Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
Drexel University or The Math Forum.



Re: Data division problem in neural network
Posted:
Apr 27, 2013 6:17 PM


"srishti" wrote in message <kl6731$8a8$1@newscl01ah.mathworks.com>... > Hello, > I am using neural network pattern recognition for classification purpose.My problem is I am not getting whether for input and target both I have to write individual command "trainInd,valInd,testInd] = divideind(Q,trainInd,valInd,testInd)" > and if yes then how to define parameters(net.divideparm)? I have written the following code.I have read the documentation but I am not able to clear my doubt. Please help. > > s = RandStream('mcg16807','Seed', 0); > RandStream.setDefaultStream(s) > x=input; % size of x is 10x70 > t=target;% size of t is 3x70
[I N ] = size(x) % [ 10 70 ] [O N ] = size(t) % [ 3 70 ]
% The default division ratios are 0.7/0.15/0.15 Ntst =round(0.15*N) % 11 Nval = Ntst % 11 Ntrn = N2*Ntst % 48 trainind = 1:Ntst valind = Ntrn+1:Ntrn+Nval tstind = Ntrn+Nval+1:N Ntrneq = Ntrn*O % 144 No. of training equations
% No. of unknown weights % Nw = (I+1)*H+(H+1)*O % For more equations than unknowns H <= Hub
Hub= 1+ ceil( (Ntrneq O) / (I+O+1) ) % 10
> net = patternnet(22);
No. Choose H by trial and error for 0 <= H <= 10
H = 5 % Prefer a loop H = 0:1:10 net = patternnet(H); % remove semicolon to see default parameter values
> net.divideFcn='divideInd'; > [trainInd,valInd,testInd] = divideind(x,1:20,35:45,54:65);
Replace x with N Not sure why you are not using all of the data [trainInd,valInd,testInd] = divideind(N,trnind,valind,tstind);
> net.divideParam.trainInd = trainInd > net.divideParamvalInd = valInd > net.divideParam.testInd = testInd > net= train(net,x,t);
[ [ net tr y ] = train(net,x,t); % y is output
net = net % See all of the input parameters
tr = tr % See all of the training results.
Hope this helps.
Greg



