Efficiently Calculating a Euclidean Distance Matrix Using Numpy

Question

I have a set of points in 2-dimensional space and need to calculate the distance from each point to each other point.

I have a relatively small number of points, ma

Answer 1

Jake Vanderplas gives this example using broadcasting in Python Data Science Handbook, which is very similar to what @shx2 proposed.

import numpy as np
rand = random.RandomState(42)
X = rand.rand(3, 2)  
dist_sq = np.sum((X[:, np.newaxis, :] - X[np.newaxis, :, :]) ** 2, axis = -1)

dist_sq
array([[0.        , 0.18543317, 0.81602495],
       [0.18543317, 0.        , 0.22819282],
       [0.81602495, 0.22819282, 0.        ]])

Answer 2

Here is how you can do it using numpy:

import numpy as np

x = np.array([0,1,2])
y = np.array([2,4,6])

# take advantage of broadcasting, to make a 2dim array of diffs
dx = x[..., np.newaxis] - x[np.newaxis, ...]
dy = y[..., np.newaxis] - y[np.newaxis, ...]
dx
=> array([[ 0, -1, -2],
          [ 1,  0, -1],
          [ 2,  1,  0]])

# stack in one array, to speed up calculations
d = np.array([dx,dy])
d.shape
=> (2, 3, 3)

Now all is left is computing the L2-norm along the 0-axis (as discussed here):

(d**2).sum(axis=0)**0.5
=> array([[ 0.        ,  2.23606798,  4.47213595],
          [ 2.23606798,  0.        ,  2.23606798],
          [ 4.47213595,  2.23606798,  0.        ]])

Answer 3

If you don't need the full distance matrix, you will be better off using kd-tree. Consider scipy.spatial.cKDTree or sklearn.neighbors.KDTree. This is because a kd-tree kan find k-nearnest neighbors in O(n log n) time, and therefore you avoid the O(n**2) complexity of computing all n by n distances.

Answer 4

You can take advantage of the complex type :

# build a complex array of your cells
z = np.array([complex(c.m_x, c.m_y) for c in cells])

First solution

# mesh this array so that you will have all combinations
m, n = np.meshgrid(z, z)
# get the distance via the norm
out = abs(m-n)

Second solution

Meshing is the main idea. But numpy is clever, so you don't have to generate m & n. Just compute the difference using a transposed version of z. The mesh is done automatically :

out = abs(z[..., np.newaxis] - z)

Third solution

And if z is directly set as a 2-dimensional array, you can use z.T instead of the weird z[..., np.newaxis]. So finally, your code will look like this :

z = np.array([[complex(c.m_x, c.m_y) for c in cells]]) # notice the [[ ... ]]
out = abs(z.T-z)

Example

>>> z = np.array([[0.+0.j, 2.+1.j, -1.+4.j]])
>>> abs(z.T-z)
array([[ 0.        ,  2.23606798,  4.12310563],
       [ 2.23606798,  0.        ,  4.24264069],
       [ 4.12310563,  4.24264069,  0.        ]])

As a complement, you may want to remove duplicates afterwards, taking the upper triangle :

>>> np.triu(out)
array([[ 0.        ,  2.23606798,  4.12310563],
       [ 0.        ,  0.        ,  4.24264069],
       [ 0.        ,  0.        ,  0.        ]])

Some benchmarks

>>> timeit.timeit('abs(z.T-z)', setup='import numpy as np;z = np.array([[0.+0.j, 2.+1.j, -1.+4.j]])')
4.645645342274779
>>> timeit.timeit('abs(z[..., np.newaxis] - z)', setup='import numpy as np;z = np.array([0.+0.j, 2.+1.j, -1.+4.j])')
5.049334864854522
>>> timeit.timeit('m, n = np.meshgrid(z, z); abs(m-n)', setup='import numpy as np;z = np.array([0.+0.j, 2.+1.j, -1.+4.j])')
22.489568296184686