NP Euclidean Distance Python
import numpy as np
a = np.array((1,1,1))
b = np.array((2,2,2))
dist = np.linalg.norm(a-b)
Lovely Lark
import numpy as np
a = np.array((1,1,1))
b = np.array((2,2,2))
dist = np.linalg.norm(a-b)
# Use numpy.linalg.norm:
import numpy as np
a = np.array([1.0, 3.5, -6.3])
b = np.array([4.5, 1.6, 1.2])
dist = np.linalg.norm(a-b)# I hope to be of help and to have understood the request
from math import sqrt # import square root from the math module
# the x and y coordinates are the points on the Cartesian plane
pointA = (x, y) # first point
pointB = (x, y) # second point
distance = calc_distance(pointA, pointB) # here your beautiful result
def calc_distance(p1, p2): # simple function, I hope you are more comfortable
return sqrt((p1[0]-p2[0])**2+(p1[1]-p2[1])**2) # Pythagorean theorem# Python code to find Euclidean distance
# using distance.euclidean() method
# Import SciPi Library
from scipy.spatial import distance
# initializing points in
# numpy arrays
point1 = (4, 4, 2)
point2 = (1, 2, 1)
# print Euclidean distance
print(distance.euclidean(point1,point2))dist = numpy.linalg.norm(a-b)
# Python code to find Euclidean distance
# using dot() and sqrt() methods
# Import NumPy Library
import numpy as np
# initializing points in
# numpy arrays
point1 = np.array((4, 4, 2))
point2 = np.array((1, 2, 1))
# subtracting both the vectors
temp = point1 - point2
# Perform dot product
# and do the square root
dist = np.sqrt(np.dot(temp.T, temp))
# printing Euclidean distance
print(dist)