python - 编号 : No speed-up for Matrix Multiplication

Question

在过去的几天里，我一直在努力理解为什么 Numbapro(来自 Continuum Analytics, Inc. 的加速器；我正在运行 30 天试用版)在我的 MacBook Pro(Intel Core i7、2.6GHz、 16GB RAM，NVIDIA GeForce GT 650M，1GB PCI 总线)。

我从 (NxM)x(MxN) 矩阵乘法的代码中获取了一个示例，其中 Continuum Analytics, Inc. 声称通过 CUDA 加速了计算，我比较了 CUDA.JIT 和 numpy 之间的时间。我的想法是运行例如 1e4 iterations 并且矩阵 B 在每次迭代中随机化。在我使用的以下代码下方，我引用了我获得的时间。有什么解决办法吗？谢谢!

from numbapro import *
from numba import *
import numpy as np
import math
from timeit import default_timer as timer

m=1000
n=1000
A = np.array(np.random.random((n,m)), dtype=np.float32)
C = np.empty([n,n])

iterations = 10000

start = timer()
for i in range(iterations):
    B = np.array(np.random.random((m,n)), dtype=np.float32)
    X=np.dot(A,B)
numpy_time=(timer() - start)

@cuda.jit(void(float32[:,:],float32[:,:],float32[:,:]))
def cu_square_matrix_mul(A, B, C):

    tx = cuda.threadIdx.x
    ty = cuda.threadIdx.y
    bx = cuda.blockIdx.x
    by = cuda.blockIdx.y
    bw = cuda.blockDim.x
    bh = cuda.blockDim.y
    x = tx + bx * bw
    y = ty + by * bh
    n = C.shape[0]

    if x >= n or y >= n:
        return

    cs = 0
    for i in range(n):
        cs += A[y,i]*B[i,x]
    C[y,x]= cs

    cuda.syncthreads()

blockdim = 256,3
griddim = 10,3

stream = cuda.stream()
dA = cuda.to_device(A, stream)
dC = cuda.to_device(C, stream)

start = timer()    
for i in range(iterations):
    B = np.array(np.random.random((m,n)), dtype=np.float32)
    dB = cuda.to_device(B, stream)
    cu_square_matrix_mul[griddim,blockdim,stream](dA, dB, dC) 
    dC.to_host()
    stream.synchronize()
cuda_time = (timer() - start)    

print
print("Numpy took    %f seconds" % numpy_time)
print("CUDA JIT took %f seconds, %.5fx speedup" % (cuda_time, numpy_time / cuda_time))

结果:

Vendor:  Continuum Analytics, Inc.
Package: mkl
Message: trial mode expires in 30 days
Vendor:  Continuum Analytics, Inc.
Package: mkl
Message: trial mode expires in 30 days
Vendor:  Continuum Analytics, Inc.
Package: numbapro
Message: trial mode expires in 30 days

Numpy took    378.328881 seconds
CUDA JIT took 342.723757 seconds, 1.10389x speedup

Answer 1

这是 GPU 上一个完全朴素的矩阵乘法例程，而 numpy 例程实际上是一个库调用:

X=np.dot(A,B)

可能会被高度优化。 GPU 的速度更快给我留下了深刻的印象。

“解决方案”是 make a call to CUBLAS用于矩阵乘法，而不是编写自己的内核。