如何使用numpy查找两个非常大的矩阵的行之间的成对差异？

如何使用numpy查找两个非常大的矩阵的行之间的成对差异？

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 npnp.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 outdef broadcasting_app(A,B):    return ((A[:,np.newaxis,:] - B)**2).sum(-1)# @Paul Panzer's solndef 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 solndef einsum_all_app(A,B):    return np.einsum('ijk,ijk->ij', A[:,None,:] - B[None,:,:],        A[:,None,:] - B[None,:,:])# Proposed in this postdef 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 loop1 loops, best of 3: 302 ms per loop100 loops, best of 3: 8.51 ms per loop1 loops, best of 3: 341 ms per loop100 loops, best of 3: 8.38 ms per loop