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Optimize a Cross-Validated SVM Classifier Using Bayesian Optimization

Plot Posterior Probability Regions for SVM Classification Models

You can use a support vector machine (SVM) when your data has
exactly two classes. An SVM classifies data by finding the best hyperplane
that separates all data points of one class from those of the other
class. The *best* hyperplane for an SVM means
the one with the largest *margin* between the
two classes. Margin means the maximal width of the slab parallel to
the hyperplane that has no interior data points.

The *support vectors* are the data points
that are closest to the separating hyperplane; these points are on
the boundary of the slab. The following figure illustrates these definitions,
with + indicating data points of type 1, and – indicating data
points of type –1.

**Mathematical Formulation: Primal. **This discussion follows Hastie, Tibshirani, and Friedman [1] and Christianini and
Shawe-Taylor [2].

The data for training is a set of points (vectors)
*x _{j}* along with their categories

$$f(x)=x\prime \beta +b=0$$

where *β* ∊ *R ^{d}* and

The following problem defines the *best* separating hyperplane
(i.e., the decision boundary). Find *β* and *b* that
minimize ||*β*|| such that for all data points
(*x _{j}*,

$${y}_{j}f\left({x}_{j}\right)\ge 1.$$

The support vectors are the *x _{j}* on the
boundary, those for which $${y}_{j}f\left({x}_{j}\right)=1.$$

For mathematical convenience, the problem is usually given as the equivalent problem
of minimizing $$\Vert \beta \Vert $$. This is a quadratic programming problem. The optimal solution $$\left(\widehat{\beta},\widehat{b}\right)$$ enables classification of a vector *z* as
follows:

$$\text{class}(z)=\text{sign}\left(z\prime \widehat{\beta}+\widehat{b}\right)=\text{sign}\left(\widehat{f}\left(z\right)\right).$$

$$\widehat{f}\left(z\right)$$ is the *classification score* and represents the
distance *z* is from the decision boundary.

**Mathematical Formulation: Dual. **It is computationally simpler to solve the dual quadratic programming
problem. To obtain the dual, take positive Lagrange multipliers *α _{j}* multiplied
by each constraint, and subtract from the objective function:

$${L}_{P}=\frac{1}{2}\beta \prime \beta -{\displaystyle \sum _{j}{\alpha}_{j}\left({y}_{j}\left({x}_{j}\prime \beta +b\right)-1\right)},$$

where you look for a stationary point of *L _{P}* over

$$\begin{array}{c}\beta ={\displaystyle \sum _{j}{\alpha}_{j}{y}_{j}{x}_{j}}\\ 0={\displaystyle \sum _{j}{\alpha}_{j}{y}_{j}}.\end{array}$$ | (18-1) |

Substituting into *L _{P}*,
you get the dual

$${L}_{D}={\displaystyle \sum _{j}{\alpha}_{j}}-\frac{1}{2}{\displaystyle \sum _{j}{\displaystyle \sum _{k}{\alpha}_{j}{\alpha}_{k}{y}_{j}{y}_{k}{x}_{j}\prime {x}_{k}}},$$

which you maximize over *α _{j}* ≥ 0. In general,
many

The derivative of *L _{D}* with
respect to a nonzero

$${y}_{j}f\left({x}_{j}\right)-1=0.$$

In particular, this gives the value of *b* at
the solution, by taking any *j* with nonzero *α _{j}*.

The dual is a standard quadratic programming problem. For example,
the Optimization
Toolbox™ `quadprog`

solver
solves this type of problem.

Your data might not allow for a separating hyperplane. In that
case, SVM can use a *soft margin*, meaning a
hyperplane that separates many, but not all data points.

There are two standard formulations of soft margins. Both involve
adding slack variables *ξ _{j}* and
a penalty parameter

The

*L*^{1}-norm problem is:$$\underset{\beta ,b,\xi}{\mathrm{min}}\left(\frac{1}{2}\beta \prime \beta +C{\displaystyle \sum _{j}{\xi}_{j}}\right)$$

such that

$$\begin{array}{c}{y}_{j}f\left({x}_{j}\right)\ge 1-{\xi}_{j}\\ {\xi}_{j}\ge 0.\end{array}$$

The

*L*^{1}-norm refers to using*ξ*as slack variables instead of their squares. The three solver options_{j}`SMO`

,`ISDA`

, and`L1QP`

of`fitcsvm`

minimize the*L*^{1}-norm problem.The

*L*^{2}-norm problem is:$$\underset{\beta ,b,\xi}{\mathrm{min}}\left(\frac{1}{2}\beta \prime \beta +C{\displaystyle \sum _{j}{\xi}_{j}^{2}}\right)$$

subject to the same constraints.

