performing a least squares with regularisation in matlab

25 visualizzazioni (ultimi 30 giorni)
cgo
cgo il 13 Lug 2018
Spostato: Bruno Luong il 26 Ott 2024 alle 7:44
I have data sets X (2n by 8) and Y(2n by 1). I want to find the coefficients a so that Y = Xa. So we can perform a = X\Y (as a least squares minimisation).
I wanted to ask if it possible to proceed with a form of regularisation (L1 or something simple) from this?
Please help.
  1 Commento
SAKO
SAKO il 25 Ott 2024 alle 21:41
Spostato: Bruno Luong il 26 Ott 2024 alle 7:44
bonjours,je n'écris pas pour repondre a une question mais pour poser ma préoccupation.j'ai utiliser le package TOOL BOX de Per Christian Hansen pour faire une reconstruction de force.Avec la regularisation de Tikhonov pour le critère L_curve,le paramètre de regularisation qu'il me renvoi ne me permet pas de reconstruire ma force(ma courbe L_curve presente deux coins).Pouvez vous m'aider ?

Accedi per commentare.

Risposte (2)

Diwakar
Diwakar il 13 Lug 2018
My understanding of your problem is that you want to find the coefficient a. So in order to implement optimization you can implement average of sum of least squares as shown below.
Loss= ((Y-X*a)'*(Y-X*a))/(2*n);
The above shown function is a vectorized implementation of the squared error loss function. So this can be minimized in order to get the optimal value of a. If you want to fit a curve to this then any form of regularization should be fine.
Hope this helps
Cheers!

Bruno Luong
Bruno Luong il 23 Set 2020
Modificato: Bruno Luong il 23 Set 2020
Simpless method:
n = size(X,2); % 8
lambda = 1e-6; % <= regularization parameter, 0 no regularization, larger value stronger regularized solution
a = [X; lambda*eye(n)] \ [Y; zeros(n,1)]

Categorie

Scopri di più su Mathematics and Optimization in Help Center e File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by