Calculate Euclidean distance between 4-dimensional vectors

Question

Let's say I have two 4-dimensional vectors (i.e. a and b) as follows:

a = {a1, a2, a3, a4}
b= {b1, b2, b3, b4}

How do I compute the Euclidean distance between these vectors?

Ahwaz

Same as for a 2D or 3D vector. You just keep adding +(a-b)^2 terms to the sqrt. I.E. sqrt((a1-b1)^2+(a2-b2)^2...) – Clothespin

I'm voting to close this question as off-topic because it is about math, not programming. – Dribble

I'm voting to close this question as off-topic because it is about Mathematics instead of programming or software development. – Determined

Answer 1

The euclidian distance calculus is independent of dimensions.

In your case, the euclidian distance between a and b can be written as:

d(a,b) = sqrt( sum_{ i=1 } ^ { 4 } (a[ i ] - b[ i ])^2 )

Or, more specifically:

d(a,b) = sqrt( (a1 - b1)^2 + (a2 - b2)^2 + (a3 -b3)^2 + (a4 - b4)^2 )

Elbowroom

Easy to think of other distance measures: Pearson, Manhattan, etc. – Magnesia

Answer 2

public static float ndistance(float[] a, float[] b) {
    float total = 0, diff;
    for (int i = 0; i < a.length; i++) {
        diff = b[i] - a[i];
        total += diff * diff;
    }
    return (float) Math.sqrt(total);
}

The function/method/code above will calculate the distance in n-dimensional space. a and b are arrays of floating point number and have the same length/size or simply the n. Since you want a 4-dimension, you simply pass a 4-length array representing the data of your 4-D vector.

Vala

Answer 3

Old questions but I thought I'd share a one liner:

import numpy as np
from functools import reduce

def euclidean_distance(arr: np.ndarray):
    """
        d = √((x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)² + ... + (xn₂ - xn₁)²)
    """
    return np.sqrt(sum(np.power(reduce(np.subtract, arr), 2)))

Gaona