In these formulations, you can see that increasing *C* places
more weight on the slack variables *ξ _{j}*,
meaning the optimization attempts to make a stricter separation between
classes. Equivalently, reducing

**Mathematical Formulation: Dual. **For easier calculations, consider the *L*^{1} dual
problem to this soft-margin formulation. Using Lagrange multipliers *μ _{j}*,
the function to minimize for the

$${L}_{P}=\frac{1}{2}\beta \prime \beta +C{\displaystyle \sum _{j}{\xi}_{j}}-{\displaystyle \sum _{j}{\alpha}_{j}\left({y}_{i}f\left({x}_{j}\right)-\left(1-{\xi}_{j}\right)\right)}-{\displaystyle \sum _{j}{\mu}_{j}{\xi}_{j}},$$

where you look for a stationary point of *L _{P}* over

$$\begin{array}{c}\beta ={\displaystyle \sum _{j}{\alpha}_{j}{y}_{j}{x}_{j}}\\ {\displaystyle \sum _{j}{\alpha}_{j}{y}_{j}}=0\\ {\alpha}_{j}=C-{\mu}_{j}\\ {\alpha}_{j},{\mu}_{j},{\xi}_{j}\ge 0.\end{array}$$

These equations lead directly to the dual formulation:

$$\underset{\alpha}{\mathrm{max}}{\displaystyle \sum _{j}{\alpha}_{j}}-\frac{1}{2}{\displaystyle \sum _{j}{\displaystyle \sum _{k}{\alpha}_{j}{\alpha}_{k}{y}_{j}{y}_{k}{x}_{j}\prime {x}_{k}}}$$

subject to the constraints

$$\begin{array}{l}{\displaystyle \sum _{j}{y}_{j}{\alpha}_{j}}=0\\ 0\le {\alpha}_{j}\le C.\end{array}$$

The final set of inequalities, 0 ≤ *α _{j}* ≤

The gradient equation for *b* gives the solution *b* in
terms of the set of nonzero *α _{j}*,
which correspond to the support vectors.

You can write and solve the dual of the *L*^{2}-norm
problem in an analogous manner. For details, see Christianini and Shawe-Taylor [2], Chapter 6.

**fitcsvm Implementation. **Both dual soft-margin problems are quadratic programming problems.
Internally, `fitcsvm`

has several
different algorithms for solving the problems.

For one-class or binary classification, if you do not set a fraction of expected outliers in the data (see

`OutlierFraction`

), then the default solver is Sequential Minimal Optimization (SMO). SMO minimizes the one-norm problem by a series of two-point minimizations. During optimization, SMO respects the linear constraint $$\sum _{i}{\alpha}_{i}}{y}_{i}=0,$$ and explicitly includes the bias term in the model. SMO is relatively fast. For more details on SMO, see [3].For binary classification, if you set a fraction of expected outliers in the data, then the default solver is the Iterative Single Data Algorithm. Like SMO, ISDA solves the one-norm problem. Unlike SMO, ISDA minimizes by a series on one-point minimizations, does not respect the linear constraint, and does not explicitly include the bias term in the model. For more details on ISDA, see [4].

For one-class or binary classification, and if you have an Optimization Toolbox license, you can choose to use

`quadprog`

to solve the one-norm problem.`quadprog`

uses a good deal of memory, but solves quadratic programs to a high degree of precision. For more details, see Quadratic Programming Definition (Optimization Toolbox).

Some binary classification problems do not have a simple hyperplane as a useful separating criterion. For those problems, there is a variant of the mathematical approach that retains nearly all the simplicity of an SVM separating hyperplane.

This approach uses these results from the theory of reproducing kernels:

There is a class of functions

*G*(*x*_{1},*x*_{2}) with the following property. There is a linear space*S*and a function*φ*mapping*x*to*S*such that*G*(*x*_{1},*x*_{2}) = <*φ*(*x*_{1}),*φ*(*x*_{2})>.The dot product takes place in the space

*S*.This class of functions includes:

Polynomials: For some positive integer

*p*,*G*(*x*_{1},*x*_{2}) = (1 +*x*_{1}′*x*_{2})^{p}.Radial basis function (Gaussian):

*G*(*x*_{1},*x*_{2}) = exp(–∥*x*_{1}–*x*_{2})∥^{2}).Multilayer perceptron or sigmoid (neural network): For a positive number

*p*_{1}and a negative number*p*_{2},*G*(*x*_{1},*x*_{2}) = tanh(*p*_{1}*x*_{1}′*x*_{2}+*p*_{2}).### Note

Not every set of

*p*_{1}and*p*_{2}yields a valid reproducing kernel.`fitcsvm`

does not support the sigmoid kernel.

The mathematical approach using kernels relies on the computational
method of hyperplanes. All the calculations for hyperplane classification
use nothing more than dot products. Therefore, nonlinear kernels can
use identical calculations and solution algorithms, and obtain classifiers
that are nonlinear. The resulting classifiers are hypersurfaces in
some space *S*, but the space *S* does
not have to be identified or examined.

As with any supervised learning model, you first train a support vector machine, and then cross validate the classifier. Use the trained machine to classify (predict) new data. In addition, to obtain satisfactory predictive accuracy, you can use various SVM kernel functions, and you must tune the parameters of the kernel functions.

Train, and optionally cross validate, an SVM classifier using `fitcsvm`

. The most common syntax is:

SVMModel = fitcsvm(X,Y,'KernelFunction','rbf',... 'Standardize',true,'ClassNames',{'negClass','posClass'});

The inputs are:

`X`

— Matrix of predictor data, where each row is one observation, and each column is one predictor.`Y`

— Array of class labels with each row corresponding to the value of the corresponding row in`X`

.`Y`

can be a character array, categorical, logical or numeric vector, or cell vector of character vectors. Column vector with each row corresponding to the value of the corresponding row in`X`

.`Y`

can be a categorical or character array, logical or numeric vector, or cell array of character vectors.`KernelFunction`

— The default value is`'linear'`

for two-class learning, which separates the data by a hyperplane. The value`'gaussian'`

(or`'rbf'`

) is the default for one-class learning, and specifies to use the Gaussian (or radial basis function) kernel. An important step to successfully train an SVM classifier is to choose an appropriate kernel function.`Standardize`

— Flag indicating whether the software should standardize the predictors before training the classifier.`ClassNames`

— Distinguishes between the negative and positive classes, or specifies which classes to include in the data. The negative class is the first element (or row of a character array), e.g.,`'negClass'`

, and the positive class is the second element (or row of a character array), e.g.,`'posClass'`

.`ClassNames`

must be the same data type as`Y`

. It is good practice to specify the class names, especially if you are comparing the performance of different classifiers.

