如何修复使用斯库达·库夫特？

Question

我想做一个python包装的GPU fft函数，可以使用sciket计算任意大小输入的变换-库达·库夫特. （我试过只取2次方幂的PyFFT）

我模仿我的斯库达·库夫特CUDA代码：

__host__ cuDoubleComplex* FFT(cuDoubleComplex *data, int NX){

cudaEvent_t start, stop;
cudaEventCreate(&start);
cudaEventCreate(&stop);

cudaEventRecord(start, 0);

cuDoubleComplex *d_data;
cudaMalloc((void **)&d_data,NX*sizeof(cuDoubleComplex));

cufftHandle plan;
cufftPlan1d(&plan,NX,CUFFT_Z2Z,1);
cudaMemcpy(d_data, data, NX*sizeof(cuDoubleComplex), cudaMemcpyHostToDevice);
cufftExecZ2Z(plan,d_data,d_data,CUFFT_FORWARD);
cudaMemcpy(data,d_data,NX*sizeof(cuDoubleComplex),cudaMemcpyDeviceToHost);
cufftDestroy(plan);

cudaEventRecord(stop, 0);
cudaEventSynchronize(stop);

float elapsedTime;
cudaEventElapsedTime(&elapsedTime, start, stop);
printf("\n Elapsed Time:  %3.1f ms\n", elapsedTime);

cudaFree(d_data);

return data;
}

还有我的斯库达·库夫特代码如下所示：

import skcuda.cufft as ft
import pycuda.autoinit
import pycuda.gpuarray as gpuarray
import numpy as np

N=100

x=np.array(np.random.random(N),np.float32)
x_gpu=gpuarray.to_gpu(x)
xf_gpu = gpuarray.empty(N,np.complex64)
plan=ft.cufftPlan1d(N,ft.CUFFT_Z2Z,1)
ft.cufftExecZ2Z(plan,x_gpu,xf_gpu,ft.CUFFT_FORWARD)
ft.cufftDestroy(plan)

xf=x_gpu.get()

但它给出了一个错误：

runfile('/home/jesli/sk-cufft_test.py', wdir='/home/jesli') Traceback (most recent call last):

File "", line 1, in runfile('/home/jesli/sk-cufft_test.py', wdir='/home/jesli')

File "/home/jesli/anaconda/lib/python2.7/site-packages/spyderlib/widgets/externalshell/sitecustomize.py", line 580, in runfile execfile(filename, namespace)

File "/home/jesli/sk-cufft_test.py", line 19, in ft.cufftExecZ2Z(plan,x_gpu,xf_gpu,ft.CUFFT_FORWARD)

File "/home/jesli/anaconda/lib/python2.7/site-packages/skcuda/cufft.py", line 319, in cufftExecZ2Z direction)

ArgumentError: argument 2: : wrong type

变换方向（CUFFT\u FORWARD，CUFFT\u INVERSE）已经在源代码中定义。你知道吗

http://scikit-cuda.readthedocs.org/en/latest/_modules/skcuda/cufft.html

我想知道代码出了什么问题，或者函数需要什么参数。你知道吗