[PYTHON] Standardize the vector to any range

There are many opportunities to standardize an appropriate vector $ v $ into the range $ [0, 1] $ and $ [-1, 1] $. Now create $ v $ as a vector with random numbers like this:

random_vector.py

import random
random.seed(1)
v = [random.random() for _ in range(10)]

In the python code above, $ v $ is given, for example:

To standardize this vector to the range $ [0,1], [-1,1] $, respectively:

-Standardized to $ [0,1] $

\frac{v - min(v)}{max(v) - min(v)}

-Standardized to $ [-1,1] $

\frac{2v - (max(v) + min(v))}{max(v)-min(v)}

This will linearly standardize $ v $.

However, there are times when you want to ** standardize ** a vector to ** arbitrary range ** [newmin, newmax].

In such a case, first calculate two constants $ a and b $.

a = (newmax - newmin)/(max(v) - min(v))　\\
b = newmax - a*max(v)

At this time, the same is true for $ b = newmin --a * min (v) $.

Finally, create a linear function ** with ** $ a $ tilting and $ b $ as the intercept, and substitute the original vector $ v $ to get a new linearly standardized vector.

normalized\ vector = a*v + b

This is the general form for linearly standardizing vectors.

Here is the code that standardizes the vector $ v $ in the range [newmin, newmax] in Python and Matlab.

Python

normalize.py

def normalize(v,newmin,newmax):
    a = (newmax - newmin)/(max(v) - min(v))
    b = newmax - a*max(v)
        
    return [a*i + b for i in v]

Matlab

normalize.m

function [normalized_vector] = normalize(v, newmin, newmax)

a = (newmax - newmin)/(max(v) - min(v));
b = newmax - max(v)*a;

normalized_vector = a*v + b;