The resulting, trained model (`SVMModel`

) contains
the optimized parameters from the SVM algorithm, enabling you to classify
new data.

For more name-value pairs you can use to control the training,
see the `fitcsvm`

reference page.

Classify new data using `predict`

. The syntax for
classifying new data using a trained SVM classifier (`SVMModel`

)
is:

[label,score] = predict(SVMModel,newX);

The resulting vector, `label`

, represents the
classification of each row in `X`

. `score`

is
an *n*-by-2 matrix of soft scores. Each row corresponds
to a row in `X`

, which is a new observation. The
first column contains the scores for the observations being classified
in the negative class, and the second column contains the scores observations
being classified in the positive class.

To estimate posterior probabilities rather than scores, first
pass the trained SVM classifier (`SVMModel`

) to `fitPosterior`

,
which fits a score-to-posterior-probability transformation function
to the scores. The syntax is:

ScoreSVMModel = fitPosterior(SVMModel,X,Y);

The property `ScoreTransform`

of the classifier `ScoreSVMModel`

contains
the optimal transformation function. Pass `ScoreSVMModel`

to `predict`

.
Rather than returning the scores, the output argument `score`

contains
the posterior probabilities of an observation being classified in
the negative (column 1 of `score`

) or positive (column
2 of `score`

) class.

Try tuning parameters of your classifier according to this scheme:

Pass the data to

`fitcsvm`

, and set the name-value pair arguments`'KernelScale','auto'`

. Suppose that the trained SVM model is called`SVMModel`

. The software uses a heuristic procedure to select the kernel scale. The heuristic procedure uses subsampling. Therefore, to reproduce results, set a random number seed using`rng`

before training the classifier.Cross validate the classifier by passing it to

`crossval`

. By default, the software conducts 10-fold cross validation.Pass the cross-validated SVM model to

`kFoldLoss`

to estimate and retain the classification error.Retrain the SVM classifier, but adjust the

`'KernelScale'`

and`'BoxConstraint'`

name-value pair arguments.`BoxConstraint`

— One strategy is to try a geometric sequence of the box constraint parameter. For example, take 11 values, from`1e-5`

to`1e5`

by a factor of 10. Increasing`BoxConstraint`

might decrease the number of support vectors, but also might increase training time.`KernelScale`

— One strategy is to try a geometric sequence of the RBF sigma parameter scaled at the original kernel scale. Do this by:Retrieving the original kernel scale, e.g.,

`ks`

, using dot notation:`ks = SVMModel.KernelParameters.Scale`

.Use as new kernel scales factors of the original. For example, multiply

`ks`

by the 11 values`1e-5`

to`1e5`

, increasing by a factor of 10.

Choose the model that yields the lowest classification error.

You might want to further refine your parameters to obtain better
accuracy. Start with your initial parameters and perform another cross-validation
step, this time using a factor of 1.2. Alternatively, optimize your
parameters with `fminsearch`

,
as shown in Optimize a Cross-Validated SVM Classifier Using Bayesian Optimization.

This example shows how to generate a nonlinear classifier with Gaussian kernel function. First, generate one class of points inside the unit disk in two dimensions, and another class of points in the annulus from radius 1 to radius 2. Then, generates a classifier based on the data with the Gaussian radial basis function kernel. The default linear classifier is obviously unsuitable for this problem, since the model is circularly symmetric. Set the box constraint parameter to `Inf`

to make a strict classification, meaning no misclassified training points. Other kernel functions might not work with this strict box constraint, since they might be unable to provide a strict classification. Even though the rbf classifier can separate the classes, the result can be overtrained.

Generate 100 points uniformly distributed in the unit disk. To do so, generate a radius *r* as the square root of a uniform random variable, generate an angle *t* uniformly in (0, ), and put the point at (*r* cos( *t* ), *r* sin( *t* )).

rng(1); % For reproducibility r = sqrt(rand(100,1)); % Radius t = 2*pi*rand(100,1); % Angle data1 = [r.*cos(t), r.*sin(t)]; % Points

Generate 100 points uniformly distributed in the annulus. The radius is again proportional to a square root, this time a square root of the uniform distribution from 1 through 4.

r2 = sqrt(3*rand(100,1)+1); % Radius t2 = 2*pi*rand(100,1); % Angle data2 = [r2.*cos(t2), r2.*sin(t2)]; % points

Plot the points, and plot circles of radii 1 and 2 for comparison.

figure; plot(data1(:,1),data1(:,2),'r.','MarkerSize',15) hold on plot(data2(:,1),data2(:,2),'b.','MarkerSize',15) ezpolar(@(x)1);ezpolar(@(x)2); axis equal hold off

Put the data in one matrix, and make a vector of classifications.

data3 = [data1;data2]; theclass = ones(200,1); theclass(1:100) = -1;

Train an SVM classifier with `KernelFunction`

set to `'rbf'`

and `BoxConstraint`

set to `Inf`

. Plot the decision boundary and flag the support vectors.

