Shortest distance between one point and a group of others? [duplicate]

Question

I have a problem. I want to write a function in python which will receive a coordinate X and a group of coordinates S. I need to return the closest coordinate to x from group s. So when you call a function it will return this:

closest((9, 2), {(0, 0), (10, 0), (10, 10)}) # calling a function
(10, 0) 

Because its the closest to both points. I already have a function which calculates the distance between two points

def distance(s,t):
    v = 0
    for i in range(len(t)):
        v = v+(s[i]-t[i])**2
    return (sqrt(v))

But now I'm stuck on how to return the closest tuple of coordinates to the one given in x. My English is not that good so if you do not understand my question, please say it and I will try to explain somehow.

Answer 1

First you can make a distance function that just returns the distance between two points

import math
def distance(p1, p2):
    return math.sqrt((p2[0] - p1[0])**2 + (p2[1] - p1[1])**2)

Then closest can use the min function with the key parameter to use your distance function with each element from others

def closest(pt, others):
    return min(others, key = lambda i: distance(pt, i))

Example

>>> closest((9, 2), {(0, 0), (10, 0), (10, 10)})
(10, 0)

To calculate the average distance

def avgDistance(pt, others):
    dists = [distance(pt, i) for i in others]
    return sum(dists) / len(dists)

>>> avgDistance((9, 2), {(0, 0), (10, 0), (10, 10)})
6.505956727697075

Answer 2

Don't reinvent the wheel. This functionality exists in python's scientific libraries (scipy) and if you're going to be performing more mathematics on your data, you should really start using those libraries as they are much faster than doing it with regular python, because the code is vectorized and is often a thin wrapper to optimized C/fortran libraries.

>>> import numpy as np
>>> from scipy.spatial.distance import cdist
>>> a = np.array([[9, 2]])
>>> others = np.array([[0, 0], [10, 0], [10, 10]])
>>> def closest_point(pt, others):
...     distances = cdist(pt, others)
...     return others[distances.argmin()]
... 
>>> closest_point(a, others)
array([10,  0])

To compute the mean distance over all distances to that one point:

>>> distances = cdist(a, others)
>>> distances
array([[ 9.21954446,  2.23606798,  8.06225775]])
>>> distances.mean()
6.5059567276970753

Finally, to find the point that is "most centered", in the sense that it has the smallest average distance to all the others, you'll have to compute the distance between all nodes:

>>> all_nodes = np.array([[0,0], [10,0], [0,10], [10, 10], [5,4]])
>>> from scipy.spatial.distance import pdist, squareform
>>> avg_dists = squareform(pdist(all_nodes)).mean(axis=1)
>>> avg_dists
array([ 8.10905197,  8.10905197,  8.39047706,  8.39047706,  5.68534957])
>>> all_nodes[avg_dists.argmin()]
array([5, 4])