Plz, Edit the NEWFF according to the latest version of MATLAB.
Afficher commentaires plus anciens
when i simulate the below code it is showing some errors.
like obsolete way of using NEWFF.
what is the new model for it ?
Can some one edit the NEWFF according to the latest version.
- load data.txt
- P = data(1:15,1);
- T = data(16:30,1);
- a = data(31:45,1);
- s = data(46:60,1);
- [py, pys] = mapminmax(P');
- [ay, ays] = mapminmax(a');
- [ty, tys] = mapminmax(T');
- [sy, sys] = mapminmax(s');
- net = newff(minmax(py),[6 1], {'logsig','logsig'}, 'triangdm')
- net.trainParam.epochs = 3000;
- net.trainParam.lr = 0.5;
- net.trainParm.mc = 0.8;
- net = train(net,py,ty);
- y = sim(net,ay);
6 commentaires
Greg Heath
le 7 Août 2013
Modifié(e) : Greg Heath
le 7 Août 2013
0. help newff
doc newff
type newff
1. Do you think it is inconsiderate of you to not include one of the MATLAB data sets for us to use as an example?
>> help nndatasets
2. Do you think it is inconsiderate of you to not include your error messages"
3. Why not use the defaults as in the help newff example?
4. Why not use the transpose in statements 2-5?
5. traingdm is misspelled
6. net.trainParam is misspelled
7. mapminmax is a default
8. {'tansig' 'tansig' } is the natural combination to use with the mapminmax transformations to [ -1, 1 ]
net = newff(P,T, 6 ,{'tansig' 'tansig'},'traingdm')
9. The defaults settings for 'traingdm' are
net.trainParam.epochs = 1000
net.trainParam.lr = 0.01
net.trainParam.mc = 0.9
Anjireddy Thatiparthy
le 10 Août 2013
Greg Heath
le 10 Août 2013
How many hidden nodes did you need to get a satisfactory answer?
Anjireddy Thatiparthy
le 13 Août 2013
Greg Heath
le 13 Août 2013
Modifié(e) : Greg Heath
le 13 Août 2013
1. That is not a clear explanation AND it seems to have little to do with your original post.
2. Why are you posting an equation that
a. is obsolete
b. has inappropriate transfer functions
c. has a misspelled training function (to which you were alerted earlier)
3. If you have 2012a, why are you trying to use the obsolete newff?
4. Now it seems that you might want the simple classifier
output = hardlim(input-5663)
4. Please clarify.
a. Single output y(t) = ( 566x.xx or 0/1?)
b. Corresponding input y( t-d:t-1)
Anjireddy Thatiparthy
le 19 Août 2013
Réponse acceptée
Plus de réponses (1)
Greg Heath
le 7 Août 2013
if true
% code
end
clear all, clc
[ inputs, targets ] = simplefit_dataset;
P = inputs(1:2:end);
T = targets(1:2:end);
[ I N ] = size(P)
[ O N ] = size(T)
MSE00 = var(T,1) % 8.3328 Reference MSE
Neq = N*O % No. of equations = prod(size(T)
a = inputs(2:2:end);
s = targets(2:2:end);
% Nw = (I+1)*H+(H+1)*O % No. of weights = Nw
{Hub = -1+ceil( (Neq-O)/(I+O+1)) % 15 (Neq >= Nw)
Hmin = 0
dH = 2
Hmax =ceil(Hub/2)
Ntrials = 10
MSEgoal = MSE00/100
MinGrad = MSEgoal/10
rng(0)
j = 0
for h = Hmin:dH:Hmax
j=j+1
if h ==0
net = newff(P,T, []);
else
net=newff(P,T,h);
end
for i = 1:Ntrials
hidden = h
ntrials = i
net.trainParam.goal = MSEgoal;
net.trainParam.min_grad = MinGrad;
[ net tr Y E ]= train(net,P,T);
NMSE(i,j) = mse(E)/MSE00;
end
end
NMSEtst = mse(s-net(a))/var(s,1) %4.0567e-005
H = Hmin:dH:Hmax
NMSE=NMSE
2 commentaires
Anjireddy Thatiparthy
le 13 Août 2013
Greg Heath
le 24 Oct 2013
Sorry I missed your comment. If you have any SPECIFIC questions on the code,
please post.
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