How to avoid quadratic computation resulting from double 'for loop' when computing distances between vectors

问题

Given the following code that computes distances between vectors in list 'vect’:

import numpy as np
vect=([0.123, 0.345, 0.789], [0.234, 0.456, 0.567],[0.134, 0.246, 0.831])
def kn():
    for j in vect:
        c=np.array(j)
        for z in vect:
            p=np.array(z)
            space = np.linalg.norm(c-p)
            print space
kn()

Is there a way to avoid the quadratic complexity that will result from the double ‘for loop’?

Computation as a result of the double for loop (quadratic) ==3X3=9

Ideal computation (what I want) should be (3):
[0.123, 0.345, 0.789] and [0.234, 0.456, 0.567] =
[0.123, 0.345, 0.789] and [0.134, 0.246, 0.831] =
[0.234, 0.456, 0.567] and [0.134, 0.246, 0.831] =

Thanks in advance for your suggestion(s).

回答1:

To avoid duplicate pairs nested loop should go upwards from the index of the outer loop, i.e.:

for i, v1 in enumerate(vect):
    for j in xrange(i + 1, len(vect)):
        a = np.array(v1)
        b = np.array(vect[j])
        space = np.linalg.norm(b - a)
        print space

Or use a solution provided by the standard library:

import itertools

for v1, v2 in itertools.combinations(vect, 2):
    a = np.array(v1)
    b = np.array(v2)
    space = np.linalg.norm(b - a)
    print space

回答2:

Others have noted that you can't avoid the quadratic complexity. But if your concern is performance, you can speed things up considerably by using scipy.spatial.distance.pdist, which will have all the looping and function calling happen in C, not Python. The following code returns a square, symmetrical array, but it only makes n *(n-1)/2 calculations:

from scipy.spatial.distance import pdist

pairwise_distances = pdist(vect, metric='euclidean', p=2)

If you want a flattened array with only the unique, off-diagonal values, you can use scipy.spatial.distance.squareform:

from scipy.spatial.distance import pdist, squareform

pairwise_distances = squareform(pdist(vect, metric='euclidean', p=2))

In your case, pairwise_distances would be a 3 element long array.

来源：https://stackoverflow.com/questions/18014889/how-to-avoid-quadratic-computation-resulting-from-double-for-loop-when-computi

标签

python

numpy

complexity-theory

distance

quadratic