## calculate distance of 2 list of points in numpy

I have 2 lists of points as numpy.ndarray, each row is the coordinate of a point, like:

a = np.array([[1,0,0],[0,1,0],[0,0,1]]) b = np.array([[1,1,0],[0,1,1],[1,0,1]])

Here I want to calculate the euclidean distance between all pairs of points in the 2 lists, for each point p_a in a, I want to calculate the distance between it and every point p_b in b. So the result is

d = np.array([[1,sqrt(3),1],[1,1,sqrt(3)],[sqrt(3),1,1]])

How to use matrix multiplication in numpy to compute the distance matrix?

Using direct numpy broadcasting, you can do this:

dist = np.sqrt(((a[:, None] - b[:, :, None]) ** 2).sum(0))

Alternatively, `scipy`

has a routine that will compute this slightly more efficiently (particularly for large matrices)

from scipy.spatial.distance import cdist dist = cdist(a, b)

I would avoid solutions that depend on factoring-out matrix products (of the form A^2 + B^2 - 2AB), because they can be numerically unstable due to floating point roundoff errors.

**High-Performance computation in Python,** Euclidean Distance is a termbase in mathematics; therefore I won't discuss it at length. Generally speaking, it is a straight-line distance between two points in� mindist=numpy.zeros(len(xy1)) minid=numpy.zeros(len(xy1)) for i,xy in enumerate(xy1): dists=numpy.sqrt(numpy.sum((xy-xy2)**2,axis=1)) mindist[i],minid[i]=dists.min(),dists.argmin() Is there a way to eliminate the for loop and somehow do element-by-element calculations between the two arrays?

To compute the squared euclidean distance for each pair of elements off them - x and y, we need to find :

(Xik-Yjk)**2 = Xik**2 + Yjk**2 - 2*Xik*Yjk

and then sum along k to get the distance at coressponding point as dist(Xi,Yj).

Using associativity, it reduces to :

dist(Xi,Yj) = sum_k(Xik**2) + sum_k(Yjk**2) - 2*sum_k(Xik*Yjk)

Bringing in matrix-multiplication for the last part, we would have all the distances, like so -

dist = sum_rows(X^2), sum_rows(Y^2), -2*matrix_multiplication(X, Y.T)

Hence, putting into NumPy terms, we would end up with the euclidean distances for our case with `a`

and `b`

as the inputs, like so -

np.sqrt((a**2).sum(1)[:,None] + (b**2).sum(1) - 2*a.dot(b.T))

Leveraging `np.einsum`

, we could replace the first two summation-reductions with -

np.einsum('ij,ij->i',a,a)[:,None] + np.einsum('ij,ij->i',b,b)

More info could be found on `eucl_dist`

package's `wiki page`

(disclaimer: I am its author).

**NumPy: Calculate the Euclidean distance,** From Wikipedia: In mathematics, the Euclidean distance or Euclidean metric is the "ordinary" straight-line distance between two points in� Euclidean Distance Euclidean metric is the “ordinary” straight-line distance between two points. if p = (p1, p2) and q = (q1, q2) then the distance is given by For three dimension1, formula is ##### # name: eudistance_samples.py # desc: Simple scatter plot # date: 2018-08-28 # Author: conquistadorjd ##### from scipy import spatial import numpy …

If you have 2 each 1-dimensional arrays, x and y, you can convert the arrays into matrices with repeating columns, transpose, and apply the distance formula. This assumes that x and y are coordinated pairs. The result is a symmetrical distance matrix.

x = [1, 2, 3] y = [4, 5, 6] xx = np.repeat(x,3,axis = 0).reshape(3,3) yy = np.repeat(y,3,axis = 0).reshape(3,3) dist = np.sqrt((xx-xx.T)**2 + (yy-yy.T)**2) dist Out[135]: array([[0. , 1.41421356, 2.82842712], [1.41421356, 0. , 1.41421356], [2.82842712, 1.41421356, 0. ]])

**scipy.spatial.distance.cdist — SciPy v1.5.2 Reference Guide,** Extra arguments to metric: refer to each metric documentation for a list of all possible arguments. Computes the city block or Manhattan distance between the points. The standardized Euclidean distance between two n-vectors u and v is. numpy.linalg.norm¶ numpy.linalg.norm (x, ord=None, axis=None, keepdims=False) [source] ¶ Matrix or vector norm. This function is able to return one of eight different matrix norms, or one of an infinite number of vector norms (described below), depending on the value of the ord parameter.

**How to find euclidean distance in Python,** The euclidean distance is the shortest distance between two points. For example, the euclidean distance between (0,0,0) and (1,1,1) is 1.73205� To compute the squared euclidean distance for each pair of elements off them - x and y, we need to find : (Xik-Yjk)**2 = Xik**2 + Yjk**2 - 2*Xik*Yjk and then sum along k to get the distance at coressponding point as dist(Xi,Yj). Using associativity, it reduces to : dist(Xi,Yj) = sum_k(Xik**2) + sum_k(Yjk**2) - 2*sum_k(Xik*Yjk)

**How can the Euclidean distance be calculated with NumPy ,** I have two points in 3D: (xa, ya, za) (xb, yb, zb) And I want to calculate the To calculate Euclidean distance with NumPy you can use numpy.linalg.norm: You can use the following piece of code to calculate the distance:- a = (1, 2, 3). If you just want the distances between each pair of points, then you don't need to calculate a full distance matrix. Instead, calculate it directly: import numpy as np x = np.array([[[1,2,3,4,5], [5,6,7,8,5], [5,6,7,8,5]], [[11,22,23,24,5], [25,26,27,28,5], [5,6,7,8,5]]]) y = np.array([[[31,32,33,34,5], [35,36,37,38,5], [5,6,7,8,5]], [[41,42,43,44,5], [45,46,47,48,5], [5,6,7,8,5]]]) xx = x.reshape(2, -1) yy

**Python Numpy Tutorial (with Jupyter and Colab),** SciPy. Image operations; MATLAB files; Distance between points. Matplotlib. Plotting xs = [3, 1, 2] # Create a list print(xs, xs[2]) # Prints "[3, 1, 2] 2" print(xs[-1 ])� Vincenty’s formulae are two related iterative methods used in geodesy to calculate the distance between two points on the surface of a spheroid, developed by Thaddeus Vincenty (1975a).