How to find the closests points of two numpy arrays in python

Question

I have two numpy arrays: ones is some coordinates (x, y, z) with an id and another one is only some coordinates (x, y, z):

cord_id=np.array([[0.,0.,0.,8.],[0.,0.,10.,8.],[0.,0.,20.,8.],[0.,0.,30.,8.],\
                  [2.,1.,0.,9.],[2.,1.,10.,7.],[2.,1.,20.,7.],[2.,1.,30.,7.]])
cord_only=np.array([[0.,0.,.1],[0.,0.,20.1],[2.1,1.,0.],[2.,1.,11.1]])

I want to find out each point of cord_only is closer to which point of cord_id and then add the related id (last column of cord_id) to the point. For example, the first point of cord_only is closest to first point of cord_id, so I add 8. to it. Second point of cord_only is closest to third of cord_id, third is closest to fifth and fourth is also closet to sixth point. Finally, I want to get it as:

new_arr=np.array([[0.,0.,.1,8.],[0.,0.,20.1,8.],[2.1,1.,0.,9.],[2.,1.,11.1,7.]])

I tried the following code but could not find out the closet points:

from scipy.spatial import distance
cord_id[np.where(np.min(distance.cdist(cord_only, cord_id[:,:-1]),axis=0))]

i do appreciate any help in advance to do it in python.

user3235916 · Accepted Answer · 2021-06-10 08:55:05Z

2

using np.where(np.min(X)) doesn't give you the correct answer as min returns the minimum value (rather than the index of the minimum value) and where will return all nonzeros. I think what you are looking for is argmin:

from scipy.spatial import distance
ids = cord_id[np.argmin(distance.cdist(cord_only, cord_id[:,:-1]),axis=1)][:,-1]
new_arr = np.hstack([cord_only,ids.reshape(-1,1)])

answered Jun 10, 2021 at 8:55

user3235916

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