python – Anaconda的NumbaPro CUDA断言错误

我正在尝试使用NumbaPro的cuda扩展来增加大型阵列矩阵.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,a.dot(numpy.diagflat(b))我发现它是一个同义词* b).但是,我收到的断言错误没有提供任何信息.

如果我将两个1D阵列矩阵相乘,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

解决方法:

问题是你在一个带有非标量参数的函数上使用了vectorize.使用NumbaPro的矢量化的想法是它将标量函数作为输入,并生成一个函数,将标量运算并行应用于矢量的所有元素.见NumbaPro documentation.

你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包装器或编写自己的简单内核函数,在GPU上执行您想要的操作.这是一个演示两者的例子.注意将需要NumbaPro 0.12.2或更高版本(刚刚在此编辑时发布).

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执行完整的矩阵 – 矩阵乘法(O(n ^ 3)),并将对角线扩展为完整矩阵. diagproduct功能更智能.它只是对每个矩阵元素进行单次乘法运算,并且从不将对角线扩展为完整矩阵.以下是我的NVIDIA Tesla K20c GPU上N = 1000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GPU的所有副本,这是小型矩阵的一个重要瓶颈.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,因此它更接近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

标签：python,numpy,anaconda,numba,numba-pro
来源： https://codeday.me/bug/20190620/1243530.html