j=1; j<jtot-1; j++)
for (int i=1; i<itot-1; i++)
{
const int ijk = i + j*jj + k*kk;
at[ijk] += visc * (
+ ( (a[ijk+ii] - a[ijk ])
- (a[ijk ] - a[ijk-ii]) ) * dxidxi
+ ( (a[ijk+jj] - a[ijk ])
- (a[ijk ] - a[ijk-jj]) ) * dyidyi
+ ( (a[ijk+kk] - a[ijk ])
- (a[ijk ] - a[ijk-kk]) ) * dzidzi
);
}
}
int main()
{
const int nloop = 20;
const int itot = 256;
const int jtot = 256;
const int ktot = 256;
const int ncells = itot*jtot*ktot;
double *a = new double[ncells];
double *at = new double[ncells];
init(a, at, ncells);
// Check results
diff(at, a, 0.1, 0.1, 0.1, 0.1, itot, jtot, ktot);
printf("at=%.20f\n",at[itot*jtot+itot+itot/2]);
// Time performance
std::clock_t start = std::clock();
for (int i=0; i<nloop; ++i)
diff(at, a, 0.1, 0.1, 0.1, 0.1, itot, jtot, ktot);
double duration = (std::clock() - start ) / (double)CLOCKS_PER_SEC;
printf("time/iter = %f s (%i iters)\n",duration/(double)nloop, nloop);
return 0;
}