# python – How can the Euclidean distance be calculated with NumPy?

## The Question :

584 people think this question is useful

I have two points in 3D:

(xa, ya, za)
(xb, yb, zb)



And I want to calculate the distance:

dist = sqrt((xa-xb)^2 + (ya-yb)^2 + (za-zb)^2)



What’s the best way to do this with NumPy, or with Python in general? I have:

import numpy
a = numpy.array((xa ,ya, za))
b = numpy.array((xb, yb, zb))



991 people think this answer is useful
dist = numpy.linalg.norm(a-b)



You can find the theory behind this in Introduction to Data Mining

This works because the Euclidean distance is the l2 norm, and the default value of the ord parameter in numpy.linalg.norm is 2. 181 people think this answer is useful

There’s a function for that in SciPy. It’s called Euclidean.

Example:

from scipy.spatial import distance
a = (1, 2, 3)
b = (4, 5, 6)
dst = distance.euclidean(a, b)



122 people think this answer is useful

For anyone interested in computing multiple distances at once, I’ve done a little comparison using perfplot (a small project of mine).

The first advice is to organize your data such that the arrays have dimension (3, n) (and are C-contiguous obviously). If adding happens in the contiguous first dimension, things are faster, and it doesn’t matter too much if you use sqrt-sum with axis=0, linalg.norm with axis=0, or

a_min_b = a - b
numpy.sqrt(numpy.einsum('ij,ij->j', a_min_b, a_min_b))



which is, by a slight margin, the fastest variant. (That actually holds true for just one row as well.)

The variants where you sum up over the second axis, axis=1, are all substantially slower. Code to reproduce the plot:

import numpy
import perfplot
from scipy.spatial import distance

def linalg_norm(data):
a, b = data
return numpy.linalg.norm(a - b, axis=1)

def linalg_norm_T(data):
a, b = data
return numpy.linalg.norm(a - b, axis=0)

def sqrt_sum(data):
a, b = data
return numpy.sqrt(numpy.sum((a - b) ** 2, axis=1))

def sqrt_sum_T(data):
a, b = data
return numpy.sqrt(numpy.sum((a - b) ** 2, axis=0))

def scipy_distance(data):
a, b = data
return list(map(distance.euclidean, a, b))

def sqrt_einsum(data):
a, b = data
a_min_b = a - b
return numpy.sqrt(numpy.einsum("ij,ij->i", a_min_b, a_min_b))

def sqrt_einsum_T(data):
a, b = data
a_min_b = a - b
return numpy.sqrt(numpy.einsum("ij,ij->j", a_min_b, a_min_b))

def setup(n):
a = numpy.random.rand(n, 3)
b = numpy.random.rand(n, 3)
out0 = numpy.array([a, b])
out1 = numpy.array([a.T, b.T])
return out0, out1

perfplot.save(
"norm.png",
setup=setup,
n_range=[2 ** k for k in range(22)],
kernels=[
linalg_norm,
linalg_norm_T,
scipy_distance,
sqrt_sum,
sqrt_sum_T,
sqrt_einsum,
sqrt_einsum_T,
],
logx=True,
logy=True,
xlabel="len(x), len(y)",
)



44 people think this answer is useful

I want to expound on the simple answer with various performance notes. np.linalg.norm will do perhaps more than you need:

dist = numpy.linalg.norm(a-b)



Firstly – this function is designed to work over a list and return all of the values, e.g. to compare the distance from pA to the set of points sP:

sP = set(points)
pA = point
distances = np.linalg.norm(sP - pA, ord=2, axis=1.)  # 'distances' is a list



Remember several things:

• Python function calls are expensive.
• [Regular] Python doesn’t cache name lookups.

So

def distance(pointA, pointB):
dist = np.linalg.norm(pointA - pointB)
return dist



isn’t as innocent as it looks.

>>> dis.dis(distance)
10 BINARY_SUBTRACT
12 CALL_FUNCTION            1
14 STORE_FAST               2 (dist)

18 RETURN_VALUE



Firstly – every time we call it, we have to do a global lookup for “np”, a scoped lookup for “linalg” and a scoped lookup for “norm”, and the overhead of merely calling the function can equate to dozens of python instructions.

