I use linear SVM from scikit learn (LinearSVC) for binary classification problem. I understand that LinearSVC can give me the predicted labels, and the decision scores but I wanted probability estimates (confidence in the label). I want to continue using LinearSVC because of speed (as compared to sklearn.svm.SVC with linear kernel) Is it reasonable to use a logistic function to convert the decision scores to probabilities?
import sklearn.svm as suppmach
# Fit model:
I took a look at the apis in sklearn.svm.* family. All below models, e.g.,
have a common interface that supplies a
probability: boolean, optional (default=False)
parameter to the model. If this parameter is set to True, libsvm will train a probability transformation model on top of the SVM's outputs based on idea of Platt Scaling. The form of transformation is similar to a logistic function as you pointed out, however two specific constants
B are learned in a post-processing step. Also see this stackoverflow post for more details.
I actually don't know why this post-processing is not available for LinearSVC. Otherwise, you would just call
predict_proba(X) to get the probability estimate.
Of course, if you just apply a naive logistic transform, it will not perform as well as a calibrated approach like Platt Scaling. If you can understand the underline algorithm of platt scaling, probably you can write your own or contribute to the scikit-learn svm family. :) Also feel free to use the above four SVM variations that support