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.
而是直接计算:
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
为了进一步说明正在发生的事情,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):
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,将得到一个2xN数组:
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:
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等于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:
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更改为您要使用的函数.例如,对于曼哈顿/城市街区距离:
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])