Lastly, we wasted two operations on to store the result and reload it for return…

First pass at improvement: make the lookup faster, skip the store

def distance(pointA, pointB, _norm=np.linalg.norm):
return _norm(pointA - pointB)



We get the far more streamlined:

>>> dis.dis(distance)
6 BINARY_SUBTRACT
8 CALL_FUNCTION            1
10 RETURN_VALUE



The function call overhead still amounts to some work, though. And you’ll want to do benchmarks to determine whether you might be better doing the math yourself:

def distance(pointA, pointB):
return (
((pointA.x - pointB.x) ** 2) +
((pointA.y - pointB.y) ** 2) +
((pointA.z - pointB.z) ** 2)
) ** 0.5  # fast sqrt



On some platforms, **0.5 is faster than math.sqrt. Your mileage may vary.

Why are you calculating distance? If the sole purpose is to display it,

 print("The target is %.2fm away" % (distance(a, b)))



move along. But if you’re comparing distances, doing range checks, etc., I’d like to add some useful performance observations.

Let’s take two cases: sorting by distance or culling a list to items that meet a range constraint.

# Ultra naive implementations. Hold onto your hat.

def sort_things_by_distance(origin, things):
return things.sort(key=lambda thing: distance(origin, thing))

def in_range(origin, range, things):
things_in_range = []
for thing in things:
if distance(origin, thing) <= range:
things_in_range.append(thing)



The first thing we need to remember is that we are using Pythagoras to calculate the distance (dist = sqrt(x^2 + y^2 + z^2)) so we’re making a lot of sqrt calls. Math 101:

dist = root ( x^2 + y^2 + z^2 )
:.
dist^2 = x^2 + y^2 + z^2
and
sq(N) < sq(M) iff M > N
and
sq(N) > sq(M) iff N > M
and
sq(N) = sq(M) iff N == M



In short: until we actually require the distance in a unit of X rather than X^2, we can eliminate the hardest part of the calculations.

# Still naive, but much faster.

def distance_sq(left, right):
""" Returns the square of the distance between left and right. """
return (
((left.x - right.x) ** 2) +
((left.y - right.y) ** 2) +
((left.z - right.z) ** 2)
)

def sort_things_by_distance(origin, things):
return things.sort(key=lambda thing: distance_sq(origin, thing))

def in_range(origin, range, things):
things_in_range = []

# Remember that sqrt(N)**2 == N, so if we square
# range, we don't need to root the distances.
range_sq = range**2

for thing in things:
if distance_sq(origin, thing) <= range_sq:
things_in_range.append(thing)



Great, both functions no-longer do any expensive square roots. That’ll be much faster. We can also improve in_range by converting it to a generator:

def in_range(origin, range, things):
range_sq = range**2
yield from (thing for thing in things
if distance_sq(origin, thing) <= range_sq)



This especially has benefits if you are doing something like:

if any(in_range(origin, max_dist, things)):
...



But if the very next thing you are going to do requires a distance,

for nearby in in_range(origin, walking_distance, hotdog_stands):
print("%s %.2fm" % (nearby.name, distance(origin, nearby)))



consider yielding tuples:

def in_range_with_dist_sq(origin, range, things):
range_sq = range**2
for thing in things:
dist_sq = distance_sq(origin, thing)
if dist_sq <= range_sq: yield (thing, dist_sq)



This can be especially useful if you might chain range checks (‘find things that are near X and within Nm of Y’, since you don’t have to calculate the distance again).

But what about if we’re searching a really large list of things and we anticipate a lot of them not being worth consideration?

There is actually a very simple optimization:

def in_range_all_the_things(origin, range, things):
range_sq = range**2
for thing in things:
dist_sq = (origin.x - thing.x) ** 2
if dist_sq <= range_sq:
dist_sq += (origin.y - thing.y) ** 2
if dist_sq <= range_sq:
dist_sq += (origin.z - thing.z) ** 2
if dist_sq <= range_sq:
yield thing



Whether this is useful will depend on the size of ‘things’.

def in_range_all_the_things(origin, range, things):
range_sq = range**2
if len(things) >= 4096:
for thing in things:
dist_sq = (origin.x - thing.x) ** 2
if dist_sq <= range_sq:
dist_sq += (origin.y - thing.y) ** 2
if dist_sq <= range_sq:
dist_sq += (origin.z - thing.z) ** 2
if dist_sq <= range_sq:
yield thing
elif len(things) > 32:
for things in things:
dist_sq = (origin.x - thing.x) ** 2
if dist_sq <= range_sq:
dist_sq += (origin.y - thing.y) ** 2 + (origin.z - thing.z) ** 2
if dist_sq <= range_sq:
yield thing
else:
... just calculate distance and range-check it ...



