MATLAB: Do I need to normalize data when using patternnet or Matlab will take care of it

anncross-validationDeep Learning Toolboxnormalization

I am working on Electromyogram signals to classify 11 different hand movements using ANN. My code is as follow:
    load FeatureSet;
p=input;  %[16 by 1342 ]
t=target;  %[11 by 1342]
trainFcn = 'trainbr';  % Use Bayesian Regularization to prevent overtraining
net = patternnet([32 32], trainFcn);
net.layers{1}.transferFcn = 'tansig';    % hidden layer 1
net.layers{2}.transferFcn = 'logsig';    % hidden layer 2
% %------------------------parameter
net.trainParam.lr = 0.1;   %learning rate
net.trainParam.mc = 0.1;   %momentum  
%------------------------ train

[net,tr]= train(net,p,t);
%------------------------ test

outputs= sim(net,P);
[c,cm] = confusion(t,outputs);
pct=100*(1-c);   %correction rate
1) I am trying to normalize the input between 0 to +1 but I am not sure if it is needed. Dose patternnet do anything regarding the normalization of the input? my data range is between 0.1 to 120.
    normalized_p=[];
    for ii=1:length(p(1,:))
        norm_p=[];
        Max_p(ii)=max(p(:,ii));
        Min_p(ii)=min(p(:,ii));
        norm_p = (p(:,ii) - Min_p(ii))./(Max_p(ii) - Min_p(ii));
        normalized_p=[normalized_p, norm_p];
    end
2) I get the following warning when using patternnet:
   Warning:  
Performance function replaced with squared error performance.
    > In trainbr>formatNet (line 160)
  In trainbr (line 69)
  In nntraining.setup (line 14)
  In network/train (line 335)
  In ANN_demo (line 95)
I also used newpr, but I did not get any warning.
3) Patternnet divides data into three sets which are training, validation and testing. I found the following code to get the performance of the classifier but I do not know how to calculate the correction rate of each set.
RandStream.setGlobalStream(RandStream('mt19937ar','seed',1)); % to get constant result
net.divideFcn = 'divideblock'; % Divide targets into three sets using blocks of indices
net.divideParam.trainRatio = 0.6;
net.divideParam.valRatio = 0.2;
net.divideParam.testRatio = 0.2;
%------------------------ train
[net,tr]= train(net,p,t);
%------------------------ test
outputs = net(p);
performance  = perform(net,t,outputs) 
trainTargets = t .* tr.trainMask{1};
valTargets   = t .* tr.valMask{1};
testTargets  = t .* tr.testMask{1};
trainPerformance = perform(net,trainTargets,outputs)
valPerformance   = perform(net,valTargets,outputs)
testPerformance  = perform(net,testTargets,outputs)
4) I want to compare the result of ANN with some other classifiers such as LDA, SVM , etc. and I am using 10-fold cross validation (9 for training and 1 for testing) for those classifiers. In ANN, the data is divided to three parts (0.6 training 0.2 validation 0.2 testing). How can I evaluate all the methods in the same way. For example 10-fold cross validation for all of them including ANN.

Best Answer

BY DEFAULT: All of the neural network training algorithms normalize inputs and targets to [ -1 1 ] AND denormalizes the output using the parameters of the target normalization.
Why in the world would you substantially deviate from the examples in the documentation
 help patternnet
and
 doc patternnet ?
The first thing you should do is to use the documentation code in a loop with multiple random weight initializations.
Next use a double loop approach where the outer loop systematically increases the number of hidden nodes.
I have posted zillions of single and double loop examples in both the NEWSGROUP and ANSWERS. You can begin searches with
                                      HITS
                              NEWGROUP    ANSWERS
 greg patternet                    60       451
and
 greg patternnet tutorial          14        44
In particular note that most cases only involve searches over number of hidden nodes and initial random weights.
Although there are some cases of m-fold cross validation, the index bookkeeping is so tricky it tends to be far less fruitful than straightforward multiple uses of the default DIVIDERAND.
Hope this helps.
Thank you for formally accepting my answer
Greg
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