PLS:利用PLS(两个主成分的贡献率就可达100%)提高测试集辛烷值含量预测准确度并《测试集辛烷值含量预测结果对比》—Jason niu

load spectra;

temp = randperm(size(NIR, 1));

P_train = NIR(temp(1:50),:);
T_train = octane(temp(1:50),:);

P_test = NIR(temp(51:end),:);
T_test = octane(temp(51:end),:);

k = 2;  
[Xloadings,Yloadings,Xscores,Yscores,betaPLS,PLSPctVar,MSE,stats] = plsregress(P_train,T_train,k);

figure
percent_explained = 100 * PLSPctVar(2,:) / sum(PLSPctVar(2,:));
pareto(percent_explained)
xlabel('主成分')
ylabel('贡献率(%)')
title('PLS:各个主成分的贡献率—Jason niu')

N = size(P_test,1);
T_sim = [ones(N,1) P_test] * betaPLS;

error = abs(T_sim - T_test) ./ T_test;

R2 = (N * sum(T_sim .* T_test) - sum(T_sim) * sum(T_test))^2 / ((N * sum((T_sim).^2) - (sum(T_sim))^2) * (N * sum((T_test).^2) - (sum(T_test))^2)); 

result = [T_test T_sim error]

figure
plot(1:N,T_test,'b:*',1:N,T_sim,'r-o')
legend('真实值','预测值','location','best')
xlabel('预测样本')
ylabel('辛烷值')
string = {'PLS:利用PLS(两个主成分的贡献率就可达100%)提高《测试集辛烷值含量预测结果对比》的准确度—Jason niu';['R^2=' num2str(R2)]};
title(string)

 

posted @ 2018-03-03 22:34  一个处女座的程序猿  阅读(1077)  评论(0编辑  收藏  举报