%Train the SVM Classifier cl = fitcsvm(data3,theclass,'KernelFunction','rbf',... 'BoxConstraint',Inf,'ClassNames',[-1,1]); % Predict scores over the grid d = 0.02; [x1Grid,x2Grid] = meshgrid(min(data3(:,1)):d:max(data3(:,1)),... min(data3(:,2)):d:max(data3(:,2))); xGrid = [x1Grid(:),x2Grid(:)]; [~,scores] = predict(cl,xGrid); % Plot the data and the decision boundary figure; h(1:2) = gscatter(data3(:,1),data3(:,2),theclass,'rb','.'); hold on ezpolar(@(x)1); h(3) = plot(data3(cl.IsSupportVector,1),data3(cl.IsSupportVector,2),'ko'); contour(x1Grid,x2Grid,reshape(scores(:,2),size(x1Grid)),[0 0],'k'); legend(h,{'-1','+1','Support Vectors'}); axis equal hold off

`fitcsvm`

generates a classifier that is close to a circle of radius 1. The difference is due to the random training data.

Training with the default parameters makes a more nearly circular classification boundary, but one that misclassifies some training data. Also, the default value of `BoxConstraint`

is `1`

, and, therefore, there are more support vectors.

cl2 = fitcsvm(data3,theclass,'KernelFunction','rbf'); [~,scores2] = predict(cl2,xGrid); figure; h(1:2) = gscatter(data3(:,1),data3(:,2),theclass,'rb','.'); hold on ezpolar(@(x)1); h(3) = plot(data3(cl2.IsSupportVector,1),data3(cl2.IsSupportVector,2),'ko'); contour(x1Grid,x2Grid,reshape(scores2(:,2),size(x1Grid)),[0 0],'k'); legend(h,{'-1','+1','Support Vectors'}); axis equal hold off

This example shows how to use a custom kernel function, such as the sigmoid kernel, to train SVM classifiers, and adjust custom kernel function parameters.

Generate a random set of points within the unit circle. Label points in the first and third quadrants as belonging to the positive class, and those in the second and fourth quadrants in the negative class.

rng(1); % For reproducibility n = 100; % Number of points per quadrant r1 = sqrt(rand(2*n,1)); % Random radii t1 = [pi/2*rand(n,1); (pi/2*rand(n,1)+pi)]; % Random angles for Q1 and Q3 X1 = [r1.*cos(t1) r1.*sin(t1)]; % Polar-to-Cartesian conversion r2 = sqrt(rand(2*n,1)); t2 = [pi/2*rand(n,1)+pi/2; (pi/2*rand(n,1)-pi/2)]; % Random angles for Q2 and Q4 X2 = [r2.*cos(t2) r2.*sin(t2)]; X = [X1; X2]; % Predictors Y = ones(4*n,1); Y(2*n + 1:end) = -1; % Labels

Plot the data.

```
figure;
gscatter(X(:,1),X(:,2),Y);
title('Scatter Diagram of Simulated Data')
```

Write a function that accepts two matrices in the feature space as inputs, and transforms them into a Gram matrix using the sigmoid kernel.

function G = mysigmoid(U,V) % Sigmoid kernel function with slope gamma and intercept c gamma = 1; c = -1; G = tanh(gamma*U*V' + c); end

Save this code as a file named `mysigmoid`

on your MATLAB® path.

Train an SVM classifier using the sigmoid kernel function. It is good practice to standardize the data.

Mdl1 = fitcsvm(X,Y,'KernelFunction','mysigmoid','Standardize',true);

`Mdl1`

is a `ClassificationSVM`

classifier containing the estimated parameters.

Plot the data, and identify the support vectors and the decision boundary.

% Compute the scores over a grid d = 0.02; % Step size of the grid [x1Grid,x2Grid] = meshgrid(min(X(:,1)):d:max(X(:,1)),... min(X(:,2)):d:max(X(:,2))); xGrid = [x1Grid(:),x2Grid(:)]; % The grid [~,scores1] = predict(Mdl1,xGrid); % The scores figure; h(1:2) = gscatter(X(:,1),X(:,2),Y); hold on h(3) = plot(X(Mdl1.IsSupportVector,1),... X(Mdl1.IsSupportVector,2),'ko','MarkerSize',10); % Support vectors contour(x1Grid,x2Grid,reshape(scores1(:,2),size(x1Grid)),[0 0],'k'); % Decision boundary title('Scatter Diagram with the Decision Boundary') legend({'-1','1','Support Vectors'},'Location','Best'); hold off

You can adjust the kernel parameters in an attempt to improve the shape of the decision boundary. This might also decrease the within-sample misclassification rate, but, you should first determine the out-of-sample misclassification rate.

Determine the out-of-sample misclassification rate by using 10-fold cross validation.

CVMdl1 = crossval(Mdl1); misclass1 = kfoldLoss(CVMdl1); misclass1

misclass1 = 0.1350

The out-of-sample misclassification rate is 13.5%.

