Plz, Edit the NEWFF according to the latest version of MATLAB.

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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.
  1. load data.txt
  2. P = data(1:15,1);
  3. T = data(16:30,1);
  4. a = data(31:45,1);
  5. s = data(46:60,1);
  6. [py, pys] = mapminmax(P');
  7. [ay, ays] = mapminmax(a');
  8. [ty, tys] = mapminmax(T');
  9. [sy, sys] = mapminmax(s');
  10. net = newff(minmax(py),[6 1], {'logsig','logsig'}, 'triangdm')
  11. net.trainParam.epochs = 3000;
  12. net.trainParam.lr = 0.5;
  13. net.trainParm.mc = 0.8;
  14. net = train(net,py,ty);
  15. y = sim(net,ay);
  6 comentarios
Greg Heath
Greg Heath el 13 de Ag. de 2013
Editada: Greg Heath el 13 de Ag. de 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a. is obsolete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
Anjireddy Thatiparthy el 19 de Ag. de 2013
I got it from old records. i want to re-build it, so i edited few commands with my knowledge. i'm not able to do it further.
4.
a. Single output y(t) = ( 566x.xx or 0/1?) it is in the range of -1 to 1(includes both boundaries).

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Respuesta aceptada

Greg Heath
Greg Heath el 11 de Ag. de 2013
This is a Time-Series Problem that can be solved using NARNET with a feedback delay of 15.
help NARNET
doc NARNET
Search NARNET in the NEWSGROUP and ANSWERS
Thank you for formally accepting my answer
Greg

Más respuestas (1)

Greg Heath
Greg Heath el 7 de Ag. de 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 comentarios
Anjireddy Thatiparthy
Anjireddy Thatiparthy el 13 de Ag. de 2013
I didn't get it.
Can you explain.
Thanks
@#?!
Greg Heath
Greg Heath el 24 de Oct. de 2013
Sorry I missed your comment. If you have any SPECIFIC questions on the code,
please post.

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