How can the Euclidean distance be calculated with NumPy?

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

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

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)))

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)

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)))

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([1], dtype='uint8')
>>> a2 = np.array([2], dtype='uint8')
>>> a1 - a2
array([255], 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)

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

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

Well best way would be safest and also the fastest

I would suggest hypot usage for reliable results for chances of underflow and overflow are very little compared to writing own sqroot calculator

Lets see math.hypot, np.hypot vs vanilla np.sqrt(np.sum((np.array([i, j, k])) ** 2, axis=1))

i, j, k = 1e+200, 1e+200, 1e+200
math.hypot(i, j, k)
# 1.7320508075688773e+200

np.sqrt(np.sum((np.array([i, j, k])) ** 2))
# RuntimeWarning: overflow encountered in square

Speed wise math.hypot look better

%%timeit
math.hypot(i, j, k)
# 100 ns ± 1.05 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)

%%timeit
np.sqrt(np.sum((np.array([i, j, k])) ** 2))
# 6.41 µs ± 33.3 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

Underflow

i, j = 1e-200, 1e-200
np.sqrt(i**2+j**2)
# 0.0

Overflow

i, j = 1e+200, 1e+200
np.sqrt(i**2+j**2)
# inf

No Underflow

i, j = 1e-200, 1e-200
np.hypot(i, j)
# 1.414213562373095e-200

No Overflow

i, j = 1e+200, 1e+200
np.hypot(i, j)
# 1.414213562373095e+200

Refer