我一直在玩numba并尝试实现一个简单的元素矩阵乘法。使用＆＃39; vectorize＆＃39;我得到了与numpy乘法相同的结果，但是当我使用＆c 39. cuda.jit＆＃39;他们不一样。其中许多都是零。我为此提供了最低限度的工作示例。任何有关此问题的帮助将不胜感激。我使用的是numba o.35.0和python 2.7
from __future__ import division
from __future__ import print_function
import numpy as np
from numba import vectorize, cuda, jit
M = 80
N = 40
P = 40
# Set the number of threads in a block
threadsperblock = 32
# Calculate the number of thread blocks in the grid
blockspergrid = (M*N*P + (threadsperblock - 1)) // threadsperblock
@vectorize(['float32(float32,float32)'], target='cuda')
def VectorMult3d(a, b):
return a*b
@cuda.jit('void(float32[:, :, :], float32[:, :, :], float32[:, :, :])')
def mult_gpu_3d(a, b, c):
[x, y, z] = cuda.grid(3)
if x < c.shape[0] and y < c.shape[1] and z < c.shape[2]:
c[x, y, z] = a[x, y, z] * b[x, y, z]
if __name__ == '__main__':
A = np.random.normal(size=(M, N, P)).astype(np.float32)
B = np.random.normal(size=(M, N, P)).astype(np.float32)
numpy_C = A*B
A_gpu = cuda.to_device(A)
B_gpu = cuda.to_device(B)
C_gpu = cuda.device_array((M,N,P), dtype=np.float32) # cuda.device_array_like(A_gpu)
mult_gpu_3d[blockspergrid,threadsperblock](A_gpu,B_gpu,C_gpu)
cudajit_C = C_gpu.copy_to_host()
print('------- using cuda.jit -------')
print('Is close?: {}'.format(np.allclose(numpy_C,cudajit_C)))
print('{} of {} elements are close'.format(np.sum(np.isclose(numpy_C,cudajit_C)), M*N*P))
print('------- using cuda.jit -------\n')
vectorize_C_gpu = VectorMult3d(A_gpu, B_gpu)
vectorize_C = vectorize_C_gpu.copy_to_host()
print('------- using vectorize -------')
print('Is close?: {}'.format(np.allclose(numpy_C,vectorize_C)))
print('{} of {} elements are close'.format(np.sum(np.isclose(numpy_C,vectorize_C)), M*N*P))
print('------- using vectorize -------\n')
import numba; print("numba version: "+numba.__version__)
答案 0 :(得分：2)
以下是调试方法。
考虑一个较小的简化示例：
from __future__ import (division, print_function)
import numpy as np
from numba import cuda
M = 2
N = 3
P = 1
threadsperblock = 1
blockspergrid = (M * N * P + (threadsperblock - 1)) // threadsperblock
@cuda.jit
def mult_gpu_3d(a, b, c, grid_ran, grid_multed):
grid = cuda.grid(3)
x, y, z = grid
grid_ran[x] = 1
if (x < c.shape[0]) and (y < c.shape[1]) and (z < c.shape[2]):
grid_multed[x] = 1
c[grid] = a[grid] * b[grid]
if __name__ == '__main__':
A = np.ones((M, N, P), np.int32)
B = np.ones((M, N, P), np.int32)
A_gpu = cuda.to_device(A)
B_gpu = cuda.to_device(B)
C_gpu = cuda.to_device(np.zeros_like(A))
# Tells whether thread at index i have ran
grid_ran = cuda.to_device(np.zeros([blockspergrid], np.int32))
# Tells whether thread at index i have performed multiplication
grid_multed = cuda.to_device(np.zeros(blockspergrid, np.int32))
mult_gpu_3d[blockspergrid, threadsperblock](
A_gpu, B_gpu, C_gpu, grid_ran, grid_multed)
print("grid_ran.shape : ", grid_ran.shape)
print("grid_multed.shape : ", grid_multed.shape)
print("C_gpu.shape : ", C_gpu.shape)
print("grid_ran : ", grid_ran.copy_to_host())
print("grid_multed : ", grid_multed.copy_to_host())
C = C_gpu.copy_to_host()
print("C transpose flat : ", C.T.flatten())
print("C : \n", C)
输出：
grid_ran.shape : (6,)
grid_multed.shape : (6,)
C_gpu.shape : (2, 3, 1)
grid_ran : [1 1 1 1 1 1]
grid_multed : [1 1 0 0 0 0]
C transpose flat : [1 1 0 0 0 0]
C :
[[[1]
[0]
[0]]
[[1]
[0]
[0]]]
您可以看到设备网格形状与数组的形状不对应：网格是平面(M*N*P)，而数组都是三维(M, N, P)。也就是说，网格的第一维具有范围0..M*N*P-1（0..5中的索引，在我的示例中总共6个值），而数组的第一维仅在0..M-1中{{1在我的例子中总计2个值）。这个错误通常导致越界访问，但是您使用条件保护内核，从而减少了违规的线程：
0..1这一行不允许索引高于if (x <= c.shape[0])
（在我的例子中为M-1）的线程运行（好吧，排序[1]），这就是为什么没有写入值并且你得到很多结果数组中的零。
可能的解决方案：
1的3D矢量而不是标量[2]。参考文献：