为什么a.dot(b)比a @ b快，尽管Numpy建议a @ b [英] Why is a.dot(b) faster than a@b although Numpy recommends a@b

问题描述

根据此问题的答案并根据 numpy ，与a.dot(b)相比，最好通过a @ b或numpy.matmul(a,b)完成二维数组的矩阵乘法.

According to the answers from this question and also according to numpy, matrix multiplication of 2-D arrays is best done via a @ b, or numpy.matmul(a,b) as compared to a.dot(b).

如果a和b均为二维数组，则为矩阵乘法，但使用 最好使用matmul或a @ b.

If both a and b are 2-D arrays, it is matrix multiplication, but using matmul or a @ b is preferred.

我进行了以下基准测试，发现相反的结果.

I did the following benchmark and found contrary results.

问题:我的基准测试有问题吗?如果不是，为什么当Numpy比a@b或numpy.matmul(a,b)快时，为什么不推荐a.dot(b)?

Questions: Is there an issue with my benchmark? If not, why does Numpy not recommend a.dot(b) when it is faster than a@b or numpy.matmul(a,b)?

基准使用python 3.5 numpy 1.15.0.

Benchmark used python 3.5 numpy 1.15.0.

$ pip3 list | grep numpy
numpy                         1.15.0
$ python3 --version
Python 3.5.2

基准代码:

import timeit

setup = '''
import numpy as np
a = np.arange(16).reshape(4,4)
b = np.arange(16).reshape(4,4)
''' 
test = '''
for i in range(1000):
    a @ b
'''
test1 = '''
for i in range(1000):
    np.matmul(a,b)
'''
test2 = '''
for i in range(1000):
    a.dot(b)
'''

print( timeit.timeit(test, setup, number=100) )
print( timeit.timeit(test1, setup, number=100) )
print( timeit.timeit(test2, setup, number=100) )

结果:

test  : 0.11132473500038031
test1 : 0.10812476599676302
test2 : 0.06115105600474635

添加结果:

>>> a = np.arange(16).reshape(4,4)
>>> b = np.arange(16).reshape(4,4)
>>> a@b
array([[ 56,  62,  68,  74],
       [152, 174, 196, 218],
       [248, 286, 324, 362],
       [344, 398, 452, 506]])
>>> np.matmul(a,b)
array([[ 56,  62,  68,  74],
       [152, 174, 196, 218],
       [248, 286, 324, 362],
       [344, 398, 452, 506]])
>>> a.dot(b)
array([[ 56,  62,  68,  74],
       [152, 174, 196, 218],
       [248, 286, 324, 362],
       [344, 398, 452, 506]])

推荐答案

您的前提不正确.您应该使用更大的矩阵来衡量性能，以避免函数调用使无关紧要的计算相形见

Your premise is incorrect. You should use larger matrices to measure performance to avoid function calls dwarfing insignificant calculations.

您会发现使用Python 3.60/NumPy 1.11.3，如

Using Python 3.60 / NumPy 1.11.3 you will find, as explained here, that @ calls np.matmul and both outperform np.dot.

import numpy as np

n = 500
a = np.arange(n**2).reshape(n, n)
b = np.arange(n**2).reshape(n, n)

%timeit a.dot(b)        # 134 ms per loop
%timeit a @ b           # 71 ms per loop
%timeit np.matmul(a,b)  # 70.6 ms per loop

还要注意，如文档中所述，np.dot在功能上与@/np.matmul不同.特别是，它们在处理尺寸大于2的矩阵方面有所不同.

Also note, as explained in the docs, np.dot is functionally different to @ / np.matmul. In particular, they differ in treatment of matrices with dimensions greater than 2.

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