Bor*_*rys 4 python numpy scipy
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。
http://docs.scipy.org/doc/scipy-0.14.0/reference/generated/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
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但是,上述结果会产生太多不需要的结果。如何仅针对我需要的结果限制它。
我想计算 [1,11] 和 [31,41] 之间的距离;[2,22] 和 [32,42] 等等。
如果你只想要每对点之间的距离,那么你不需要计算一个完整的距离矩阵。
相反,直接计算它:
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
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为了更多地解释发生了什么,首先我们重塑数组,使它们具有 2xN 形状(-1是一个占位符,它告诉 numpy 自动计算沿该轴的正确大小):
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]])
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因此,当我们减去xxand 时yy,我们将得到一个 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]])
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然后我们可以将其解压到dx和dy组件中:
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])
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并计算距离(np.hypot相当于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. ])
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或者我们可以自动完成解包并一步完成所有操作:
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. ])
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如果您想计算其他类型的距离,只需更改np.hypot为您要使用的函数即可。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])
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