在python中使用scipy.spatial.distance.cdist（X，Y）查找一组点之间的距离

I have an array of data, called A that looks something like:

我有一个数组，称为A，看起来像：

array([[0.59, 1.23], [0.89, 1.67], [0.21,0.99]...])

and has about 400 sets of [x,y] points in it. I want to find the distance between every set of points in A to each sets of points in B, which is another array which looks exactly the same as A but is half the length (So about 200 sets of[x,y] points). So if I wanted to find the distance between the q-th pair of [x,y] values in B against all [x,y] values in A, I've tried doing something along the lines of

并且其中有大约400套[x，y]点。我想找到A中每组点与B中每组点之间的距离，这是另一个看起来与A完全相同但是长度的一半的数组（所以大约有200组[x，y]点）。因此，如果我想找到B中第q对[x，y]值与A中所有[x，y]值之间的距离，我尝试按照

import scipy.spatial.distance
for q in range(0,len(B)):
    y=scipy.spatial.distance.cdist(A,B[:q,:])

but I don't think this is working. I just want an output that shows the distance between the q-th row of B against all points in A.

但我认为这不起作用。我只想要一个输出，显示B的第q行与A中所有点之间的距离。

1 个解决方案

#1

Two solutions:

两种解决方案

calculate the complete matrix directly, and the access the q-th column for the values between A and B[q].

直接计算完整矩阵，并访问第q列，得到A和B [q]之间的值。

d = scipy.spatial.distance.cdist(A,B)

for q in range(len(B)):
    y = d[:,q]
    print y

If the resulting matrix is too big to hold in memory. You could do this.

如果结果矩阵太大而无法保存在内存中。你可以做到这一点。

for q in range(len(B)):
    y = scipy.spatial.distance.cdist(A,[B[q]])
    print y

#1