此错误消息告诉您发生了什么问题:
小地毯pyx:17:27:无法将类型“double complex*”分配给“double*”
也就是说,您有一个来自numpy的双复数指针(指向complex128 numpy dtype的指针),您试图使用双指针将其传递到C++函数中。C++需要能够处理复数,因此如果您更改double*->复杂这应该可以解决你的问题
void mult(double *M, double *N, double *Q)
成为
#include <complex>
void mult(std::complex<double> *M, std::complex<double> *N, std::complex<double> *Q)
numpy矩阵乘法是否不足以满足您的用例?Cython可能有点过头了。
编辑:好的,我终于找到了一些东西,处理C++std::复杂和C double \u复杂类型有点奇怪。
cppmul公司。pyx:
import numpy as np
cimport numpy as np
cdef extern from "./matmult.h" nogil:
void mult(np.complex128_t* M, np.complex128_t* N, np.complex128_t* Q)
def sim():
cdef:
np.ndarray[np.complex128_t,ndim=2] N = np.zeros(( 2 , 2 ), dtype=np.complex128)
np.ndarray[np.complex128_t,ndim=2] Q = np.zeros(( 2 , 2 ), dtype=np.complex128)
np.ndarray[np.complex128_t,ndim=2] M = np.zeros(( 2 , 2 ), dtype=np.complex128)
N = np.array([[1.1 + 2j,2.2],[3.3,4.4]])
Q = np.array([[3.3,4.4+5j],[5.5,6.6]])
mult(&M[0,0], &N[0,0], &Q[0,0])
print M
马特穆尔。c:
#include "matmult.h"
void mult(complex_t *M, complex_t *N, complex_t *Q)
{
complex_t P[2][2], A[2][2], B[2][2];
for (int i=0; i<2; i++)
{
for (int j=0; j<2; j++)
{
A[i][j] = *( N + ((2*i) + j) );
B[i][j] = *( Q + ((2*i) + j) );
P[i][j] = 0;
}
}
for (int i=0; i<2; i++)
{
for (int j=0; j<2; j++)
{
for (int k=0; k<2; k++)
{
P[i][j] += A[i][k]*B[k][i];
}
}
}
for (int i=0; i<2; i++)
{
for (int j=0; j<2; j++)
{
*( M + ((2*i) + j) ) = P[i][j];
}
}
}
马特穆特。h:
#include <complex.h>
typedef double _Complex complex_t;
void mult(complex_t *M, complex_t *N, complex_t *Q);
设置。py:
from distutils.core import setup
from Cython.Build import cythonize
from distutils.extension import Extension
import numpy as np
sourcefiles = ['cppmul.pyx', 'matmult.c']
extensions = [Extension("cppmul",
sourcefiles,
include_dirs=[np.get_include()],
extra_compile_args=['-O3']
)]
setup(
ext_modules = cythonize(extensions)
)
运行后
python setup.py build_ext --inplace
它按预期导入和运行
import cppmul
cppmul.sim()
结果:
[[15.73 +6.6j 15.73 +6.6j]
[43.56+16.5j 43.56+16.5j]]