python – einsum和距离计算

我已经搜索了一个解决方案来确定使用einsum的numpy数组的距离,这些数组的行数不相同,但列数相等.我尝试了各种组合,但我能成功的唯一方法是使用以下代码.我显然缺少一些东西,文献和众多线程并没有让我更接近解决方案.我希望找到一个通用性,使得起源可以是任何数字的目标数组的任何数字.我只使用2D数组,无意将其扩展到其他维度.我也熟悉pdist和cdist以及其他达到我想要的解决方案的方法,但是,我只对einsum感兴趣,因为我想要完善我的一些例子.任何帮助,将不胜感激.

import numpy as np
origs = np.array([[0.,0.],[1.,0.],[0.,1.],[1.,1.]])
dests = np.asarray([[4.,0.],[1.,1.],[2.,2.],[2.,3.],[0.,5.]])
for i in origs:
    d =np.sqrt(np.einsum("ij,ij->i", i-dests, i-dests))
    print("orig {}...dist: {}".format(i,d))

我正在寻找以下结果……

orig [ 0.  0.]...dist: [ 4.          1.41421356  2.82842712  3.60555128  5.        ]
orig [ 1.  0.]...dist: [ 3.          1.          2.23606798  3.16227766  5.09901951]
orig [ 0.  1.]...dist: [ 4.12310563  1.          2.23606798  2.82842712  4.        ]
orig [ 1.  1.]...dist: [ 3.16227766  0.          1.41421356  2.23606798  4.12310563]

解决方法:

如果我正确理解了这个问题,那么在考虑2D数组时,你发布的for循环代码对我来说是通用的.现在,如果您希望通过一次调用np.einsum来获得通用的矢量化解决方案,那么您可以将broadcasting引入游戏,就像这样 –

d_all = np.sqrt(np.einsum('ijk->ij',(origs[:,None,:] - dests)**2))

样品运行 –

In [85]: origs = np.array([[0.,0.],[1.,0.],[0.,1.],[1.,1.]])
    ...: dests = np.asarray([[4.,0.],[1.,1.],[2.,2.],[2.,3.],[0.,5.]])
    ...: 

In [86]: for i in origs:
    ...:     d =np.sqrt(np.einsum("ij,ij->i", i-dests, i-dests))
    ...:     print(d)
    ...:     
[ 4.          1.41421356  2.82842712  3.60555128  5.        ]
[ 3.          1.          2.23606798  3.16227766  5.09901951]
[ 4.12310563  1.          2.23606798  2.82842712  4.        ]
[ 3.16227766  0.          1.41421356  2.23606798  4.12310563]

In [87]: np.sqrt(np.einsum('ijk->ij',(origs[:,None,:] - dests)**2))
Out[87]: 
array([[ 4.        ,  1.41421356,  2.82842712,  3.60555128,  5.        ],
       [ 3.        ,  1.        ,  2.23606798,  3.16227766,  5.09901951],
       [ 4.12310563,  1.        ,  2.23606798,  2.82842712,  4.        ],
       [ 3.16227766,  0.        ,  1.41421356,  2.23606798,  4.12310563]])

根据comments by @hpaulj,你也可以用np.einsum本身进行平方,就像这样 –

subts = origs[:,None,:] - dests
d_all = np.sqrt(np.einsum('ijk,ijk->ij',subts,subts))

这是一个运行时测试,将其与之前在np.einsum之外进行平方的方法进行比较 –

In [7]: def all_einsum(origs,dests):
   ...:     subts = origs[:,None,:] - dests
   ...:     return np.sqrt(np.einsum('ijk,ijk->ij',subts,subts))
   ...: 
   ...: def partial_einsum(origs,dests):
   ...:     return np.sqrt(np.einsum('ijk->ij',(origs[:,None,:] - dests)**2))
   ...: 

In [8]: origs = np.random.rand(400,100)

In [9]: dests = np.random.rand(500,100)

In [10]: %timeit all_einsum(origs,dests)
10 loops, best of 3: 139 ms per loop

In [11]: %timeit partial_einsum(origs,dests)
1 loops, best of 3: 251 ms per loop

标签：euclidean-distance,python,arrays,numpy
来源： https://codeday.me/bug/20190727/1557868.html