Efficient and precise calculation of the euclidean distance

Question

Following some online research (1, 2, numpy, scipy, scikit, math), I have found several ways for calculating the Euclidean Distance in Python:

# 1
numpy.linalg.norm(a-b)

# 2
distance.euclidean(vector1, vector2)

# 3
sklearn.metrics.pairwise.euclidean_distances  

# 4
sqrt((xa-xb)^2 + (ya-yb)^2 + (za-zb)^2)

# 5
dist = [(a - b)**2 for a, b in zip(vector1, vector2)]
dist = math.sqrt(sum(dist))

# 6
math.hypot(x, y)

I was wondering if someone could provide an insight on which of the above (or any other that I have not found) is considered the best in terms of efficiency and precision. If someone is aware of any resource(s) which discusses the subject that would also be great.

The context I am interesting in is in calculating the Euclidean Distance between pairs of number-tuples, e.g. the distance between (52, 106, 35, 12) and (33, 153, 75, 10).

Answer 1

Conclusion first:

From the test result by using timeit for efficiency test, we can conclude that regarding the efficiency:

Method5 (zip, math.sqrt) > Method1 (numpy.linalg.norm) > Method2 (scipy.spatial.distance) > Method3 (sklearn.metrics.pairwise.euclidean_distances )

While I didn't really test your Method4 as it is not suitable for general cases and it is generally equivalent to Method5.

For the rest, quite surprisingly, Method5 is the fastest one. While for Method1 which uses numpy, as what we expected, which is heavily optimized in C, is the second fastest.

For scipy.spatial.distance, if you go directly to the function definition, you will see that it is actually using numpy.linalg.norm, except it will perform the validation on the two input vectors before the actual numpy.linalg.norm. That's why it is slightly slower thant numpy.linalg.norm.

Finally for sklearn, according to the documentation:

This formulation has two advantages over other ways of computing distances. First, it is computationally efficient when dealing with sparse data. Second, if one argument varies but the other remains unchanged, then dot(x, x) and/or dot(y, y) can be pre-computed. However, this is not the most precise way of doing this computation, and the distance matrix returned by this function may not be exactly symmetric as required

Since in your question you would like to use a fixed set of data, the advantage of this implementation is not reflected. And due to the trade off between the performance and precision, it also gives the worst precision among all of the methods.

Regarding the precision, Method5=Metho1=Method2>Method3

Efficiency Test Script:

import numpy as np
from scipy.spatial import distance
from sklearn.metrics.pairwise import euclidean_distances
import math

# 1
def eudis1(v1, v2):
    return np.linalg.norm(v1-v2)

# 2
def eudis2(v1, v2):
    return distance.euclidean(v1, v2)

# 3
def eudis3(v1, v2):
    return euclidean_distances(v1, v2)

# 5
def eudis5(v1, v2):
    dist = [(a - b)**2 for a, b in zip(v1, v2)]
    dist = math.sqrt(sum(dist))
    return dist

dis1 = (52, 106, 35, 12)
dis2 = (33, 153, 75, 10)
v1, v2 = np.array(dis1), np.array(dis2)

import timeit

def wrapper(func, *args, **kwargs):
    def wrapped():
        return func(*args, **kwargs)
    return wrapped

wrappered1 = wrapper(eudis1, v1, v2)
wrappered2 = wrapper(eudis2, v1, v2)
wrappered3 = wrapper(eudis3, v1, v2)
wrappered5 = wrapper(eudis5, v1, v2)
t1 = timeit.repeat(wrappered1, repeat=3, number=100000)
t2 = timeit.repeat(wrappered2, repeat=3, number=100000)
t3 = timeit.repeat(wrappered3, repeat=3, number=100000)
t5 = timeit.repeat(wrappered5, repeat=3, number=100000)

print('\n')
print('t1: ', sum(t1)/len(t1))
print('t2: ', sum(t2)/len(t2))
print('t3: ', sum(t3)/len(t3))
print('t5: ', sum(t5)/len(t5))

Efficiency Test Output:

t1:  0.654838958307
t2:  1.53977598714
t3:  6.7898791732
t5:  0.422228400305

Precision Test Script & Result:

In [8]: eudis1(v1,v2)
Out[8]: 64.60650122085238

In [9]: eudis2(v1,v2)
Out[9]: 64.60650122085238

In [10]: eudis3(v1,v2)
Out[10]: array([[ 64.60650122]])

In [11]: eudis5(v1,v2)
Out[11]: 64.60650122085238