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

1 Ansicht (letzte 30 Tage)
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 Kommentare
Greg Heath
Greg Heath am 13 Aug. 2013
Bearbeitet: Greg Heath am 13 Aug. 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a. is obsolete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 am 19 Aug. 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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Akzeptierte Antwort

Greg Heath
Greg Heath am 11 Aug. 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

Weitere Antworten (1)

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

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