Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
620 views
in Technique[技术] by (71.8m points)

python - calculating distance between two numpy arrays

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/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

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

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

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

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

Instead, calculate it directly:

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

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]])

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]])

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])

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.        ])

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.        ])

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])

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...