Matlab至python转换矩阵运算

Question

Hi I am trying to covert this distance formula for rectilinear distance from matlab to python. X1 and X2 are two matrices of two dimensional points and could be differing lengths.

nd = size(X1); n = nd(1); 
d = nd(2);  
m = size(X2,1);

D = abs(X1(:,ones(1,m)) - X2(:,ones(1,n))') + ...
     abs(X1(:,2*ones(1,m)) - X2(:,2*ones(1,n))');

I think the problem I am having most in python is appending the ones matrices with X1 and X2 since they are np.arrays.

Answer 1

First your code:

octave:1> X1=[0,1,2,3;2,3,1,1]'
octave:2> X2=[2,3,2;4,2,4]'
<your code>
octave:21> D
D =
   4   3   4
   2   3   2
   3   2   3
   4   1   4

Matching numpy code:

X1=np.array([[0,1,2,3],[2,3,1,1]]).T
X2=np.array([[2,3,2],[4,2,4]]).T
D=np.abs(X1[:,None,:]-X2[None,:,:]).sum(axis=-1)

produces, D :

array([[4, 3, 4],
       [2, 3, 2],
       [3, 2, 3],
       [4, 1, 4]])

numpy broadcasts automatically, so it doesn't need the ones() to expand the dimensions. Instead I use None (same as np.newaxis ) to create new dimensions. The difference is then 3D , which is then summed on the last axis.

I forgot how spoilt we are with the numpy broadcasting . Though newer Octave has something similar:

D = sum(abs(reshape(X1,[],1,2)-reshape(X2,1,[],2)),3)