Extract decision boundary with scikit-learn linear SVM

Question

I have a very simple 1D classification problem: a list of values [0, 0.5, 2] and their associated classes [0, 1, 2]. I would like to get the classification boundaries between th

Answer 1

Exact boundary calculated from coef_ and intercept_

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I think this is a great question and haven't been able to find a general answer to it anywhere in the documentation. This site really needs Latex, but anyway, I'll try to do my best without...

In general, a hyperplane is defined by its unit normal and an offset from the origin. So we hope to find some decision function of the form: x dot n + d > 0 (where the > may of course be replaced with >=).

In the case of the SVM Margins Example, we can manipulate the equation they start with to clarify its conceptual significance. First, let's establish the notational convenience of writing coef to represent coef_[0] and intercept to represent intercept_[0], since these arrays only have 1 value. Then some simple substitution yields the equation:

y + coef[0]*x/coef[1] + intercept/coef[1] = 0

Multiplying through by coef[1], we obtain

coef[1]*y + coef[0]*x + intercept = 0

And so we see that the coefficients and intercept function roughly as their names would imply. Applying one quick generalization of notation should make the answer clear - we will replace x and y with a single vector x.

coef[0]*x[0] + coef[1]*x[1] + intercept = 0

In general, the coef_ and intercept_ members of the svm classifier will have dimension matching the data set it was trained on, so we can extrapolate this equation to data of arbitrary dimension. And to avoid leading anyone astray, here is the final generalized decision boundary using the original variable names from the svm:

coef_[0][0]*x[0] + coef_[0][1]*x[1] + coef_[0][2]*x[2] + ... + coef_[0][n-1]*x[n-1] + intercept_[0] = 0

where the dimension of the data is n.

Or more tersely:

sum(coef_[0][i]*x[i]) + intercept_[0] = 0

where i sums over the range of the dimension of the input data.