Write another sigmoid function, but Set `gamma = 0.5;`

.

function G = mysigmoid2(U,V) % Sigmoid kernel function with slope gamma and intercept c gamma = 0.5; c = -1; G = tanh(gamma*U*V' + c); end

Save this code as a file named `mysigmoid2`

on your MATLAB® path.

Train another SVM classifier using the adjusted sigmoid kernel. Plot the data and the decision region, and determine the out-of-sample misclassification rate.

Mdl2 = fitcsvm(X,Y,'KernelFunction','mysigmoid2','Standardize',true); [~,scores2] = predict(Mdl2,xGrid); figure; h(1:2) = gscatter(X(:,1),X(:,2),Y); hold on h(3) = plot(X(Mdl2.IsSupportVector,1),... X(Mdl2.IsSupportVector,2),'ko','MarkerSize',10); title('Scatter Diagram with the Decision Boundary') contour(x1Grid,x2Grid,reshape(scores2(:,2),size(x1Grid)),[0 0],'k'); legend({'-1','1','Support Vectors'},'Location','Best'); hold off CVMdl2 = crossval(Mdl2); misclass2 = kfoldLoss(CVMdl2); misclass2

misclass2 = 0.0450

After the sigmoid slope adjustment, the new decision boundary seems to provide a better within-sample fit, and the cross-validation rate contracts by more than 66%.

This example shows how to optimize an SVM classification. The classification works on locations of points from a Gaussian mixture model. In *The Elements of Statistical Learning*, Hastie, Tibshirani, and Friedman (2009), page 17 describes the model. The model begins with generating 10 base points for a "green" class, distributed as 2-D independent normals with mean (1,0) and unit variance. It also generates 10 base points for a "red" class, distributed as 2-D independent normals with mean (0,1) and unit variance. For each class (green and red), generate 100 random points as follows:

Choose a base point

*m*of the appropriate color uniformly at random.Generate an independent random point with 2-D normal distribution with mean

*m*and variance I/5, where I is the 2-by-2 identity matrix. In this example, use a variance I/50 to show the advantage of optimization more clearly.

After generating 100 green and 100 red points, classify them using `fitcsvm`

. Then use `bayesopt`

to optimize the parameters of the resulting SVM model with respect to cross validation.

**Generate the Points and Classifier**

Generate the 10 base points for each class.

```
rng default
grnpop = mvnrnd([1,0],eye(2),10);
redpop = mvnrnd([0,1],eye(2),10);
```

View the base points.

plot(grnpop(:,1),grnpop(:,2),'go') hold on plot(redpop(:,1),redpop(:,2),'ro') hold off

Since some red base points are close to green base points, it can be difficult to classify the data points based on location alone.

Generate the 100 data points of each class.

redpts = zeros(100,2);grnpts = redpts; for i = 1:100 grnpts(i,:) = mvnrnd(grnpop(randi(10),:),eye(2)*0.02); redpts(i,:) = mvnrnd(redpop(randi(10),:),eye(2)*0.02); end

View the data points.

figure plot(grnpts(:,1),grnpts(:,2),'go') hold on plot(redpts(:,1),redpts(:,2),'ro') hold off

**Prepare Data For Classification**

Put the data into one matrix, and make a vector `grp`

that labels the class of each point.

```
cdata = [grnpts;redpts];
grp = ones(200,1);
% Green label 1, red label -1
grp(101:200) = -1;
```

**Prepare Cross-Validation**

Set up a partition for cross-validation. This step fixes the train and test sets that the optimization uses at each step.

```
c = cvpartition(200,'KFold',10);
```

**Prepare Variables for Bayesian Optimization**

Set up a function that takes an input `z = [rbf_sigma,boxconstraint]`

and returns the cross-validation loss value of `z`

. Take the components of `z`

as positive, log-transformed variables between `1e-5`

and `1e5`

. Choose a wide range, because you don't know which values are likely to be good.

sigma = optimizableVariable('sigma',[1e-5,1e5],'Transform','log'); box = optimizableVariable('box',[1e-5,1e5],'Transform','log');

**Objective Function**

This function handle computes the cross-validation loss at parameters `[sigma,box]`

. For details, see `ClassificationPartitionedModel.kfoldLoss`

.

`bayesopt`

passes the variable `z`

to the objective function as a one-row table.

minfn = @(z)kfoldLoss(fitcsvm(cdata,grp,'CVPartition',c,... 'KernelFunction','rbf','BoxConstraint',z.box,... 'KernelScale',z.sigma));

**Optimize Classifier**

Search for the best parameters `[sigma,box]`

using `bayesopt`

. For reproducibility, choose the `'expected-improvement-plus'`

acquisition function. The default acquisition function depends on run time, and so can give varying results.

