我正在尝试使用 NumbaPro 的 cuda 扩展来乘以大型数组矩阵。我最终想要的是将大小为 NxN 的矩阵乘以一个对角矩阵，该矩阵将作为一维矩阵输入(因此，我发现 a.dot(numpy.diagflat(b)) 与 a * b)。但是，我收到一个断言错误，没有提供任何信息。
如果我将两个一维数组矩阵相乘，我只能避免这个断言错误，但这不是我想要做的。
from numbapro import vectorize, cuda
from numba import f4,f8
import numpy as np
def generate_input(n):
import numpy as np
A = np.array(np.random.sample((n,n)))
B = np.array(np.random.sample(n) + 10)
return A, B
def product(a, b):
return a * b
def main():
cu_product = vectorize([f4(f4, f4), f8(f8, f8)], target='gpu')(product)
N = 1000
A, B = generate_input(N)
D = np.empty(A.shape)
stream = cuda.stream()
with stream.auto_synchronize():
dA = cuda.to_device(A, stream)
dB = cuda.to_device(B, stream)
dD = cuda.to_device(D, stream, copy=False)
cu_product(dA, dB, out=dD, stream=stream)
dD.to_host(stream)
if __name__ == '__main__':
main()
这是我的终端吐出来的:
Traceback (most recent call last):
File "cuda_vectorize.py", line 32, in <module>
main()
File "cuda_vectorize.py", line 28, in main
cu_product(dA, dB, out=dD, stream=stream)
File "/opt/anaconda1anaconda2anaconda3/lib/python2.7/site-packages/numbapro/_cudadispatch.py", line 109, in __call__
File "/opt/anaconda1anaconda2anaconda3/lib/python2.7/site-packages/numbapro/_cudadispatch.py", line 191, in _arguments_requirement
AssertionError
最佳答案
问题是您在采用非标量参数的函数上使用 vectorize。 NumbaPro 的 vectorize 的想法是它将标量函数作为输入，并生成一个函数，该函数将标量运算并行应用于向量的所有元素。请参阅 NumbaPro documentation 。
您的函数采用矩阵和向量，它们绝对不是标量。 [编辑] 您可以使用 NumbaPro 的 cuBLAS 包装器在 GPU 上执行您想要的操作，或者编写您自己的简单内核函数。这是一个演示两者的示例。注意将需要 NumbaPro 0.12.2 或更高版本(在本次编辑时刚刚发布)。
from numbapro import jit, cuda
from numba import float32
import numbapro.cudalib.cublas as cublas
import numpy as np
from timeit import default_timer as timer
def generate_input(n):
A = np.array(np.random.sample((n,n)), dtype=np.float32)
B = np.array(np.random.sample(n), dtype=A.dtype)
return A, B
@cuda.jit(argtypes=[float32[:,:], float32[:,:], float32[:]])
def diagproduct(c, a, b):
startX, startY = cuda.grid(2)
gridX = cuda.gridDim.x * cuda.blockDim.x;
gridY = cuda.gridDim.y * cuda.blockDim.y;
height, width = c.shape
for y in range(startY, height, gridY):
for x in range(startX, width, gridX):
c[y, x] = a[y, x] * b[x]
def main():
N = 1000
A, B = generate_input(N)
D = np.empty(A.shape, dtype=A.dtype)
E = np.zeros(A.shape, dtype=A.dtype)
F = np.empty(A.shape, dtype=A.dtype)
start = timer()
E = np.dot(A, np.diag(B))
numpy_time = timer() - start
blas = cublas.api.Blas()
start = timer()
blas.gemm('N', 'N', N, N, N, 1.0, np.diag(B), A, 0.0, D)
cublas_time = timer() - start
diff = np.abs(D-E)
print("Maximum CUBLAS error %f" % np.max(diff))
blockdim = (32, 8)
griddim = (16, 16)
start = timer()
dA = cuda.to_device(A)
dB = cuda.to_device(B)
dF = cuda.to_device(F, copy=False)
diagproduct[griddim, blockdim](dF, dA, dB)
dF.to_host()
cuda_time = timer() - start
diff = np.abs(F-E)
print("Maximum CUDA error %f" % np.max(diff))
print("Numpy took %f seconds" % numpy_time)
print("CUBLAS took %f seconds, %0.2fx speedup" % (cublas_time, numpy_time / cublas_time))
print("CUDA JIT took %f seconds, %0.2fx speedup" % (cuda_time, numpy_time / cuda_time))
if __name__ == '__main__':
main()
内核明显更快，因为 SGEMM 执行完整矩阵-矩阵乘法 (O(n^3))，并将对角线扩展为完整矩阵。
diagproduct 函数更智能。它只是对每个矩阵元素进行一次乘法运算，并且永远不会将对角线扩展为完整矩阵。以下是 N=1000 在我的 NVIDIA Tesla K20c GPU 上的结果:Maximum CUBLAS error 0.000000
Maximum CUDA error 0.000000
Numpy took 0.024535 seconds
CUBLAS took 0.010345 seconds, 2.37x speedup
CUDA JIT took 0.004857 seconds, 5.05x speedup
时序包括进出 GPU 的所有副本，这对于小矩阵来说是一个重大瓶颈。如果我们将 N 设置为 10,000 并再次运行，我们将获得更大的加速:
Maximum CUBLAS error 0.000000
Maximum CUDA error 0.000000
Numpy took 7.245677 seconds
CUBLAS took 1.371524 seconds, 5.28x speedup
CUDA JIT took 0.264598 seconds, 27.38x speedup
但是，对于非常小的矩阵，CUBLAS SGEMM 具有优化的路径，因此更接近 CUDA 性能。这里，N=100
Maximum CUBLAS error 0.000000
Maximum CUDA error 0.000000
Numpy took 0.006876 seconds
CUBLAS took 0.001425 seconds, 4.83x speedup
CUDA JIT took 0.001313 seconds, 5.24x speedup
关于python - Anaconda 的 NumbaPro CUDA 断言错误，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/17153985/