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.
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.
Best,
365 Eli
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