PyCuda中3D数组的置换

Question

我有一个3D数组，我想把它的前两个维度(x &y)，但不是第三(z)。在3D数组a上，我希望得到与numpy的A.transpose((1,0,2))相同的结果。具体来说，我想要得到"转置"的全局threadIdx。下面的代码应该在3D数组a中未转置的位置写入转置索引，它没有。

任何建议吗?

import numpy as np
from pycuda import compiler, gpuarray
import pycuda.driver as cuda
import pycuda.autoinit
kernel_code = """
__global__ void test_indexTranspose(uint*A){
    const size_t size_x = 4;
    const size_t size_y = 4;
    const size_t size_z = 3;
    // Thread position in each dimension
    const size_t tx = blockDim.x * blockIdx.x + threadIdx.x;
    const size_t ty = blockDim.y * blockIdx.y + threadIdx.y;
    const size_t tz = blockDim.z * blockIdx.z + threadIdx.z;
    if(tx < size_x && ty < size_y && tz < size_z){
        // Flat index
        const size_t ti = tz * size_x * size_y + ty * size_x + tx;
        // Transposed flat index
        const size_t tiT = tz * size_x * size_y + tx * size_x + ty;
        A[ti] = tiT;
    }
}
"""
A = np.zeros((4,4,3),dtype=np.uint32)
mod = compiler.SourceModule(kernel_code)
test_indexTranspose = mod.get_function('test_indexTranspose')
A_gpu = gpuarray.to_gpu(A)
test_indexTranspose(A_gpu, block=(2, 2, 1), grid=(2,2,3))

这是返回的结果(不是我期望的):

A_gpu.get()[:,:,0]
array([[ 0, 12,  9,  6],
       [ 3, 15, 24, 21],
       [18, 30, 27, 36],
       [33, 45, 42, 39]], dtype=uint32)
A_gpu.get()[:,:,1]
array([[ 4,  1, 13, 10],
       [ 7, 16, 28, 25],
       [22, 19, 31, 40],
       [37, 34, 46, 43]], dtype=uint32)
A_gpu.get()[:,:,2]
array([[ 8,  5,  2, 14],
       [11, 20, 17, 29],
       [26, 23, 32, 44],
       [41, 38, 35, 47]], dtype=uint32)

这是我所期望的(但没有返回):

A_gpu.get()[:,:,0]
array([[0, 4, 8,  12],
       [1, 5, 9,  13],
       [2, 6, 10, 14],
       [3, 7, 11, 15]], dtype=uint32)
A_gpu.get()[:,:,1]
array([[16, 20, 24, 28],
       [17, 21, 25, 29],
       [18, 22, 26, 30],
       [19, 23, 27, 31]], dtype=uint32)
A_gpu.get()[:,:,2]
...

谢谢,

Answer 1

创建与CUDA内核代码一致的numpy数组解决了这个问题。numpy数组的默认布局不是我的内核所假定的行、列、深度。但是，可以在创建数组时设置步长。
如果像这样创建数组，上面的内核可以正常工作:

nRows = 4
nCols = 4
nSlices = 3
nBytes = np.dtype(np.uint32).itemsize
A = np.ndarray(shape=(nRows, nCols, nSlices), 
               dtype=np.uint32, 
               strides=(nCols*nBytes, 1*nBytes, nCols*nRows*nBytes))

步长是内存连续索引需要为每个维度(以字节为单位)进行的跳跃。例如，从第1行第1个元素到第2行第1个元素有nCols * nBytes，即16个字节。