Ho to add activation function neuron?

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I am developing backpropagation Neural Network for classification, but I am confused how to determine the number of neurons in layer and the activation functions for each neuron? The input parameters are 9 and the output is 8
Scripts for training are as follows :
%clear; clc; close all; %%%
%This code to choose the .csv file as 'training input data' %clear; clc; [FileName,PathName] = uigetfile('.*','Select data as Training Input'); PD=csvread([PathName,FileName],1,11); % csv file from sensor PD=abs(PD);%% Absolute all Parameters of (.csv)
%%% This code to normalize all input parameters % Normalization of 9 Parameters(.csv) A=PD(:,1); %peak value normA = A - min(A(:)); %norm of peak value normA = normA ./ max(normA(:));
B=PD(:,6); %phase normB = B - min(B(:)); %norm of phase normB = normB ./ max(normB(:));
C=PD(:,11); %risetime normC = C - min(C(:)); %norm of risetime normC = normC ./ max(normC(:));
D=PD(:,12); %falltime normD = D - min(D(:)); %norm of falltime normD = normD ./ max(normD(:));
E=PD(:,13); %pulsewidth normE = E - min(E(:)); %pulsewidth normE = normE ./ max(normE(:));
F=PD(:,15); %pulsearea normF = F - min(F(:)); %pulsearea normF = normF ./ max(normF(:));
G=PD(:,16); %eventwidth normG = G - min(G(:)); %eventwidth normG = normG ./ max(normG(:));
H=PD(:,17); %freqofmaxampltd normH = H - min(H(:)); %freqofmaxampltd normH = normH ./ max(normH(:));
I=PD(:,19); %freqof1stmom normI = I - min(I(:)); %freqof1stmom normI = normI ./ max(normI(:));
normPD = table([normA],[normB],[normC],[normD],[normE],[normF],[normG],[normH],[normI]);
% this code to assign 'x' as a variable for training input normPD=table2array(normPD); x=normPD; x=x'; %transpose the training input %L=length(x); %count the number of rows 'x' %%%
%This code to choose the .csv file as 'target output data' [FileName,PathName] = uigetfile('.*','Select file as Target Output'); target_output_TEV=csvread([PathName,FileName],0,0); target_output_TEV=target_output_TEV'; %transpose the target output t=target_output_TEV; %%%
% Choose a Training Function trainFcn = 'trainbr'; % Bayesian Regularization
% Choose a Performance Function performFcn = 'mse'; % Mean-square error
% Create a Pattern Recognition Network hiddenLayerSize = 13; net9_tev2d = patternnet(hiddenLayerSize,trainFcn, performFcn);
% Setup Division of Data for Training, Validation, Testing % For a list of all data division functions type: help nndivide RandStream.setGlobalStream(RandStream('mt19937ar','seed',1)); % to get constant result net9_tev2d.divideFcn = 'dividerand'; % Divide data randomly net9_tev2d.divideMode = 'sample'; % Divide up every sample net9_tev2d.divideParam.trainRatio = 70/100; net9_tev2d.divideParam.valRatio = 15/100; net9_tev2d.divideParam.testRatio = 15/100;
%Training Parameters net9_tev2d.trainParam.show=50; %# of ephocs in display net9_tev2d.trainParam.lr=0.05; %learning rate net9_tev2d.trainParam.epochs=1000; %max epochs net9_tev2d.trainParam.goal=0.01^2; %training goal net9_tev2d.efficiency.memoryReduction=100; net9_tev2d.trainParam.max_fail=1000;%maximum validation fail net9_tev2d.trainParam.min_grad=1e-5;
% Choose a Performance Function % For a list of all performance functions type: help nnperformance net9_tev2d.performFcn = 'mse'; % Mean-square error
% Choose Plot Functions % For a list of all plot functions type: help nnplot net9_tev2d.plotFcns = {'plotperform','plottrainstate','ploterrhist', ... 'plotconfusion', 'plotroc'};
% Train the Network [net9_tev2d,tr] = train(net9_tev2d,x,t);
% Test the Network y = net9_tev2d(x); e = gsubtract(t,y); performance = perform(net9_tev2d,t,y) tind = vec2ind(t); yind = vec2ind(y); percentErrors = sum(tind ~= yind)/numel(tind);
% Recalculate Training, Validation and Test Performance trainTargets = t .* tr.trainMask{1}; valTargets = t .* tr.valMask{1}; testTargets = t .* tr.testMask{1}; trainPerformance = perform(net9_tev2d,trainTargets,y) valPerformance = perform(net9_tev2d,valTargets,y) testPerformance = perform(net9_tev2d,testTargets,y)
% Plots figure, plotconfusion(t,y)
Than you so much for the help.

Risposta accettata

Greg Heath
Greg Heath il 4 Ott 2018
Search the NEWSGROUP AND ANSWERS using
greg patternnet
Hope this helps.
Thank you for formally accepting my answer
Greg

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