Compute numpy array pairwise Euclidean distance except with self

Question

edit: this question is not specifically about calculating distances, rather the most efficient way to loop through a numpy array, specifying that for index i all comparisons should be made with the rest of the array, as long as the second index is not i.

I have a numpy array with columns (X, Y, ID) and want to compare each element to each other element, but not itself. So, for each X, Y coordinate, I want to calculate the distance to each other X, Y coordinate, but not itself (where distance = 0).

Here is what I have - there must be a more "numpy" way to write this.

import math, arcpy
# Point feature class
fc = "MY_FEATURE_CLASS"
# Load points to numpy array: (X, Y, ID)
npArray = arcpy.da.FeatureClassToNumPyArray(fc,["SHAPE@X","SHAPE@Y","OID@"])
for row in npArray:
    for row2 in npArray:
        if row[2] != row2[2]:
            # Pythagoras's theorem
            distance = math.sqrt(math.pow((row[0]-row2[0]),2)+math.pow((row[1]-row2[1]),2))

Obviously, I'm a numpy newbie. I will not be surprised to find this a duplicate, but I don't have the numpy vocabulary to search out the answer. Any help appreciated!

possible duplicate of Computing Euclidean distance for numpy in python — Oliver W.
– Oliver W., Commented Mar 18, 2015 at 21:39
Note that in many geometries d(x, y) = 0 iff x = y, so you may want to skip that check, and deal with the zeros later on. Disregard this comment if you're dealing computing distance under a non-injective mapping - I just noticed you've discussing "features" in your code, which suggests this may be the case. — Patrick McLaren
– Patrick McLaren, Commented Mar 18, 2015 at 22:19
Well, if you are willing to use scipy, then the comment by @user2357112 should do the trick. If not, you could use a list comprehension, say distances = [dist(x, y) for x in npArray for y in npArray if x != y], where dist is some metric. — Patrick McLaren
– Patrick McLaren, Commented Mar 18, 2015 at 22:36
@user2312518 I've been using cKDTree that comes in SciPy to do this kind of task very efficiently, it seems much faster than calculating all the distances — Saullo G. P. Castro
– Saullo G. P. Castro, Commented Mar 19, 2015 at 0:35

wht · Accepted Answer · 2019-06-14 20:51:53Z

0

Using SciPy's pdist, you could write something like

from scipy.spatial.distance import pdist, squareform

distances = squareform(pdist(npArray, lambda a,b: np.sqrt((a[0]-b[0])**2 + (a[1]-b[1])**2)))

pdist will compute the pair-wise distances using the custom metric that ignores the 3rd coordinate (which is your ID in this case). squareform turns this into a more readable matrix such that distances[0,1] gives the distance between the 0th and 1st rows.

answered Jun 14, 2019 at 20:51

wht

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Andronicus · Accepted Answer · 2019-12-29 16:03:36Z

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Each row of X is a 3 dimensional data instance or point. The output pairwisedist[i, j] is distance of X[i, :] and X[j, :]

X = np.array([[6,1,7],[10,9,4],[13,9,3],[10,8,15],[14,4,1]])
a = np.sum(X*X,1)
b = np.repeat( a[:,np.newaxis],5,axis=1)
pairwisedist = b + b.T -2* X.dot(X.T)

edited Dec 29, 2019 at 16:03

Andronicus

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answered Dec 29, 2019 at 15:40

Mohammad R. Moosavi

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eroot163pi · Accepted Answer · 2021-09-18 11:32:06Z

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I wanted to point out that custom written sqrt of sum of squares are prone to overflow and underflow. Bultin math.hypot, np.hypot are way safer for no compromise on performance

from scipy.spatial.distance import pdist, squareform

distances = squareform(pdist(npArray, lambda a,b: math.hypot(*(a-b))

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answered Sep 18, 2021 at 11:32

eroot163pi

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