newnarx initialization and validation problem
4 visualizzazioni (ultimi 30 giorni)
Mostra commenti meno recenti
Dear all,
I am working on the problem: y(k)=0.3*y(k-1)+0.6*y(k-2)+u(k)^3+0.3*u(k)^2-0.4*u(k) using narx. the code is listed below.
net = newnarx(u,yn,0,1:2,[15 10],{'tansig','tansig','purelin'},'trainscg');
net.trainParam.lr = 0.05; net.trainParam.lr_inc = 1.05;
net.trainParam.lr_dec = 0.7; net.trainParam.hide = 50;
net.trainParam.mc = 0.9; net.trainParam.epochs = s;
net.trainParam.goal = 1e-8; net.trainParam.time = 5*3600; [trainP,valP,testV,trainInd,valInd,testInd] = divideblock(u,0.6,0.2,0.2);
[trainT,valT,testT] = divideind(yn,trainInd,valInd,testInd);
net.divideFcn='divideblock';
net = init(net);
My training stops after 6 validation checks. Why? The trained net gives different outputs each time. Is there any mistake in the initialization?
Please help.
0 Commenti
Risposta accettata
Greg Heath
il 27 Ott 2012
See my previous post re your last program. Most of the comments are relevant for this post.
In particular, do not overwrite defaults unless you have a darned good reason.
Also, it is very seldom that a 2nd hidden layer is necessary.
Validation stopping was created to make sure the trained net is useful for nontraining data. Stopping after MSEval increases for 6 consecutive epochs is reasonable.
You will get different results every time because the initial weights are random.
For each different setting of delays and hidden nodes I usually loop over Ntrials = 10 designs.
You would be surprised how many times only 7 or 8 out of 10 are acceptable.
However, I do initialize the random number generator before the double loop over hidden nodes and random weight initializations.
Hope this helps.
Thank you for formally accepting my answer.
Greg
Più risposte (0)
Vedere anche
Categorie
Scopri di più su Sequence and Numeric Feature Data Workflows in Help Center e File Exchange
Prodotti
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!