chet - 8 months ago 68

Python Question

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:

svmmodel=suppmach.LinearSVC(penalty='l1',C=1)

predicted_test= svmmodel.predict(x_test)

predicted_test_scores= svmmodel.decision_function(x_test)

I want to check if it makes sense to obtain Probability estimates simply as [1 / (1 + exp(-x)) ] where x is the decision score.

Alternately, are there other options wrt classifiers that I can use to do this efficiently?

Thanks.

Answer

I took a look at the apis in sklearn.svm.* family. All below models, e.g.,

- sklearn.svm.SVC
- sklearn.svm.NuSVC
- sklearn.svm.SVR
- sklearn.svm.NuSVR

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 `A`

and `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 `predict_proba`

.

Source (Stackoverflow)