Reputation: 13624
Suppose I have an matrix nxm accommodating row vectors. I want to have an distance matrix nxn that presents the distance of each vector to each other. How can I do it in Python as I am using Numpy. I know Scipy does it but I want to dirst my hands. I already write a cosine similarity function cos_dist(a,b)
where a and b two different vectors. Now I need a caller function that is doing it for each couple of items efficiently. How would I do it?
Upvotes: 2
Views: 6027
Reputation: 67507
The following code shows two option to do what you are after. One looping over the array twice and using a Python function to calculate the cos_dist. The second uses a vectorized approach and broadcasting to get the same result x1000 faster.
from __future__ import division
import numpy as np
def cos_dist(a, b):
mod_a = np.sqrt(a.dot(a))
mod_b = np.sqrt(b.dot(b))
return a.dot(b) / mod_a / mod_b
a = np.random.rand(100, 4)
# Slow option
def slow_dist(a):
items = a.shape[0]
out_slow = np.ones((items,items))
for j in xrange(items):
for k in xrange(j+1, items):
out_slow[j, k] = cos_dist(a[j], a[k])
out_slow[k, j] = out_slow[j, k]
return out_slow
# Faster option
from numpy.core.umath_tests import inner1d
def fast_dist(a):
mod_a = np.sqrt(inner1d(a ,a))
norm_a = a / mod_a[:, None]
out_fast = inner1d(norm_a[:, None, :],
norm_a[None, :, :])
return out_fast
And here are the timings:
In [2]: %timeit slow_dist(a)
10 loops, best of 3: 67.6 ms per loop
In [3]: %timeit fast_dist(a)
10000 loops, best of 3: 60.5 us per loop
In [4]: np.allclose(slow_dist(a), fast_dist(a))
Out[4]: True
Upvotes: 3
Reputation: 3119
Why don't you check on scipy's spatial.distance.pdist()
, which computes pairwise distances between observations in n-dimensional space and has a vast number of distance functions to choose from?
Since you don't have scipy installed and want to code this using numpy, I suggest you study its source code, which is linked at the top-left of its documentation page.
Upvotes: 2