results = bayesopt(minfn,[sigma,box],'IsObjectiveDeterministic',true,... 'AcquisitionFunctionName','expected-improvement-plus')

|=====================================================================================================| | Iter | Eval | Objective | Objective | BestSoFar | BestSoFar | sigma | box | | | result | | runtime | (observed) | (estim.) | | | |=====================================================================================================| | 1 | Best | 0.61 | 1.3956 | 0.61 | 0.61 | 0.00013375 | 13929 | | 2 | Best | 0.345 | 0.87505 | 0.345 | 0.345 | 24526 | 1.936 | | 3 | Accept | 0.61 | 0.7367 | 0.345 | 0.345 | 0.0026459 | 0.00084929 | | 4 | Accept | 0.345 | 0.99966 | 0.345 | 0.345 | 3506.3 | 6.7427e-05 | | 5 | Accept | 0.345 | 0.59343 | 0.345 | 0.345 | 9135.2 | 571.87 | | 6 | Accept | 0.345 | 0.94224 | 0.345 | 0.345 | 99701 | 10223 | | 7 | Best | 0.295 | 0.55381 | 0.295 | 0.295 | 455.88 | 9957.4 | | 8 | Best | 0.24 | 14.115 | 0.24 | 0.24 | 31.56 | 99389 | | 9 | Accept | 0.24 | 18.316 | 0.24 | 0.24 | 10.451 | 64429 | | 10 | Accept | 0.35 | 0.93913 | 0.24 | 0.24 | 17.331 | 1.0264e-05 | | 11 | Best | 0.23 | 11.57 | 0.23 | 0.23 | 16.005 | 90155 | | 12 | Best | 0.1 | 1.5758 | 0.1 | 0.1 | 0.36562 | 80878 | | 13 | Accept | 0.115 | 0.92054 | 0.1 | 0.1 | 0.1793 | 68459 | | 14 | Accept | 0.105 | 1.5327 | 0.1 | 0.1 | 0.2267 | 95421 | | 15 | Best | 0.095 | 0.92109 | 0.095 | 0.095 | 0.28999 | 0.0058227 | | 16 | Best | 0.075 | 0.89971 | 0.075 | 0.075 | 0.30554 | 8.9017 | | 17 | Accept | 0.085 | 0.8331 | 0.075 | 0.075 | 0.41122 | 4.4476 | | 18 | Accept | 0.085 | 0.65763 | 0.075 | 0.075 | 0.25565 | 7.8038 | | 19 | Accept | 0.075 | 0.67846 | 0.075 | 0.075 | 0.32869 | 18.076 | | 20 | Accept | 0.085 | 0.58156 | 0.075 | 0.075 | 0.32442 | 5.2118 | |=====================================================================================================| | Iter | Eval | Objective | Objective | BestSoFar | BestSoFar | sigma | box | | | result | | runtime | (observed) | (estim.) | | | |=====================================================================================================| | 21 | Accept | 0.3 | 0.90433 | 0.075 | 0.075 | 1.3592 | 0.0098067 | | 22 | Accept | 0.12 | 0.43337 | 0.075 | 0.075 | 0.17515 | 0.00070913 | | 23 | Accept | 0.175 | 0.75403 | 0.075 | 0.075 | 0.1252 | 0.010749 | | 24 | Accept | 0.105 | 0.86415 | 0.075 | 0.075 | 1.1664 | 31.13 | | 25 | Accept | 0.1 | 0.60418 | 0.075 | 0.075 | 0.57465 | 2013.8 | | 26 | Accept | 0.12 | 0.79191 | 0.075 | 0.075 | 0.42922 | 1.1602e-05 | | 27 | Accept | 0.12 | 0.87006 | 0.075 | 0.075 | 0.42956 | 0.00027218 | | 28 | Accept | 0.095 | 0.81047 | 0.075 | 0.075 | 0.4806 | 13.452 | | 29 | Accept | 0.105 | 0.79476 | 0.075 | 0.075 | 0.19755 | 943.87 | | 30 | Accept | 0.205 | 0.62004 | 0.075 | 0.075 | 3.5051 | 93.492 | __________________________________________________________ Optimization completed. MaxObjectiveEvaluations of 30 reached. Total function evaluations: 30 Total elapsed time: 304.9426 seconds. Total objective function evaluation time: 67.0858 Best observed feasible point: sigma box _______ ______ 0.30554 8.9017 Observed objective function value = 0.075 Estimated objective function value = 0.075 Function evaluation time = 0.89971 Best estimated feasible point (according to models): sigma box _______ ______ 0.32869 18.076 Estimated objective function value = 0.075 Estimated function evaluation time = 0.75114 results = BayesianOptimization with properties: ObjectiveFcn: [function_handle] VariableDescriptions: [1x2 optimizableVariable] Options: [1x1 struct] MinObjective: 0.0750 XAtMinObjective: [1x2 table] MinEstimatedObjective: 0.0750 XAtMinEstimatedObjective: [1x2 table] NumObjectiveEvaluations: 30 TotalElapsedTime: 304.9426 NextPoint: [1x2 table] XTrace: [30x2 table] ObjectiveTrace: [30x1 double] ConstraintsTrace: [] UserDataTrace: {30x1 cell} ObjectiveEvaluationTimeTrace: [30x1 double] IterationTimeTrace: [30x1 double] ErrorTrace: [30x1 double] FeasibilityTrace: [30x1 logical] FeasibilityProbabilityTrace: [30x1 double] IndexOfMinimumTrace: [30x1 double] ObjectiveMinimumTrace: [30x1 double] EstimatedObjectiveMinimumTrace: [30x1 double]

Use the results to train a new, optimized SVM classifier.

z(1) = results.XAtMinObjective.sigma; z(2) = results.XAtMinObjective.box; SVMModel = fitcsvm(cdata,grp,'KernelFunction','rbf',... 'KernelScale',z(1),'BoxConstraint',z(2));

Plot the classification boundaries. To visualize the support vector classifier, predict scores over a grid.

