Last answered:

14 Jun 2022

Posted on:

12 Jun 2022


Resolved: Do you mean C parameter as described in sklearn?

Hi, I am kinda confused!
I guess what's explained in the lecture is lambda (lambda=1/C) not C because in SVM
error = (C* classification error) + distance error

Which means high C penalizes classification error more heavily and therefore means less distance (less regularization)

Here's the definition from sklearn.svm class:

Cfloat, default=1.0
Regularization parameter. The strength of the regularization is inversely proportional to C. Must be strictly positive. The penalty is a squared l2 penalty.

Lemme know if I am wrong.
Thank you.

1 answers ( 1 marked as helpful)
Posted on:

14 Jun 2022


Hi Ahmed,
thahnks for reaching out and for the great question! You're absolutely right, you can look at the parameter as alpha, and I believe it's also how it's implemented in sk-learn. In the literature, you can sometimes find both variants, so it's important to make the distinction.

365 Eli

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