Score calculation in ClassificationSVM using linear kernel function

I have binary ClassificationSVM classifier: svmModel. It's trained using linear kernel. I know the score of test data can be obtained through predict(svmModel, testdata). I want to imitate the actual score calculation in the function predict, so I followed the documentation of ClassificationSVM which says:

The linear SVM score function is f(x)=(x/s)′β+b where: x is an observation (corresponding to a row of X). s is the kernel scale and β is the vector of fitted linear coefficients. b is the bias term (corresponding to SVMModel.Bias).

However, when I calculate the score using f(x)=(x/s)′β+b, the score is different from what is returned by the function predict.

My svmModel:

Beta is [-0.9608 0.4401 -1.8665 -0.0358 -1.2389 0.9508 -1.9353 -2.9381 2.2893 1.4051 1.4547] svmModel.KernelParameters.Scale is 1.8839

My test data (1 observation) is [8.939 8.497 7.899 6.755 5.674 7.433 8.600 10.355 5.017 7.442 9.668]. Score from the function predict is 3.2217 -3.2217. Score from f(x)=(x/s)′β+b is -10.9064.

Is there any other steps required besides f(x)=(x/s)′β+b or I am using some parameters wrong ?

Thanks.

Answer 1

[Ilya]

In 14a and 14b, the Beta coefficients of an SVM model need to be divided by KernelParameters.Scale to get correct predictions. In 15a, dividing by the scale is no longer necessary.

[Yancey Li]

The predictions are the same now. Thanks.

[liang shuaibing]

hi ,I also have this problem ,but the Scale in my model is 1 . so I still can not get the same predictions. I use the example at matlab website .here is my matlab code and result .can you teach me your calculate way?thanks

load fisheriris

inds = ~strcmp(species,'setosa');

X = meas(inds,3:4);

y = species(inds);

SVMModel = fitcsvm(X,y)

[~,score] = predict(SVMModel,X);

score(1:3,:)

# run result!!!!!!!!!!!!!!!!!!!!

ans =

1.0000 -1.0000

1.2112 -1.2112

0.3380 -0.3380

calculate_by_me=transpose(X/SVMModel.KernelParameters.Scale).*SVMModel.Beta+SVMModel.Bias;

calculate_by_me(:,1:3)

# run result!!!!!!!!!!!!!!!!!!!!

ans =

-4.1551 -4.5917 -3.7185

-11.2598 -11.0344 -11.0344

[liang shuaibing]

ok ,get same score ,need to normalize it at first