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<!--This HTML is auto-generated from an m-file.Your changes will be overwritten.--><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" style="color:#990000; font-weight:bold; font-size:x-large">Outlier Input Vectors</p><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd">A 2-input hard limit neuron is trained to classify 5 input vectors into twocategories. However, because 1 input vector is much larger than all of theothers, training takes a long time.</p><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd">Copyright 1992-2002 The MathWorks, Inc.$Revision: 1.18 $ $Date: 2002/03/29 19:36:10 $</p><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" style="color:#990000; font-weight:bold; font-size:medium; page-break-before: auto;"><a name=""></a></p><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd">Each of the five column vectors in P defines a 2-element input vectors, and arow vector T defines the vector's target categories. Plot these vectors withPLOTPV.</p><pre xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" style="position: relative; left:30px">P = [-0.5 -0.5 +0.3 -0.1 -40; -0.5 +0.5 -0.5 +1.0 50];T = [1 1 0 0 1];plotpv(P,T);</pre><img xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" src="demop4_img02.gif"><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" style="color:#990000; font-weight:bold; font-size:medium; page-break-before: auto;"><a name=""></a></p><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd">Note that 4 input vectors have much smaller magnitudes than the fifth vectorin the upper left of the plot. The perceptron must properly classify the 5input vectors in P into the two categories defined by T.</p><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd">NEWP creates aperceptron. The first argument specifies the expected rangesof two inputs. The second argument determines that there is only one neuronin the layer.</p><pre xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" style="position: relative; left:30px">net = newp([-40 1;-1 50],1);</pre><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" style="color:#990000; font-weight:bold; font-size:medium; page-break-before: auto;"><a name=""></a></p><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd">Add the the neuron's initial attempt at classification to the plot.</p><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd">The initial weights are set to zero, so any input gives the same output andthe classification line does not even appear on the plot. Fear not... we aregoing to train it!</p><pre xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" style="position: relative; left:30px">hold onlinehandle = plotpc(net.IW{1},net.b{1});</pre><img xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" src="demop4_img04.gif"><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" style="color:#990000; font-weight:bold; font-size:medium; page-break-before: auto;"><a name=""></a></p><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd">ADAPT returns a new network object that performs as a better classifier, thenetwork output, and the error. This loop allows the network to adapt for 3passes, plots the classification line, and continues until the error is zero.</p><pre xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" style="position: relative; left:30px">E = 1;net.adaptParam.passes = 3;<span style="color:blue">while</span> (sse(E)) [net,Y,E] = adapt(net,P,T); linehandle = plotpc(net.IW{1},net.b{1},linehandle); drawnow;<span style="color:blue">end</span></pre><img xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" src="demop4_img05.gif"><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" style="color:#990000; font-weight:bold; font-size:medium; page-break-before: auto;"><a name=""></a></p><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd">Note that it took the perceptron many epochs to train. This a very long timefor such a simple problem. The reason for the long training time is theoutlier vector. Despite the long training time, the perceptron still learnsproperly and can be used to classify other inputs.</p><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" style="color:#990000; font-weight:bold; font-size:medium; page-break-before: auto;"><a name=""></a></p><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd">Now SIM can be used to classify any other input vector. For example, classifyan input vector of [0.7; 1.2].</p><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd">A plot of this new point with the original training set shows how the networkperforms. To distinguish it from the training set, color it red.</p><pre xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" style="position: relative; left:30px">p = [0.7; 1.2];a = sim(net,p);plotpv(p,a);circle = findobj(gca,<span style="color:#B20000">'type'</span>,<span style="color:#B20000">'line'</span>);set(circle,<span style="color:#B20000">'Color'</span>,<span style="color:#B20000">'red'</span>);</pre><img xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" src="demop4_img07.gif"><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" style="color:#990000; font-weight:bold; font-size:medium; page-break-before: auto;"><a name=""></a></p><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd">Turn on "hold" so the previous plot is not erased. Add the training setand the classification line to the plot.</p><pre xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" style="position: relative; left:30px">hold on;plotpv(P,T);plotpc(net.IW{1},net.b{1});hold off;</pre><img xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" src="demop4_img08.gif"><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" style="color:#990000; font-weight:bold; font-size:medium; page-break-before: auto;"><a name=""></a></p><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd">Finally, zoom into the area of interest.</p><p xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd">The perceptron correctly classified our new point (in red) as category "zero"(represented by a circle) and not a "one" (represented by a plus). Despitethe long training time, the perceptron still learns properly. To see how toreduce training times associated with outlier vectors, see the "NormalizedPerceptron Rule" demo.</p><pre xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" style="position: relative; left:30px">axis([-2 2 -2 2]);</pre><img xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" src="demop4_img09.gif"><originalCode xmlns:mwsh="http://www.mathworks.com/namespace/mcode/v1/syntaxhighlight.dtd" code="%% Outlier Input Vectors
% A 2-input hard limit neuron is trained to classify 5 input vectors into two
% categories. However, because 1 input vector is much larger than all of the
% others, training takes a long time.
%
% Copyright 1992-2002 The MathWorks, Inc.
% $Revision: 1.18 $ $Date: 2002/03/29 19:36:10 $

