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)
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
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