如何使用numpy查找两个非常大的矩阵的行之间的成对差异？

Question

给定两个矩阵，我想计算所有行之间的成对差异。每个矩阵都有1000行和100列，因此它们相当大。我尝试过使用for循环和纯广播，但是for循环的运行速度似乎更快。难道我做错了什么？这是代码：

from numpy import *
A = random.randn(1000,100)
B = random.randn(1000,100)

start = time.time()
for a in A:
   sum((a - B)**2,1)
print time.time() - start

# pure broadcasting
start = time.time()
((A[:,newaxis,:] - B)**2).sum(-1)
print time.time() - start

广播方法大约需要1秒钟的时间，对于大型矩阵来说甚至更长。任何想法如何纯粹使用numpy加快速度吗？

Answer 1

这是另一种执行方式：

（ab）^ 2 = a ^ 2 + b ^ 2-2ab

与np.einsum前两个条款dot- product的第三个-

import numpy as np

np.einsum('ij,ij->i',A,A)[:,None] + np.einsum('ij,ij->i',B,B) - 2*np.dot(A,B.T)

运行时测试

方法-

def loopy_app(A,B):
    m,n = A.shape[0], B.shape[0]
    out = np.empty((m,n))
    for i,a in enumerate(A):
       out[i] = np.sum((a - B)**2,1)
    return out

def broadcasting_app(A,B):
    return ((A[:,np.newaxis,:] - B)**2).sum(-1)

# @Paul Panzer's soln
def outer_sum_dot_app(A,B):
    return np.add.outer((A*A).sum(axis=-1), (B*B).sum(axis=-1)) - 2*np.dot(A,B.T)

# @Daniel Forsman's soln
def einsum_all_app(A,B):
    return np.einsum('ijk,ijk->ij', A[:,None,:] - B[None,:,:], \
                                        A[:,None,:] - B[None,:,:])

# Proposed in this post
def outer_einsum_dot_app(A,B):
    return np.einsum('ij,ij->i',A,A)[:,None] + np.einsum('ij,ij->i',B,B) - \
                                                            2*np.dot(A,B.T)

时间-

In [51]: A = np.random.randn(1000,100)
    ...: B = np.random.randn(1000,100)
    ...:

In [52]: %timeit loopy_app(A,B)
    ...: %timeit broadcasting_app(A,B)
    ...: %timeit outer_sum_dot_app(A,B)
    ...: %timeit einsum_all_app(A,B)
    ...: %timeit outer_einsum_dot_app(A,B)
    ...: 
10 loops, best of 3: 136 ms per loop
1 loops, best of 3: 302 ms per loop
100 loops, best of 3: 8.51 ms per loop
1 loops, best of 3: 341 ms per loop
100 loops, best of 3: 8.38 ms per loop