%%
% Each of the five column vectors in P defines a 2-element input vectors, and a
% row vector T defines the vector's target categories. Plot these vectors with
% PLOTPV.

P = [-0.5 -0.5 +0.3 -0.1 -40; -0.5 +0.5 -0.5 +1.0 50];
T = [1 1 0 0 1];
plotpv(P,T);

%%
% Note that 4 input vectors have much smaller magnitudes than the fifth vector
% in the upper left of the plot. The perceptron must properly classify the 5
% input vectors in P into the two categories defined by T. 
% 
% NEWP creates aperceptron. The first argument specifies the expected ranges
% of two inputs. The second argument determines that there is only one neuron
% in the layer.

net = newp([-40 1;-1 50],1);

%%
% Add the the neuron's initial attempt at classification to the plot.
%
% The initial weights are set to zero, so any input gives the same output and
% the classification line does not even appear on the plot. Fear not... we are
% going to train it!

hold on
linehandle = plotpc(net.IW{1},net.b{1});

%%
% ADAPT returns a new network object that performs as a better classifier, the
% network output, and the error. This loop allows the network to adapt for 3
% passes, plots the classification line, and continues until the error is zero.

E = 1;
net.adaptParam.passes = 3;
while (sse(E))
 [net,Y,E] = adapt(net,P,T);
 linehandle = plotpc(net.IW{1},net.b{1},linehandle);
 drawnow;
end

%%
% Note that it took the perceptron many epochs to train. This a very long time
% for such a simple problem. The reason for the long training time is the
% outlier vector. Despite the long training time, the perceptron still learns
% properly and can be used to classify other inputs.

%%
% Now SIM can be used to classify any other input vector. For example, classify
% an input vector of [0.7; 1.2].
%
% A plot of this new point with the original training set shows how the network
% performs. To distinguish it from the training set, color it red.

p = [0.7; 1.2];
a = sim(net,p);
plotpv(p,a);
circle = findobj(gca,'type','line');
set(circle,'Color','red');

%%
% Turn on "hold" so the previous plot is not erased. Add the training set
% and the classification line to the plot.

hold on;
plotpv(P,T);
plotpc(net.IW{1},net.b{1});
hold off;

%%
% Finally, zoom into the area of interest.
%
% The perceptron correctly classified our new point (in red) as category "zero"
% (represented by a circle) and not a "one" (represented by a plus). Despite
% the long training time, the perceptron still learns properly. To see how to
% reduce training times associated with outlier vectors, see the "Normalized
% Perceptron Rule" demo.

axis([-2 2 -2 2]);
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