我对计算两个numpy数组(x和y)之间的各种空间距离感兴趣。
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。我如何仅将其限制为所需的结果。
我想计算[1,11]和[31,41]之间的距离;[2,22]和[32,42]等。
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(这-1是一个占位符,告诉numpy自动沿该轴计算正确的大小):
-1
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和时yy,我们将得到一个2xN的数组:
xx
yy
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到dx和dy组件中:
dx
dy
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)):
np.hypot
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箱并一步一步完成:
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,只需更改np.hypot为要使用的函数即可。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])