?? demsvm1.m
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function demsvm1()% DEMSVM1 - Demonstrate basic Support Vector Machine classification% % DEMSVM1 demonstrates the classification of a simple artificial data% set by a Support Vector Machine classifier, using different kernel% functions.%% See also% SVM, SVMTRAIN, SVMFWD, SVMKERNEL, DEMSVM2%% % Copyright (c) Anton Schwaighofer (2001) % This program is released unter the GNU General Public License.% X = [2 7; 3 6; 2 2; 8 1; 6 4; 4 8; 9 5; 9 9; 9 4; 6 9; 7 4];Y = [ +1; +1; +1; +1; +1; -1; -1; -1; -1; -1; -1];% define a simple artificial data setx1ran = [0 10];x2ran = [0 10];% range for plotting the data set and the decision boundarydisp(' ');disp('This demonstration illustrates the use of a Support Vector Machine');disp('(SVM) for classification. The data is a set of 2D points, together');disp('with target values (class labels) +1 or -1.');disp(' ');disp('The data set consists of the points');ind = [1:length(Y)]';fprintf('X%2i = (%2i, %2i) with label Y%2i = %2i\n', [ind, X, ind, Y]');disp(' ')disp('Press any key to plot the data set');pausef1 = figure;plotdata(X, Y, x1ran, x2ran);title('Data from class +1 (squares) and class -1 (crosses)');fprintf('\n\n\n\n');fprintf('The data is plotted in figure %i, where\n', f1);disp(' squares stand for points with label Yi = +1');disp(' crosses stand for points with label Yi = -1');disp(' ')disp(' ');disp('Now we train a Support Vector Machine classifier on this data set.');disp('We use the most simple kernel function, namely the inner product');disp('of points Xi, Xj (linear kernel K(Xi,Xj) = Xi''*Xj )');disp(' ');disp('Press any key to start training')pausenet = svm(size(X, 2), 'linear', [], 10);net = svmtrain(net, X, Y);f2 = figure;plotboundary(net, x1ran, x2ran);plotdata(X, Y, x1ran, x2ran);plotsv(net, X, Y);title(['SVM with linear kernel: decision boundary (black) plus Support' ... ' Vectors (red)']);fprintf('\n\n\n\n');fprintf('The resulting decision boundary is plotted in figure %i.\n', f2);disp('The contour plotted in black separates class +1 from class -1');disp('(this is the actual decision boundary)');disp('The contour plotted in red are the points at distance +1 from the');disp('decision boundary, the blue contour are the points at distance -1.');disp(' ');disp('All examples plotted in red are found to be Support Vectors.');disp('Support Vectors are the examples at distance +1 or -1 from the ');disp('decision boundary and all the examples that cannot be classified');disp('correctly.');disp(' ');disp('The data set shown can be correctly classified using a linear');disp('kernel. This can be seen from the coefficients alpha associated');disp('with each example: The coefficients are');ind = [1:length(Y)]';fprintf(' Example %2i: alpha%2i = %5.2f\n', [ind, ind, net.alpha]');disp('The upper bound C for the coefficients has been set to');fprintf('C = %5.2f. None of the coefficients are at the bound,\n', ... net.c(1));disp('this means that all examples in the training set can be correctly');disp('classified by the SVM.')disp(' ');disp('Press any key to continue')pauseX = [X; [4 4]];Y = [Y; -1];net = svm(size(X, 2), 'linear', [], 10);net = svmtrain(net, X, Y);f3 = figure;plotboundary(net, x1ran, x2ran);plotdata(X, Y, x1ran, x2ran);plotsv(net, X, Y);title(['SVM with linear kernel: decision boundary (black) plus Support' ... ' Vectors (red)']);fprintf('\n\n\n\n');disp('Adding an additional point X12 with label -1 gives a data set');disp('that can not be linearly separated. The SVM handles this case by');disp('allowing training points to be misclassified.');disp(' ');disp('Training the SVM on this modified data set we see that the points');disp('X5, X11 and X12 can not be correctly classified. The decision');fprintf('boundary is shown in figure %i.