Python - Closest Minimum

Question

I've 2 matrices both with Nx2 elements. Any value is a float with 8-10 decimals and they represent respectively 'x' and 'y' of a point.

For any element couple (x, y) (x is in the first column, while y is in the second one) in the first array, I'd need to find the closest point in the second one. At any loop, once found, I need to remove that value from the second array.

Finally, my main objective would be to have the optimal solution so that there's a one-to-one mapping between any element of the first array with only one element of the second array, so that the closest value clause is met.

I created a NxN matrix where I computed the distance from any point of first array to any point of the second array via

scipy.spatial.distance.cdist

Code:

def find_nearest(X_start, X_end):
    mat = scipy.spatial.distance.cdist(X_start, X_end, metric='euclidean')
    new_df = pd.DataFrame(mat)
    return new_df;

The next step is to couple a starting point with a ending point and there should NOT be any intersection, that is a mapping one-to-one.

I thought to do it via a Integer programming (using this). So if m[i][j] is an element of the matrix NxN, I found those constraints

The problem is that I don't know how to write the objective function and so I'm note sure if I need to add any other constraint related to it.

Do you think this is a good path to follow? Last question seems to be not appreciated since I did not expose what I already did.

So here it is.

Such questions should be asked at Math SE or Computational Science SE. — Rodrigo de Azevedo, Jun 08 '19 at 13:20
If the question can be moved, I would move it to https://scicomp.stackexchange.com (which also has tags for Python and SciPy). — Rodrigo de Azevedo, Jun 09 '19 at 15:06
If you post the same question on two SE sites, you should notify the reader and cross-link them. — Rodrigo de Azevedo, Jun 09 '19 at 15:13

Erwin Kalvelagen · Accepted Answer · 2019-06-24T08:52:33.220

This is called an assignment problem.

   min sum((i,j), dist[i,j]*x[i,j])
   subject to
       sum(i, x[i,j]) = 1 for all j
       sum(j, x[i,j]) = 1 for all i
       x[i,j] in {0,1}

where

 i = 1..n is an element of the first matrix
 j = 1..n is an element of the second matrix
 dist[i,j] is a distance matrix

These problems can be solved with specialized solvers or it can be formulated as an LP (linear programming) problem.

Scipy has a simple assignment solver (link). This is not a very fast implementation however: a good LP solver is faster (link).

DrBwts · Answer 2 · 2019-06-08T15:27:34.867

0

OK I think this is what you are asking. The following code will go through each coordinate in p1 & calculate the distances with every coordinate in p2 (the closest_node function was from here) then return the closest coorinate to the nearest array & the corresponding element is deleted from p2

There will be a 1 to 1 correspondence between p1 & nearest ie p1[0] maps to nearest[0] etc..

import numpy as np

def closest_node(node, nodes):
    dist_2 = np.sum((nodes - node)**2, axis=1)
    return np.argmin(dist_2)

p1 = np.random.rand(10, 2)
p2 = np.random.rand(10, 2)
nearest = []

for coord in p1:
    near = closest_node(coord, p2)
    nearest.append(p2[near])
    p2 = np.delete(p2, near, 0)

nearest = np.array(nearest)

edited Jun 08 '19 at 15:27

answered Jun 08 '19 at 13:08

DrBwts

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Does it resolve the one-to-one mapping? I don’t think so – JackLametta Jun 09 '19 at 15:03
it has all the values that were in p2 but mapped to their nearest values in p1. Run it a few time & check for yourself – DrBwts Jun 09 '19 at 19:02

Python - Closest Minimum

2 Answers2