计算两个numpy数组之间的距离 [英] calculating distance between two numpy arrays

问题描述

我对计算两个numpy数组(x和y)之间的各种空间距离很感兴趣.

I was interested in calculating various spatial distances between two numpy arrays (x and y).

http://docs.scipy.org/doc/scipy-0.14.0/reference/generation/scipy.spatial.distance.cdist.html

import numpy as np
from scipy.spatial.distance import cdist

x = np.array([[[1,2,3,4,5],
               [5,6,7,8,5],
               [5,6,7,8,5]],
              [[11,22,23,24,5],
               [25,26,27,28,5],
               [5,6,7,8,5]]])
i,j,k = x.shape

xx = x.reshape(i,j*k).T

y = np.array([[[31,32,33,34,5],
               [35,36,37,38,5],
               [5,6,7,8,5]],
              [[41,42,43,44,5],
               [45,46,47,48,5],
               [5,6,7,8,5]]])

yy = y.reshape(i,j*k).T

results =  cdist(xx,yy,'euclidean')
print results

但是，以上结果会产生太多不必要的结果.我该如何仅将其限制为所需的结果.

However, above results produces too many unwanted results. How can I limit it for my required results only.

我想计算[1,11]和[31,41]之间的距离； [2,22]和[32,42]，...等等.

I want to calculate distance between [1,11] and [31,41]; [2,22] and [32,42],...and so on.

推荐答案

如果只需要每对点之间的距离，则无需计算完整的距离矩阵.

If you just want the distances between each pair of points, then you don't need to calculate a full distance matrix.

相反，直接进行计算:

import numpy as np

x = np.array([[[1,2,3,4,5],
               [5,6,7,8,5],
               [5,6,7,8,5]],
              [[11,22,23,24,5],
               [25,26,27,28,5],
               [5,6,7,8,5]]])

y = np.array([[[31,32,33,34,5],
               [35,36,37,38,5],
               [5,6,7,8,5]],
              [[41,42,43,44,5],
               [45,46,47,48,5],
               [5,6,7,8,5]]])

xx = x.reshape(2, -1)
yy = y.reshape(2, -1)
dist = np.hypot(*(xx - yy))

print dist

为进一步说明发生的情况，首先我们对数组进行整形，使其具有2xN的形状(-1是一个占位符，告诉numpy自动沿该轴计算正确的大小):

To explain a bit more about what's going on, first we reshape the arrays such that they have a 2xN shape (-1 is a placeholder that tells numpy to calculate the correct size along that axis automatically):

In [2]: x.reshape(2, -1)
Out[2]: 
array([[ 1,  2,  3,  4,  5,  5,  6,  7,  8,  5,  5,  6,  7,  8,  5],
       [11, 22, 23, 24,  5, 25, 26, 27, 28,  5,  5,  6,  7,  8,  5]])

因此，当我们减去xx和yy时，我们将得到一个2xN的数组:

Therefore, when we subtract xx and yy, we'll get a 2xN array:

In [3]: xx - yy
Out[3]: 
array([[-30, -30, -30, -30,   0, -30, -30, -30, -30,   0,   0,   0,   0,
          0,   0],
       [-30, -20, -20, -20,   0, -20, -20, -20, -20,   0,   0,   0,   0,
          0,   0]])

然后我们可以将其解压缩为dx和dy组件:

We can then unpack this in to dx and dy components:

In [4]: dx, dy = xx - yy

In [5]: dx
Out[5]: 
array([-30, -30, -30, -30,   0, -30, -30, -30, -30,   0,   0,   0,   0,
         0,   0])

In [6]: dy
Out[6]: 
array([-30, -20, -20, -20,   0, -20, -20, -20, -20,   0,   0,   0,   0,
         0,   0])

并计算距离(np.hypot等同于np.sqrt(dx**2 + dy**2)):

And calculate the distance (np.hypot is equivalent to np.sqrt(dx**2 + dy**2)):

In [7]: np.hypot(dx, dy)
Out[7]: 
array([ 42.42640687,  36.05551275,  36.05551275,  36.05551275,
         0.        ,  36.05551275,  36.05551275,  36.05551275,
        36.05551275,   0.        ,   0.        ,   0.        ,
         0.        ,   0.        ,   0.        ])

或者我们可以自动完成拆包并一步一步完成所有操作:

Or we can have the unpacking done automatically and do it all in one step:

In [8]: np.hypot(*(xx - yy))
Out[8]: 
array([ 42.42640687,  36.05551275,  36.05551275,  36.05551275,
         0.        ,  36.05551275,  36.05551275,  36.05551275,
        36.05551275,   0.        ,   0.        ,   0.        ,
         0.        ,   0.        ,   0.        ])

如果要计算其他类型的距离，只需将np.hypot更改为要使用的功能.例如，对于曼哈顿/城市街区距离:

If you want to calculate other types of distances, just change np.hypot to the function you'd like to use. For example, for Manhattan/city-block distances:

In [9]: dist = np.sum(np.abs(xx - yy), axis=0)

In [10]: dist
Out[10]: array([60, 50, 50, 50,  0, 50, 50, 50, 50,  0,  0,  0,  0,  0,  0])

这篇关于计算两个numpy数组之间的距离的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！