\n', f3);disp('The coefficients alpha associated with each example are');ind = [1:length(Y)]';fprintf(' Example %2i: alpha%2i = %5.2f\n', [ind, ind, net.alpha]');disp('The coefficients of the misclassified points are at the upper');disp('bound C.');disp(' ')disp('Press any key to continue')pausefprintf('\n\n\n\n');disp('Adding the new point X12 has lead to a more difficult data set');disp('that can no longer be separated by a simple linear kernel.');disp('We can now switch to a more powerful kernel function, namely');disp('the Radial Basis Function (RBF) kernel.');disp(' ')disp('The RBF kernel has an associated parameter, the kernel width.');disp('We will now show the decision boundary obtained from a SVM with');disp('RBF kernel for 3 different values of the kernel width.');disp(' ');disp('Press any key to continue')pausenet = svm(size(X, 2), 'rbf', [8], 100);net = svmtrain(net, X, Y);f4 = figure;plotboundary(net, x1ran, x2ran);plotdata(X, Y, x1ran, x2ran);plotsv(net, X, Y);title(['SVM with RBF kernel, width 8: decision boundary (black)' ... ' plus Support Vectors (red)']); fprintf('\n\n\n\n');fprintf('Figure %i shows the decision boundary obtained from a SVM\n', ... f4);disp('with Radial Basis Function kernel, the kernel width has been');disp('set to 8.');disp('The SVM now interprets the new point X12 as evidence for a');disp('cluster of points from class -1, the SVM builds a small ''island''');disp('around X12.');disp(' ')disp('Press any key to continue')pausenet = svm(size(X, 2), 'rbf', [1], 100);net = svmtrain(net, X, Y);f5 = figure;plotboundary(net, x1ran, x2ran);plotdata(X, Y, x1ran, x2ran);plotsv(net, X, Y);title(['SVM with RBF kernel, width 1: decision boundary (black)' ... ' plus Support Vectors (red)']); fprintf('\n\n\n\n');fprintf('Figure %i shows the decision boundary obtained from a SVM\n', ... f5);disp('with radial basis function kernel, kernel width 1.');disp('The decision boundary is now highly shattered, since a smaller');disp('kernel width allows the decision boundary to be more curved.');disp(' ')disp('Press any key to continue')pausenet = svm(size(X, 2), 'rbf', [36], 100);net = svmtrain(net, X, Y);f6 = figure;plotboundary(net, x1ran, x2ran);plotdata(X, Y, x1ran, x2ran);plotsv(net, X, Y);title(['SVM with RBF kernel, width 36: decision boundary (black)' ... ' plus Support Vectors (red)']); fprintf('\n\n\n\n');fprintf('Figure %i shows the decision boundary obtained from a SVM\n', ... f6);disp('with radial basis function kernel, kernel width 36.');disp('This gives a decision boundary similar to the one shown in');fprintf('Figure %i for the SVM with linear kernel.\n', f2);fprintf('\n\n\n\n');disp('Press any key to end the demo')pausedelete(f1);delete(f2);delete(f3);delete(f4);delete(f5);delete(f6);function plotdata(X, Y, x1ran, x2ran)% PLOTDATA - Plot 2D data set% hold on;ind = find(Y>0);plot(X(ind,1), X(ind,2), 'ks');ind = find(Y<0);plot(X(ind,1), X(ind,2), 'kx');text(X(:,1)+.2,X(:,2), int2str([1:length(Y)]'));axis([x1ran x2ran]);axis xy;function plotsv(net, X, Y)% PLOTSV - Plot Support Vectors% hold on;ind = find(Y(net.svind)>0);plot(X(net.svind(ind),1),X(net.svind(ind),2),'rs');ind = find(Y(net.svind)<0);plot(X(net.svind(ind),1),X(net.svind(ind),2),'rx');function [x11, x22, x1x2out] = plotboundary(net, x1ran, x2ran)% PLOTBOUNDARY - Plot SVM decision boundary on range X1RAN and X2RAN% hold on;nbpoints = 100;x1 = x1ran(1):(x1ran(2)-x1ran(1))/nbpoints:x1ran(2);x2 = x2ran(1):(x2ran(2)-x2ran(1))/nbpoints:x2ran(2);[x11, x22] = meshgrid(x1, x2);[dummy, x1x2out] = svmfwd(net, [x11(:),x22(:)]);x1x2out = reshape(x1x2out, [length(x1) length(x2)]);contour(x11, x22, x1x2out, [-0.99 -0.99], 'b-');contour(x11, x22, x1x2out, [0 0], 'k-');contour(x11, x22, x1x2out, [0.99 0.99], 'g-');
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