And again, consider yielding the dist_sq. Our hotdog example then becomes:

# Chaining generators
info = in_range_with_dist_sq(origin, walking_distance, hotdog_stands)
info = (stand, dist_sq**0.5 for stand, dist_sq in info)
for stand, dist in info:
print("%s %.2fm" % (stand, dist))



37 people think this answer is useful

Another instance of this problem solving method:

def dist(x,y):
return numpy.sqrt(numpy.sum((x-y)**2))

a = numpy.array((xa,ya,za))
b = numpy.array((xb,yb,zb))
dist_a_b = dist(a,b)



14 people think this answer is useful

Starting Python 3.8, the math module directly provides the dist function, which returns the euclidean distance between two points (given as tuples or lists of coordinates):

from math import dist

dist((1, 2, 6), (-2, 3, 2)) # 5.0990195135927845



And if you’re working with lists:

dist([1, 2, 6], [-2, 3, 2]) # 5.0990195135927845



12 people think this answer is useful

It can be done like the following. I don’t know how fast it is, but it’s not using NumPy.

from math import sqrt
a = (1, 2, 3) # Data point 1
b = (4, 5, 6) # Data point 2
print sqrt(sum( (a - b)**2 for a, b in zip(a, b)))



11 people think this answer is useful

A nice one-liner:

dist = numpy.linalg.norm(a-b)



However, if speed is a concern I would recommend experimenting on your machine. I’ve found that using math library’s sqrt with the ** operator for the square is much faster on my machine than the one-liner NumPy solution.

I ran my tests using this simple program:

#!/usr/bin/python
import math
import numpy
from random import uniform

def fastest_calc_dist(p1,p2):
return math.sqrt((p2 - p1) ** 2 +
(p2 - p1) ** 2 +
(p2 - p1) ** 2)

def math_calc_dist(p1,p2):
return math.sqrt(math.pow((p2 - p1), 2) +
math.pow((p2 - p1), 2) +
math.pow((p2 - p1), 2))

def numpy_calc_dist(p1,p2):
return numpy.linalg.norm(numpy.array(p1)-numpy.array(p2))

TOTAL_LOCATIONS = 1000

p1 = dict()
p2 = dict()
for i in range(0, TOTAL_LOCATIONS):
p1[i] = (uniform(0,1000),uniform(0,1000),uniform(0,1000))
p2[i] = (uniform(0,1000),uniform(0,1000),uniform(0,1000))

total_dist = 0
for i in range(0, TOTAL_LOCATIONS):
for j in range(0, TOTAL_LOCATIONS):
dist = fastest_calc_dist(p1[i], p2[j]) #change this line for testing
total_dist += dist

print total_dist



On my machine, math_calc_dist runs much faster than numpy_calc_dist: 1.5 seconds versus 23.5 seconds.

To get a measurable difference between fastest_calc_dist and math_calc_dist I had to up TOTAL_LOCATIONS to 6000. Then fastest_calc_dist takes ~50 seconds while math_calc_dist takes ~60 seconds.

You can also experiment with numpy.sqrt and numpy.square though both were slower than the math alternatives on my machine.

My tests were run with Python 2.6.6.

10 people think this answer is useful

I find a ‘dist’ function in matplotlib.mlab, but I don’t think it’s handy enough.

I’m posting it here just for reference.

import numpy as np
import matplotlib as plt

a = np.array([1, 2, 3])
b = np.array([2, 3, 4])

# Distance between a and b
dis = plt.mlab.dist(a, b)



8 people think this answer is useful

I like np.dot (dot product):

a = numpy.array((xa,ya,za))
b = numpy.array((xb,yb,zb))

distance = (np.dot(a-b,a-b))**.5



8 people think this answer is useful

You can just subtract the vectors and then innerproduct.

a = numpy.array((xa, ya, za))
b = numpy.array((xb, yb, zb))

tmp = a - b
sum_squared = numpy.dot(tmp.T, tmp)
result = numpy.sqrt(sum_squared)



6 people think this answer is useful

Having a and b as you defined them, you can use also:

distance = np.sqrt(np.sum((a-b)**2))



6 people think this answer is useful

With Python 3.8, it’s very easy.

https://docs.python.org/3/library/math.html#math.dist

math.dist(p, q)



Return the Euclidean distance between two points p and q, each given as a sequence (or iterable) of coordinates. The two points must have the same dimension.