d = 0.02; [x1Grid,x2Grid] = meshgrid(min(cdata(:,1)):d:max(cdata(:,1)),... min(cdata(:,2)):d:max(cdata(:,2))); xGrid = [x1Grid(:),x2Grid(:)]; [~,scores] = predict(SVMModel,xGrid); h = nan(3,1); % Preallocation figure; h(1:2) = gscatter(cdata(:,1),cdata(:,2),grp,'rg','+*'); hold on h(3) = plot(cdata(SVMModel.IsSupportVector,1),... cdata(SVMModel.IsSupportVector,2),'ko'); contour(x1Grid,x2Grid,reshape(scores(:,2),size(x1Grid)),[0 0],'k'); legend(h,{'-1','+1','Support Vectors'},'Location','Southeast'); axis equal hold off

**Evaluate Accuracy on New Data**

Generate and classify some new data points.

grnobj = gmdistribution(grnpop,.2*eye(2)); redobj = gmdistribution(redpop,.2*eye(2)); newData = random(grnobj,10); newData = [newData;random(redobj,10)]; grpData = ones(20,1); grpData(11:20) = -1; % red = -1 v = predict(SVMModel,newData); g = nan(7,1); figure; h(1:2) = gscatter(cdata(:,1),cdata(:,2),grp,'rg','+*'); hold on h(3:4) = gscatter(newData(:,1),newData(:,2),v,'mc','**'); h(5) = plot(cdata(SVMModel.IsSupportVector,1),... cdata(SVMModel.IsSupportVector,2),'ko'); contour(x1Grid,x2Grid,reshape(scores(:,2),size(x1Grid)),[0 0],'k'); legend(h(1:5),{'-1 (training)','+1 (training)','-1 (classified)',... '+1 (classified)','Support Vectors'},'Location','Southeast'); axis equal hold off

See which new data points are correctly classified. Circle the correctly classified points in red, and the incorrectly classified points in black.

mydiff = (v == grpData); % Classified correctly figure; h(1:2) = gscatter(cdata(:,1),cdata(:,2),grp,'rg','+*'); hold on h(3:4) = gscatter(newData(:,1),newData(:,2),v,'mc','**'); h(5) = plot(cdata(SVMModel.IsSupportVector,1),... cdata(SVMModel.IsSupportVector,2),'ko'); contour(x1Grid,x2Grid,reshape(scores(:,2),size(x1Grid)),[0 0],'k'); for ii = mydiff % Plot red squares around correct pts h(6) = plot(newData(ii,1),newData(ii,2),'rs','MarkerSize',12); end for ii = not(mydiff) % Plot black squares around incorrect pts h(7) = plot(newData(ii,1),newData(ii,2),'ks','MarkerSize',12); end legend(h,{'-1 (training)','+1 (training)','-1 (classified)',... '+1 (classified)','Support Vectors','Correctly Classified',... 'Misclassified'},'Location','Southeast'); hold off

This example shows how to predict posterior probabilities of SVM models over a grid of observations, and then plot the posterior probabilities over the grid. Plotting posterior probabilities exposes decision boundaries.

Load Fisher's iris data set. Train the classifier using the petal lengths and widths, and remove the virginica species from the data.

load fisheriris classKeep = ~strcmp(species,'virginica'); X = meas(classKeep,3:4); y = species(classKeep);

Train an SVM classifier using the data. It is good practice to specify the order of the classes.

SVMModel = fitcsvm(X,y,'ClassNames',{'setosa','versicolor'});

Estimate the optimal score transformation function.

```
rng(1); % For reproducibility
[SVMModel,ScoreParameters] = fitPosterior(SVMModel);
ScoreParameters
```

Warning: Classes are perfectly separated. The optimal score-to-posterior transformation is a step function. ScoreParameters = struct with fields: Type: 'step' LowerBound: -0.8431 UpperBound: 0.6897 PositiveClassProbability: 0.5000

The optimal score transformation function is the step function because the classes are separable. The fields `LowerBound`

and `UpperBound`

of `ScoreParameters`

indicate the lower and upper end points of the interval of scores corresponding to observations within the class-separating hyperplanes (the margin). No training observation falls within the margin. If a new score is in the interval, then the software assigns the corresonding observation a positive class posterior probability, i.e., the value in the `PositiveClassProbability`

field of `ScoreParameters`

.

Define a grid of values in the observed predictor space. Predict the posterior probabilities for each instance in the grid.

xMax = max(X); xMin = min(X); d = 0.01; [x1Grid,x2Grid] = meshgrid(xMin(1):d:xMax(1),xMin(2):d:xMax(2)); [~,PosteriorRegion] = predict(SVMModel,[x1Grid(:),x2Grid(:)]);

Plot the positive class posterior probability region and the training data.

figure; contourf(x1Grid,x2Grid,... reshape(PosteriorRegion(:,2),size(x1Grid,1),size(x1Grid,2))); h = colorbar; h.Label.String = 'P({\it{versicolor}})'; h.YLabel.FontSize = 16; caxis([0 1]); colormap jet; hold on gscatter(X(:,1),X(:,2),y,'mc','.x',[15,10]); sv = X(SVMModel.IsSupportVector,:); plot(sv(:,1),sv(:,2),'yo','MarkerSize',15,'LineWidth',2); axis tight hold off

In two-class learning, if the classes are separable, then there are three regions: one where observations have positive class posterior probability `0`

, one where it is `1`

, and the other where it is the postiive class prior probability.

