Apology for posting almost one question daily. I am trying to learn some aspects of Statistical Machine learning, so every day many questions coming and if I am not finding answer in my offline peer community, I am trying to ask to you.

You answer so nicely it encourages people like me to ask.

I was trying to go around Naive Bayes model. It seems it supports Multi-class classification. Now there are some binary classifiers also like logistic regression. I was thinking how would be the difference of estimation between binary classification and multiclass classification.( not multinomial logistic or multilabel).

If any one of the esteemed members of the group may kindly absolve my query.

Thanking you in Advance, Regards, Subhabrata Banerjee.


1 Answer 1


My advice is first to try at least to search on Internet. The wikipedia page for Multiclass classification explains in clear terms what it means, and it is not hard to find it after a search for "multiclass classification".

A multi-class classifier is able to classify into more 2 outcomes (classes). It is a synonym with multinomial classification. Thus, multinomial logistic regression is a multi-class classification.

However, multi-label classification it is not. Multi-label classification assumes that one observation can be labeled with (classified as) more than one category/label/class, while multi-class does not (only one class allowed for an instance).

  • $\begingroup$ Thank you. I generally found the answer from the page you referred, where it is termed as, one vs all approach and also rom Michael Collins' and Andrew Ng's lecture. I was waiting if the room has suggestion for another approach then I may be able to learn another. Regards, Subhabrata Banerjee. $\endgroup$
    Jun 27, 2014 at 12:01
  • $\begingroup$ If your question is how you can make the NaiveBayes to learn multinomial classification, you do not need that since NB is able to cope with that natively. There are some other algorithms which are build strictly for the binary classification which could be extended with one-vs-all technique, but, again, this is not the case for NB $\endgroup$
    – rapaio
    Jun 27, 2014 at 18:55
  • $\begingroup$ I was asking why Naive Bayes supports it. So I found the answer from Michael Collins, it supports (X(i), Y(i)). [Please read (i) as superscript]. Regards, Subhabrata Banerjee. $\endgroup$
    Jun 28, 2014 at 11:09

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