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I have two numpy lists with 3D coordinates of vertecies of two meshes. These lists are of unequal length. I would like to calculate the 3D distance between every coordinate in arrayA and arrayB.

arrayA([[X1 , Y1 , Z1], [X2 , Y2 , Z2], [X3 , Y3 , Z3]]) ... arrayB([[X1 , Y1 , Z1], [X2 , Y2 , Z2], [X3 , Y3 , Z3], [X4 , Y4 , Z4]]) ...

While searching I found this post about getting a single 3D coordinate pair distance but I am not sure how I can use it as my numpy arrays are of unequal lengths.

I would like to generate a numpy array with all the lengths. This means distances of All 3D coordinates from ArrayA with the first, second, third... (and so on) 3D coordinate of ArrayB.

---EDIT---

I feel like I have been able to make some progress but I think that something is still not right. I am experimenting with a subset of all the coordinates ant took the first 3 vertices from the two meshes. So ArrayA (vertex_array[0]) has 3 XYZ coordinates and ArrayB (vertex_array[1]) has 3 XYZ coordinates. I am expecting to get 9 output entries but it outputs 12. I have also put in a couple of print statements but I can't seem to figure it out.

import numpy as np ... iterate = 0 iterate2 = 0 ... for entries in vertex_array[1]: for vertex in vertex_array[0]: p1 = vertex p2 = vertex_array[1][0] iterate = iterate+1 squared_dist = np.sum((p1-p2)**2, axis=0) dist = np.sqrt(squared_dist) print(dist) print("loop 1: " + str(iterate)) p3 = vertex_array[0][0] p4 = entries iterate2 = iterate2 + 1 squared_dist2 = np.sum((p3-p4)**2, axis=0) dist2 = np.sqrt(squared_dist2) print(dist2) print("loop 2: " + str(iterate2))

Output:

3.677461821966065 loop 1: 1 3.36723045259329 loop 1: 2 3.7110807813512268 loop 1: 3 3.677461821966065 loop 2: 1 3.677461821966065 loop 1: 4 3.36723045259329 loop 1: 5 3.7110807813512268 loop 1: 6 4.029668579506317 loop 2: 2 3.677461821966065 loop 1: 7 3.36723045259329 loop 1: 8 3.7110807813512268 loop 1: 9 3.7166097270669267 loop 2: 3
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You can calculate this purely using Numpy, using the numpy linalg.norm (Euclidean distance) fucntion:

distances = np.array([ np.linalg.norm(B - p, axis=1) for p in A])

We're making use here of Numpy's matrix operations to calculate the distance for between each point in B and each point in A.

For the following two arrays, this will be the result (using 2D arrays for simplicity, but it will work exactly the same for 3D arrays):

A = np.array([[0.16869694, 0.18218867], [0.13406977, 0.92704429], [0.42916002, 0.70700565], [0.31897904, 0.74722291], [0.95014724, 0.37253304], [0.25335466, 0.46849091], [0.15266878, 0.69878319], [0.81076298, 0.46946201], [0.03145542, 0.62978006], [0.63422485, 0.37211463]]) B = np.array([[0.82858205, 0.88478606], [0.11420783, 0.21636374], [0.48167179, 0.68689633], [0.09455677, 0.25089711], [0.73376651, 0.05755397]]) distances = np.array([ np.linalg.norm(B - p, axis=1) for p in A]) print(distances) # Result: array([[0.9638939 , 0.06431951, 0.59387127, 0.10108223, 0.57865139], [0.69579671, 0.71095805, 0.42249049, 0.67730073, 1.05624316], [0.43719998, 0.58303034, 0.05623051, 0.56568042, 0.71733716], [0.52784359, 0.56898392, 0.17351723, 0.54470602, 0.80479308], [0.52648006, 0.85040199, 0.56417508, 0.86419347, 0.3821419 ], [0.71006211, 0.28797561, 0.31595829, 0.26937675, 0.63219042], [0.7010391 , 0.48395018, 0.32921768, 0.45164028, 0.8653609 ], [0.41570613, 0.74111256, 0.39443465, 0.7488137 , 0.4190426 ], [0.83692229, 0.42161715, 0.45382491, 0.38410164, 0.90591588], [0.54827613, 0.54284072, 0.34979988, 0.55311421, 0.32993477]])
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  • $\begingroup$ How do I rearrange the order that the distances are displayed in? I notice that the first distance value is distances[0][0]. The second distance value is distances[1][0]. Third value is distances[2][0]. This is not very intuitive to me. I'd prefer to have them coming one after the other so that I can easily find the index of a particular distance measurement. $\endgroup$ Commented Aug 11, 2022 at 16:57
  • $\begingroup$ You can rearrange the original arrays to change the order of the final results, or run (A - p, axis=1) for p in B in the norm function, to get [0][0], then [0][1], if I'm not mistaken $\endgroup$ Commented Aug 12, 2022 at 6:58

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