This example shows how to determine which quadrant of an image a shape occupies by training an error-correcting output codes (ECOC) model comprised of linear SVM binary learners. This example also illustrates the disk-space consumption of ECOC models that store support vectors, their labels, and the estimated coefficients.

**Create the Data Set**

Randomly place a circle with radius five in a 50-by-50 image. Make 5000 images. Create a label for each image indicating the quadrant that the circle occupies. Quadrant 1 is in the upper right, quadrant 2 is in the upper left, quadrant 3 is in the lower left, and quadrant 4 is in the lower right. The predictors are the intensities of each pixel.

d = 50; % Height and width of the images in pixels n = 5e4; % Sample size X = zeros(n,d^2); % Predictor matrix preallocation Y = zeros(n,1); % Label preallocation theta = 0:(1/d):(2*pi); r = 5; % Circle radius rng(1); % For reproducibility for j = 1:n; figmat = zeros(d); % Empty image c = datasample((r + 1):(d - r - 1),2); % Random circle center x = r*cos(theta) + c(1); % Make the circle y = r*sin(theta) + c(2); idx = sub2ind([d d],round(y),round(x)); % Convert to linear indexing figmat(idx) = 1; % Draw the circle X(j,:) = figmat(:); % Store the data Y(j) = (c(2) >= floor(d/2)) + 2*(c(2) < floor(d/2)) + ... (c(1) < floor(d/2)) + ... 2*((c(1) >= floor(d/2)) & (c(2) < floor(d/2))); % Determine the quadrant end

Plot an observation.

figure; imagesc(figmat); h = gca; h.YDir = 'normal'; title(sprintf('Quadrant %d',Y(end)));

**Train the ECOC Model**

Use a 25% holdout sample and specify the training and holdout sample indices.

p = 0.25; CVP = cvpartition(Y,'Holdout',p); % Cross-validation data partition isIdx = training(CVP); % Training sample indices oosIdx = test(CVP); % Test sample indices

Create an SVM template that specifies storing the support vectors of the binary learners. Pass it and the training data to `fitcecoc`

to train the model. Determine the training sample classification error.

t = templateSVM('SaveSupportVectors',true); MdlSV = fitcecoc(X(isIdx,:),Y(isIdx),'Learners',t); isLoss = resubLoss(MdlSV)

isLoss = 0

`MdlSV`

is a trained `ClassificationECOC`

multiclass model. It stores the training data and the support vectors of each binary learner. For large data sets, such as those in image analysis, the model can consume a lot of memory.

Determine the amount of disk space that the ECOC model consumes.

```
infoMdlSV = whos('MdlSV');
mbMdlSV = infoMdlSV.bytes/1.049e6
```

mbMdlSV = 763.5919

The model consumes 1477.5 MB.

**Improve Model Efficiency**

You can assess out-of-sample performance. You can also assess whether the model has been overfit with a compacted model that does not contain the support vectors, their related parameters, and the training data.

Discard the support vectors and related parameters from the trained ECOC model. Then, discard the training data from the resulting model by using `compact`

.

Mdl = discardSupportVectors(MdlSV); CMdl = compact(Mdl); info = whos('Mdl','CMdl'); [bytesCMdl,bytesMdl] = info.bytes; memReduction = 1 - [bytesMdl bytesCMdl]/infoMdlSV.bytes

memReduction = 0.0626 0.9996

In this case, discarding the support vectors reduces the memory consumption by about 3%. Compacting and discarding support vectors reduces the size by about 99.99%.

An alternative way to manage support vectors is to reduce their numbers during training by specifying a larger box constraint, such as 100. Though SVM models that use fewer support vectors are more desirable and consume less memory, increasing the value of the box constraint tends to increase the training time.

Remove `MdlSV`

and `Mdl`

from the workspace.

clear Mdl MdlSV;

**Assess Holdout Sample Performance**

Calculate the classification error of the holdout sample. Plot a sample of the holdout sample predictions.

oosLoss = loss(CMdl,X(oosIdx,:),Y(oosIdx)) yHat = predict(CMdl,X(oosIdx,:)); nVec = 1:size(X,1); oosIdx = nVec(oosIdx); figure; for j = 1:9; subplot(3,3,j) imagesc(reshape(X(oosIdx(j),:),[d d])); h = gca; h.YDir = 'normal'; title(sprintf('Quadrant: %d',yHat(j))) end text(-1.33*d,4.5*d + 1,'Predictions','FontSize',17)

oosLoss = 0

The model does not misclassify any holdout sample observations.

[1] Hastie, T., R. Tibshirani, and
J. Friedman. *The Elements of Statistical Learning*, second edition.
New York: Springer, 2008.

[2] Christianini, N., and J.
Shawe-Taylor. *An Introduction to Support Vector Machines and Other Kernel-Based
Learning Methods*. Cambridge, UK: Cambridge University Press,
2000.

[3] Fan, R.-E., P.-H. Chen, and
C.-J. Lin. “Working set selection using second order information for training support
vector machines.” *Journal of Machine Learning Research*, Vol 6,
2005, pp. 1889–1918.

[4] Kecman V., T. -M. Huang, and M.
Vogt. “Iterative Single Data Algorithm for Training Kernel Machines from Huge Data
Sets: Theory and Performance.” In *Support Vector Machines: Theory and
Applications*. Edited by Lipo Wang, 255–274. Berlin: Springer-Verlag,
2005.

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