How can I calculate the distances of a set of 3d points with each other using python?

Question

I have a set of 3D points, where I need to find the distances of every point with all other points. So far I came up with the code as below to calculate the distances between two consecutive (in order of the array) points but can't figure out how can I calculate the distance of each point with all other points. Every point will have 9 distances (with 9 other points), thus 10 points have a total of 45 distances (half of 90). So far I have found 9 distances only. Any idea how can get all the distances efficiently using python?

import numpy as np
import scipy as sp

rij_mat = np.zeros((9,1),dtype=np.float32) """created a 9x1 matrix for storing 9 distances for 10 points"""

npoints = 10

x = sp.randn(npoints,3)

print "here are the 3-d points..."
print x

for i in xrange(npoints-1):

 rij_mat[i] = np.linalg.norm(x[i+1]-x[i])

print "the distances are..."
print rij_mat

My output right now is something like this:

here are the 3-d points...
[[-0.89513316  0.05061497 -1.19045606]
 [ 0.31999847 -0.68013916 -0.14576028]
 [ 0.85442751 -0.64139512  1.70403995]
 [ 0.55855264  0.56652717 -0.17086825]
 [-1.22435021  0.25683649  0.85128921]
 [ 0.80310031  0.82236372 -0.40015387]
 [-1.34356018  1.034942    0.00878305]
 [-0.65347726  1.1697195  -0.45206805]
 [ 0.65714623 -1.07237429 -0.75204526]
 [ 1.17204207  0.89878048 -0.54657068]]
the distances are...
[[ 1.7612313 ]
 [ 1.92584431]
 [ 2.24986649]
 [ 2.07833028]
 [ 2.44877243]
 [ 2.19557977]
 [ 0.84069204]
 [ 2.61432695]
 [ 2.04763007]]

Answer 1

How about two loops instead of one?

distances = []
for i in xrange(npoints-1):
    for j in range(i+1, npoints):
        distances.append(np.linalg.norm(x[i]-x[j])

Answer 2

For each of your npoints points, there are exactly npoints - 1 other points to compare with, for distance. And, the distance between x[m] and x[n] is the same as the distance between x[n] and x[m] , so that cuts the total number of distances in half. The itertools package has a nice way to handle this:

import numpy as np
import scipy as sp
import itertools as its

npoints = 10
nCombos = (npoints * (npoints - 1))/2
x = sp.randn(npoints,3)
rij_mat = np.zeros(nCombos)

ii = 0
for i1, i2 in its.combinations(range(npoints), 2):
    rij_mat[ii] = np.linalg.norm(x[i1]-x[i2])
    ii += 1

print "the distances are..."
print rij_mat

If you're a really careful person, you might check at the end that ii == nCombos .

Since you called your output matrix rij_mat , maybe you intended it to be a 2-dimensional matrix? Then you'd want something like:

import numpy as np
import scipy as sp
import itertools as its

npoints = 10

x = sp.randn(npoints,3)
rij_mat = np.zeros((npoints, npoints))

for i1, i2 in its.combinations(range(npoints), 2):
    rij_mat[i2, i1] = rij_mat[i1, i2] = np.linalg.norm(x[i1]-x[i2])

print "the distances are..."
print rij_mat