从一个numpy距离数组中提取N个最接近的对

Question

I have a large, symmetric, 2D distance array. I want to get closest N pairs of observations.

The array is stored as a numpy condensed array, and has of the order of 100 million observations.

Here's an example to get the 100 closest distances on a smaller array (~500k observations), but it's a lot slower than I would like.

import numpy as np
import random
import sklearn.metrics.pairwise
import scipy.spatial.distance

N = 100
r = np.array([random.randrange(1, 1000) for _ in range(0, 1000)])
c = r[:, None]

dists = scipy.spatial.distance.pdist(c, 'cityblock')

# these are the indices of the closest N observations
closest = dists.argsort()[:N]

# but it's really slow to get out the pairs of observations
def condensed_to_square_index(n, c):
    # converts an index in a condensed array to the 
    # pair of observations it represents
    # modified from here: http://stackoverflow.com/questions/5323818/condensed-matrix-function-to-find-pairs
    ti = np.triu_indices(n, 1)
    return ti[0][c]+ 1, ti[1][c]+ 1

r = []
n = np.ceil(np.sqrt(2* len(dists)))
for i in closest:
    pair = condensed_to_square_index(n, i)
    r.append(pair)

It seems to me like there must be quicker ways to do this with standard numpy or scipy functions, but I'm stumped.

NB If lots of pairs are equidistant, that's OK and I don't care about their ordering in that case.

Answer 1

You don't need to calculate ti in each call to condensed_to_square_index . Here's a basic modification that calculates it only once:

import numpy as np
import random
import sklearn.metrics.pairwise
import scipy.spatial.distance

N = 100
r = np.array([random.randrange(1, 1000) for _ in range(0, 1000)])
c = r[:, None]

dists = scipy.spatial.distance.pdist(c, 'cityblock')

# these are the indices of the closest N observations
closest = dists.argsort()[:N]

# but it's really slow to get out the pairs of observations
def condensed_to_square_index(ti, c):
    return ti[0][c]+ 1, ti[1][c]+ 1

r = []
n = np.ceil(np.sqrt(2* len(dists)))
ti = np.triu_indices(n, 1)

for i in closest:
    pair = condensed_to_square_index(ti, i)
    r.append(pair)

You can also vectorize the creation of r :

r  = zip(ti[0][closest] + 1, ti[1][closest] + 1)

or

r = np.vstack(ti)[:, closest] + 1

Answer 2

You can speed up the location of the minimum values very notably if you are using numpy 1.8 using np.partition :

def smallest_n(a, n):
    return np.sort(np.partition(a, n)[:n])

def argsmallest_n(a, n):
    ret = np.argpartition(a, n)[:n]
    b = np.take(a, ret)
    return np.take(ret, np.argsort(b))

dists = np.random.rand(1000*999//2) # a pdist array

In [3]: np.all(argsmallest_n(dists, 100) == np.argsort(dists)[:100])
Out[3]: True

In [4]: %timeit np.argsort(dists)[:100]
10 loops, best of 3: 73.5 ms per loop

In [5]: %timeit argsmallest_n(dists, 100)
100 loops, best of 3: 5.44 ms per loop

And once you have the smallest indices, you don't need a loop to extract the indices, do it in a single shot:

closest = argsmallest_n(dists, 100)
tu = np.triu_indices(1000, 1)
pairs = np.column_stack((np.take(tu[0], closest),
                         np.take(tu[1], closest))) + 1

Answer 3

The best solution probably won't generate all of the distances.

Proposal:

1. Make a heap of max size 100 (if it grows bigger, reduce it).
2. Use the Closest Pair algorithm to find the closest pair.
3. Add the pair to the heap (priority queue).
4. Choose one of that pair. Add its 99 closest neighbors to the heap.
5. Remove the chosen point from the list.
6. Find the next closest pair and repeat. The number of neighbors added is 100 minus the number of times you ran the Closest Pair algorithm.