Euclidean distances (python3, sklearn): efficiently compute closest pairs and their corresponding distances

Question

I'm given a 2-D numpy array X consisting of floating values and need to compute the euclidean distances between all pairs of rows, then compute the top k row indices with the smallest distances and return them (where k > 0). I'm testing with a small array and this is what I have so far...

import numpy as np
from sklearn.metrics.pairwise import euclidean_distances

X_testing = np.asarray([[1,2,3.5],[4,1,2],[0,0,2],[3.4,1,5.6]])
test = euclidean_distances(X_testing, X_testing)
print(test)  

The resulting printout is:

[[ 0.          3.5         2.6925824   3.34215499]
 [ 3.5         0.          4.12310563  3.64965752]
 [ 2.6925824   4.12310563  0.          5.05173238]
 [ 3.34215499  3.64965752  5.05173238  0.        ]]

Next, I need to efficiently compute the top k smallest distances between all pairs of rows, and return the corresponding k tuples of (row1, row2, distance_value) in order in the form of a list.

So in the above test case, if k = 2, then I would need to return the following:

[(0, 2, 2.6925824), (0, 3, 3.34215499)]

Is there a built-in way (in either scipy, sklearn, numpy, etc.), or any other way to help compute this efficiently? Although the above test case is small, in reality the 2-D array is very large so memory and time is a concern. Thanks

Answer 1

Using scipy.spatial instead of sklearn (which I haven't installed yet) I can get the same distance matrix:

In [623]: from scipy import spatial
In [624]: pdist=spatial.distance.pdist(X_testing)
In [625]: pdist
Out[625]: 
array([ 3.5       ,  2.6925824 ,  3.34215499,  4.12310563,  3.64965752,
        5.05173238])
In [626]: D=spatial.distance.squareform(pdist)
In [627]: D
Out[627]: 
array([[ 0.        ,  3.5       ,  2.6925824 ,  3.34215499],
       [ 3.5       ,  0.        ,  4.12310563,  3.64965752],
       [ 2.6925824 ,  4.12310563,  0.        ,  5.05173238],
       [ 3.34215499,  3.64965752,  5.05173238,  0.        ]])

pdist is in condensed form, whose indicies in the squareform can be found with

In [629]: np.triu_indices(4,1)
Out[629]: 
(array([0, 0, 0, 1, 1, 2], dtype=int32),
 array([1, 2, 3, 2, 3, 3], dtype=int32))

The 2 smallest distances are the 1st 2 values of

In [630]: idx=np.argsort(pdist)
In [631]: idx
Out[631]: array([1, 2, 0, 4, 3, 5], dtype=int32)

So we want [1,2] from pdist and the corresponding elements of the triu:

In [633]: pdist[idx[:2]]
Out[633]: array([ 2.6925824 ,  3.34215499])
In [634]: np.transpose(np.triu_indices(4,1))[idx[:2],:]
Out[634]: 
array([[0, 2],
       [0, 3]], dtype=int32)

and to collect those values as a list of tuples:

In [636]: I,J = np.triu_indices(4,1)
In [637]: kbig = idx[:2]
In [638]: [(i,j,d) for i,j,d in zip(I[kbig], J[kbig], pdist[kbig])]
Out[638]: [(0, 2, 2.6925824035672519), (0, 3, 3.3421549934136805)]

Numpy array of distances to list of (row,col,distance)

Answer 2

This is by example, but incorporates a list comprehension so you can see the slicing. Obviously not a speed demon, but more for understanding.

>>> import numpy as np
>>> a = np.random.randint(0,10, size=(5,5))
>>> a
array([[8, 3, 3, 8, 9],
       [0, 8, 6, 6, 5],
       [6, 7, 6, 5, 0],
       [4, 2, 4, 0, 3],
       [4, 1, 3, 2, 2]])
>>> idx = np.argsort(a, axis=1)
>>> idx
array([[1, 2, 0, 3, 4],
       [0, 4, 2, 3, 1],
       [4, 3, 0, 2, 1],
       [3, 1, 4, 0, 2],
       [1, 3, 4, 2, 0]])
>>> v = np.vstack([ a[i][idx[i]] for i in range(len(idx))])
>>> v
array([[3, 3, 8, 8, 9],
       [0, 5, 6, 6, 8],
       [0, 5, 6, 6, 7],
       [0, 2, 3, 4, 4],
       [1, 2, 2, 3, 4]])
>>> 
>>> v3 = np.vstack([ a[i][idx[i]][:3] for i in range(len(idx))])
>>> v3
array([[3, 3, 8],
       [0, 5, 6],
       [0, 5, 6],
       [0, 2, 3],
       [1, 2, 2]])
>>> 

You can mess around with the slicing and put it full np if you like.