在Python / Numpy中优化许多Matrix操作(Optimizing Many Matrix Operations in Python / Numpy)

在Python / Numpy中优化许多Matrix操作(Optimizing Many Matrix Operations in Python / Numpy)

在编写一些数值分析代码时，我对需要许多Numpy调用的函数进行了瓶颈处理。 我不完全确定如何进一步进行性能优化。

问题：

该函数通过计算以下内容来确定错误，

码：

def foo(B_Mat, A_Mat): Temp = np.absolute(B_Mat) Temp /= np.amax(Temp) return np.sqrt(np.sum(np.absolute(A_Mat - Temp*Temp))) / B_Mat.shape[0]

什么是从代码中挤出一些额外性能的最佳方法？ 我最好的行动方案是使用Cython在单个for循环中执行大部分操作来减少临时阵列吗？

In writing some numerical analysis code, I have bottle-necked at a function that requires many Numpy calls. I am not entirely sure how to approach further performance optimization.

Problem:

The function determines error by calculating the following,

Code:

def foo(B_Mat, A_Mat): Temp = np.absolute(B_Mat) Temp /= np.amax(Temp) return np.sqrt(np.sum(np.absolute(A_Mat - Temp*Temp))) / B_Mat.shape[0]

What would be the best way to squeeze some extra performance out of the code? Would my best course of action be performing the majority of the operations in a single for loop with Cython to cut down on the temporary arrays?

最满意答案

从实现中可以卸载到numexpr模块的特定函数，已知这对于算术计算非常有效。 对于我们的情况，具体来说我们可以用它来执行平方，求和和绝对计算。 因此，基于numexpr的解决方案将取代原始代码中的最后一步，就像这样 -

import numexpr as ne out = np.sqrt(ne.evaluate('sum(abs(A_Mat - Temp**2))'))/B_Mat.shape[0]

通过将规范化步骤嵌入到numexpr的evaluate表达式中，可以实现进一步的性能提升。 因此，修改为使用numexpr的整个函数将是 -

def numexpr_app1(B_Mat, A_Mat): Temp = np.absolute(B_Mat) M = np.amax(Temp) return np.sqrt(ne.evaluate('sum(abs(A_Mat*M**2-Temp**2))'))/(M*B_Mat.shape[0])

运行时测试 -

In [198]: # Random arrays ...: A_Mat = np.random.randn(4000,5000) ...: B_Mat = np.random.randn(4000,5000) ...: In [199]: np.allclose(foo(B_Mat, A_Mat),numexpr_app1(B_Mat, A_Mat)) Out[199]: True In [200]: %timeit foo(B_Mat, A_Mat) 1 loops, best of 3: 891 ms per loop In [201]: %timeit numexpr_app1(B_Mat, A_Mat) 1 loops, best of 3: 400 ms per loop

There are specific functions from the implementation that could be off-loaded to numexpr module which is known to be very efficient for arithmetic computations. For our case, specifically we could perform squaring, summation and absolute computations with it. Thus, a numexpr based solution to replace the last step in the original code, would be like so -

import numexpr as ne out = np.sqrt(ne.evaluate('sum(abs(A_Mat - Temp**2))'))/B_Mat.shape[0]

A further performance boost could be achieved by embedding the normalization step into the numexpr's evaluate expression. Thus, the entire function modified to use numexpr would be -

def numexpr_app1(B_Mat, A_Mat): Temp = np.absolute(B_Mat) M = np.amax(Temp) return np.sqrt(ne.evaluate('sum(abs(A_Mat*M**2-Temp**2))'))/(M*B_Mat.shape[0])

Runtime test -

In [198]: # Random arrays ...: A_Mat = np.random.randn(4000,5000) ...: B_Mat = np.random.randn(4000,5000) ...: In [199]: np.allclose(foo(B_Mat, A_Mat),numexpr_app1(B_Mat, A_Mat)) Out[199]: True In [200]: %timeit foo(B_Mat, A_Mat) 1 loops, best of 3: 891 ms per loop In [201]: %timeit numexpr_app1(B_Mat, A_Mat) 1 loops, best of 3: 400 ms per loop