Roughly equivalent to:

sqrt(sum((px - qx) ** 2.0 for px, qx in zip(p, q)))

5 people think this answer is useful

Here’s some concise code for Euclidean distance in Python given two points represented as lists in Python.

def distance(v1,v2):
return sum([(x-y)**2 for (x,y) in zip(v1,v2)])**(0.5)



5 people think this answer is useful

## Since Python 3.8

Since Python 3.8 the math module includes the function math.dist().
See here https://docs.python.org/3.8/library/math.html#math.dist.

math.dist(p1, p2)
Return the Euclidean distance between two points p1 and p2, each given as a sequence (or iterable) of coordinates.

import math
print( math.dist( (0,0),   (1,1)   )) # sqrt(2) -> 1.4142
print( math.dist( (0,0,0), (1,1,1) )) # sqrt(3) -> 1.7321



3 people think this answer is useful

Calculate the Euclidean distance for multidimensional space:

 import math

x = [1, 2, 6]
y = [-2, 3, 2]

dist = math.sqrt(sum([(xi-yi)**2 for xi,yi in zip(x, y)]))
5.0990195135927845



2 people think this answer is useful
import numpy as np
from scipy.spatial import distance
input_arr = np.array([[0,3,0],[2,0,0],[0,1,3],[0,1,2],[-1,0,1],[1,1,1]])
test_case = np.array([0,0,0])
dst=[]
for i in range(0,6):
temp = distance.euclidean(test_case,input_arr[i])
dst.append(temp)
print(dst)



2 people think this answer is useful
import math

dist = math.hypot(math.hypot(xa-xb, ya-yb), za-zb)



2 people think this answer is useful

You can easily use the formula

distance = np.sqrt(np.sum(np.square(a-b)))



which does actually nothing more than using Pythagoras’ theorem to calculate the distance, by adding the squares of Δx, Δy and Δz and rooting the result.

1 people think this answer is useful

Find difference of two matrices first. Then, apply element wise multiplication with numpy’s multiply command. After then, find summation of the element wise multiplied new matrix. Finally, find square root of the summation.

def findEuclideanDistance(a, b):
euclidean_distance = a - b
euclidean_distance = np.sum(np.multiply(euclidean_distance, euclidean_distance))
euclidean_distance = np.sqrt(euclidean_distance)
return euclidean_distance



1 people think this answer is useful
import numpy as np
# any two python array as two points
a = [0, 0]
b = [3, 4]



You first change list to numpy array and do like this: print(np.linalg.norm(np.array(a) - np.array(b))). Second method directly from python list as: print(np.linalg.norm(np.subtract(a,b)))

0 people think this answer is useful

The other answers work for floating point numbers, but do not correctly compute the distance for integer dtypes which are subject to overflow and underflow. Note that even scipy.distance.euclidean has this issue:

>>> a1 = np.array(, dtype='uint8')
>>> a2 = np.array(, dtype='uint8')
>>> a1 - a2
array(, dtype=uint8)
>>> np.linalg.norm(a1 - a2)
255.0
>>> from scipy.spatial import distance
>>> distance.euclidean(a1, a2)
255.0



This is common, since many image libraries represent an image as an ndarray with dtype=”uint8″. This means that if you have a greyscale image which consists of very dark grey pixels (say all the pixels have color #000001) and you’re diffing it against black image (#000000), you can end up with x-y consisting of 255 in all cells, which registers as the two images being very far apart from each other. For unsigned integer types (e.g. uint8), you can safely compute the distance in numpy as:

np.linalg.norm(np.maximum(x, y) - np.minimum(x, y))



For signed integer types, you can cast to a float first:

np.linalg.norm(x.astype("float") - y.astype("float"))



For image data specifically, you can use opencv’s norm method:

import cv2
cv2.norm(x, y, cv2.NORM_L2)