# How to plot a hyper plane in 3D for the SVM results?

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Aaronne 2013 年 4 月 22 日
コメント済み: Joy 2019 年 4 月 9 日
I just wondering how to plot a hyper plane of the SVM results.
For example, here we are using two features, we can plot the decision boundary in 2D. But if how can we plot a hyper plane in 3D if we use 3 features?
features = meas(1:100,:);
featureSelcted = features(1:100,1:2); % For example, featureSelcted = features(1:100,1:3) can not be plotted
groundTruthGroup = species(1:100);
svmStruct = svmtrain(featureSelcted, groundTruthGroup, ...
'Kernel_Function', 'rbf', 'boxconstraint', Inf, 'showplot', true, 'Method', 'QP');
svmClassified = svmclassify(svmStruct,featureSelcted,'showplot',true);
A similar solution in R can be found at
but a Matlab implementation would be handy. Thanks very much.
A.

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### 採用された回答

Mohamed 2014 年 6 月 9 日

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### その他の回答 (1 件)

manan lalit 2019 年 4 月 3 日
Just putting my answer here in case someone is curious about how to find the analytical equation of the 3-D linear plane separating data belonging to two classes with the fitcsvm function in MATLAB.
You can find the coefficients ( and ) using the two equations below. Quoting from "Support-Vector Networks" by Cortes and Vapnik, 1995, "... the vector that determines the optimal hyperplane can be written as a linear combination of training vectors"
Here, l is the number of support vectors. In MATLAB version 2018b can be extracted using code such as:
SVMModel = fitcsvm(data, groups);
% "data" contains the (N * D) matrix.
% "groups" is a (N*1) matrix indicating the two groups (+1 and -1)
alpha = SVMModel.Alpha;
Similarly, can be determined using the equation below:
Once, and are available, then plotting such a plane can be done, in the following manner:
xgrid=[0:200];
ygrid=[0:200];
[X, Y]=meshgrid(xgrid, ygrid);
% w0 and b0 were determined through the two equations mentioned above.
Z=(-b0-w0(1)*X-w0(2)*Y)/w0(3);
surf(X, Y, Z)
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Joy 2019 年 4 月 9 日
I like this analytical solution . Could you please share your example code? I don't exactly get the deriving part.
Thanks!

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