用 Scipy 计算两个矩阵的行方向点积的矢量化方法 [英] Vectorized way of calculating row-wise dot product two matrices with Scipy

问题描述

我想尽快计算相同维度的两个矩阵的行式点积.我的做法是这样的:

I want to calculate the row-wise dot product of two matrices of the same dimension as fast as possible. This is the way I am doing it:

import numpy as np
a = np.array([[1,2,3], [3,4,5]])
b = np.array([[1,2,3], [1,2,3]])
result = np.array([])
for row1, row2 in a, b:
    result = np.append(result, np.dot(row1, row2))
print result

当然输出是:

[ 26.  14.]

推荐答案

查看 numpy.einsum 用于另一种方法:

Check out numpy.einsum for another method:

In [52]: a
Out[52]: 
array([[1, 2, 3],
       [3, 4, 5]])

In [53]: b
Out[53]: 
array([[1, 2, 3],
       [1, 2, 3]])

In [54]: einsum('ij,ij->i', a, b)
Out[54]: array([14, 26])

看起来 einsum 比 inner1d 快一点:

Looks like einsum is a bit faster than inner1d:

In [94]: %timeit inner1d(a,b)
1000000 loops, best of 3: 1.8 us per loop

In [95]: %timeit einsum('ij,ij->i', a, b)
1000000 loops, best of 3: 1.6 us per loop

In [96]: a = random.randn(10, 100)

In [97]: b = random.randn(10, 100)

In [98]: %timeit inner1d(a,b)
100000 loops, best of 3: 2.89 us per loop

In [99]: %timeit einsum('ij,ij->i', a, b)
100000 loops, best of 3: 2.03 us per loop

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注意:NumPy 在不断发展和改进；多年来，上述功能的相对性能可能发生了变化.如果性能对您很重要，请使用您将使用的 NumPy 版本运行您自己的测试.

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Note: NumPy is constantly evolving and improving; the relative performance of the functions shown above has probably changed over the years. If performance is important to you, run your own tests with the version of NumPy that you will be using.

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