[LINR1] - Linear regression with direct resolution

Low-level implementation, using numpy, of a direct resolution for a linear regression

Objectives :

• Just one, the illustration of a direct resolution :-)

What we're going to do :

Equation : $$Y = X.\theta + N$$
Where N is a noise vector and $\theta = (a,b)$ a vector as y = a.x + b

Step 1 - Import and init

import numpy as np
import math
import matplotlib
import matplotlib.pyplot as plt
import sys
import fidle

# Init Fidle environment
run_id, run_dir, datasets_dir = fidle.init('LINR1')

FIDLE - Environment initialization

Version              : 2.3.2
Run id               : LINR1
Run dir              : ./run/LINR1
Datasets dir         : /lustre/fswork/projects/rech/mlh/uja62cb/fidle-project/datasets-fidle
Start time           : 22/12/24 21:20:36
Hostname             : r3i5n3 (Linux)
Tensorflow log level : Info + Warning + Error  (=0)
Update keras cache   : False
Update torch cache   : False
Save figs            : ./run/LINR1/figs (True)
numpy                : 2.1.2
sklearn              : 1.5.2
yaml                 : 6.0.2
matplotlib           : 3.9.2
pandas               : 2.2.3

Step 2 - Retrieve a set of points

# ---- Paramètres
nb    = 100     # Nombre de points
xmin  = 0       # Distribution / x
xmax  = 10
a     = 4       # Distribution / y
b     = 2       # y= a.x + b (+ bruit)
noise = 7       # bruit

theta = np.array([[a],[b]])

# ---- Vecteur X  (1,x) x nb
#      la premiere colonne est a 1 afin que X.theta <=> 1.b + x.a

Xc1 = np.ones((nb,1))
Xc2 = np.random.uniform(xmin,xmax,(nb,1))
X = np.c_[ Xc1, Xc2 ]

# ---- Noise
# N = np.random.uniform(-noise,noise,(nb,1))
N = noise * np.random.normal(0,1,(nb,1))

# ---- Vecteur Y
Y = (X @ theta) + N

# print("X:\n",X,"\nY:\n ",Y)

Show it

width = 12
height = 6

fig, ax = plt.subplots()
fig.set_size_inches(width,height)
ax.plot(X[:,1], Y, ".")
ax.tick_params(axis='both', which='both', bottom=False, left=False, labelbottom=False, labelleft=False)
ax.set_xlabel('x axis')
ax.set_ylabel('y axis')
fidle.scrawler.save_fig('01-set_of_points')
plt.show()

Saved: ./run/LINR1/figs/01-set_of_points

Step 3 - Direct calculation of the normal equation

We'll try to find an optimal value of $\theta$, minimizing a cost function.
The cost function, classically used in the case of linear regressions, is the root mean square error (racine carré de l'erreur quadratique moyenne):

$$RMSE(X,h_\theta)=\sqrt{\frac1n\sum_{i=1}^n\left[h_\theta(X^{(i)})-Y^{(i)}\right]^2}$$

With the simplified variant : $$MSE(X,h_\theta)=\frac1n\sum_{i=1}^n\left[h_\theta(X^{(i)})-Y^{(i)}\right]^2$$

The optimal value of regression is : $$\hat{ \theta } =( X^{T} .X)^{-1}.X^{T}.Y$$

Démontstration : https://eli.thegreenplace.net/2014/derivation-of-the-normal-equation-for-linear-regression

theta_hat = np.linalg.inv(X.T @ X) @ X.T @ Y

print("Theta :\n",theta,"\n\ntheta hat :\n",theta_hat)

Theta :
 [[4]
 [2]] 

theta hat :
 [[1.65668594]
 [2.41198387]]

Show it

Xd = np.array([[1,xmin], [1,xmax]])
Yd = Xd @ theta_hat

fig, ax = plt.subplots()
fig.set_size_inches(width,height)
ax.plot(X[:,1], Y, ".")
ax.plot(Xd[:,1], Yd, "-")
ax.tick_params(axis='both', which='both', bottom=False, left=False, labelbottom=False, labelleft=False)
ax.set_xlabel('x axis')
ax.set_ylabel('y axis')
fidle.scrawler.save_fig('02-regression-line')
plt.show()

Saved: ./run/LINR1/figs/02-regression-line

fidle.end()

End time : 22/12/24 21:20:37
Duration : 00:00:01 091ms
This notebook ends here :-)
https://fidle.cnrs.fr

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