Let’s say you are classifying candy into “Milk”, “Dark”, “White” and “Other” types of chocolate. In this case it is not enough to simply have a function which ranks chocolates into some order, because, in contrast to the binary case, the rankings with respect to “milkness”, “darkness” and “whiteness” do not necessarily correlate in some particular way.

There are two common ways to cast a multi-class problem in a set of binary classifications – “one versus all” and “all versus all”. Those can be used both to construct multi-class classifiers from several binary classifiers and to analyze a given multi-class classifier by “decomposing” it.

In the first case (“one versus all”) you take your multi-class classifier and turn it into several binary classifiers, one for each class. E.g. “milk vs non-milk”, “dark vs non-dark”, etc. You can then study all of those classifiers separately (e.g. compute ROC AUC, precision, recall, accuracy, for each of them). If you need to have one number in the end, you can average the results (weighted by the proportion of each class).

The second case (“all vs all”) suggests you regard each pair of classes as a separate binary classification problem. E.g. you’d examine how good is your classifier at discriminating “milk vs dark”, “milk vs white”, “dark vs white”, etc. Actually, I haven’t actually seen this approach used to assess classifiers in practice, but it makes sense in theory.

Hope that relates to your question.

http://www.dirigibleflightcraft.com/prote.cs/

and full source code here:

https://github.com/lynaghk/prote.cs/

if you’re interested in this kind of visualization machine learning sort of thing.

thanks for the post. I remember learning this during my masters but couldnt remember what this thumb rule wass called or why it is so.. so I finally searched for ‘one over root n’. Apparently Google only works if you know what you’re looking for

I think it’s a super helpful tool because of its simplicity! (Not easy to explain to laymen or people in the business world why the 1/root(n) but good for a confidence estimation for oneself!)

Best
Ankur

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