Question

在过去的几天里，我一直试图理解为什么Numbapro（加速来自Continuum Analytics，Inc。;我正在运行30天试用版）在我的MacBook Pro上没有加速（英特尔酷睿i7），2.6GHz，16GB RAM，配备NVIDIA GeForce GT 650M，1GB PCI总线）。

我从（NxM）x（MxN）矩阵乘法的代码中选取了一个例子，其中Continuum Analytics，Inc。声称通过CUDA加速计算，我比较了CUDA.JIT和numpy之间的时间。我的想法是运行例如1e4 迭代，并且矩阵B在每次迭代时被随机化。在下面我使用的代码下面，我引用了我获得的时间。那有什么解决方案吗？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上一个完全天真的矩阵乘法例程，而numpy例程实际上是一个库调用：

X=np.dot(A,B)

可能会高度优化。令我印象深刻的是GPU更快。

＆＃34;解决方案＆＃34;将make a call to CUBLAS用于矩阵多重复制，而不是编写自己的内核。

Numbapro：Matrix Multiplication没有加速

1 个答案: