编写numpy ufunc时如何处理复杂的值?

时间:2019-09-08 12:26:43

标签: c numpy complex-numbers numpy-ufunc

基本上,我有一个运行在npy_cdoublenpy_cfloat数组上的numpy ufunc。例如:

static void
ufunc_H( char ** args
       , npy_intp * dimensions
       , npy_intp * steps
       , void * data)
{

    npy_cdouble * qm_in = (npy_cdouble *) (args[0]);
    npy_cdouble * qm_out = (npy_cdouble *) (args[1]);
    npy_intp i;

    for(i = 0; i < ndim; i++)
    {
        qm_out[i] = (qm_in[i] - qm_in[i ^ ipow(2, argument.act)]) * M_SQRT1_2;
    }


}

但是,这不起作用,编译器说qm_in的类型为‘npy_cdouble’ {aka ‘struct <anonymous>’}。如何正确对待npy_cdouble

1 个答案:

答案 0 :(得分:0)

结构npy_cdoublenpy_cfloat具有成员.real.imag。这些可以用来做操作:

static void
ufunc_H( char ** args
       , npy_intp * dimensions
       , npy_intp * steps
       , void * data)
{

    npy_cdouble * qm_in = (npy_cdouble *) (args[0]);
    npy_cdouble * qm_out = (npy_cdouble *) (args[1]);
    npy_intp i;

    for(i = 0; i < ndim; i++)
    {
        qm_out[i].real = (qm_in[i].real - qm_in[i ^ ipow(2, argument.act)].real) * M_SQRT1_2;
        qm_out[i].imag = (qm_in[i].imag - qm_in[i ^ ipow(2, argument.act)].imag) * M_SQRT1_2;
    }


}