Sorry if the title seems a little roundabout, but let me explain what I'm trying to do. I'm training a XGBClassifier (in python) on samples that aren't strictly in the class of 0 and 1, but have a little granularity of range-- anywhere from [0, 1], [.25, .75], [.5 .5], [.75, .25], [1, 0] for each of the two classes, where [0, 1] means it's 0% class A and 100% class B.

The reason I would rather not use regression is that the training values aren't technically continuous, but discreet within the spectrum of 0 to 1, and I'm trying to combine the power of doing multi-class classification only within the framework of all classes being simply different combinations of purely class A and class B. Perhaps regression is still a better option, or doing reg:linear as the objective-- but that doesn't exactly solve my problem.

For example, for measuring sentiment less in terms of "negative or positive" but "25% positive 75% negative" using predict_proba().

I'm trying to figure out what the best way to do this is. I know the base class of the XGBClassifier is binary:logistic, and I might try this also using multi:softmax, but I like the idea of using the predict_proba() between 0 and 1 as a measure of where a sample falls on a scale between class A and class B (or really, between 0 and 1) which would be more difficult using 5 separate "classes."

(For the following example, I'm using the letters A and B but really mean 0 and 1. It's just less confusing this way.)

My first inclination is to force classification probabilities by using ratios of A and B in the training set for each sample, essentially sending each one through four times with different classifications-- but I'm not sure if there's an easier way or if it's doing what I think it is.

For example-- if I have a sample that I want to represent as [.5, .5] so basically, a 50/50 or "neutral" sentiment (so that other samples I sent through later come out around [.5, .5], I'd train it four times with a value of A and four times with a value of B. Then for something that should be classified as [0, 1], we train it eight times with a value of 1, and for something that is [.75, .25], we'd train it six times with a value of 0 and two times with a value of 1.

Here's how I'd train each sample then, where "B B B B" would mean I train the same sample four times telling the classifier it is B, etc:

[0.00, 1.00]: B B B B 
[0.25, 0.75]: A B B B 
[0.50, 0.50]: A A B B 
[0.75, 0.25]: A A A B 
[1.00, 0.00] :A A A A

So-- barring this approach being incorrect, is there a better way to go about what I'm trying to do? Like an analog for predict_proba() but for training inputs? Knowing how the algorithm works i don't think that exists, but then again, I'm here to be schooled.

Is this a bastardization of a binary classifier parading as a regression wannabe? Or is this an alright way to do what I'm trying to do?

Thanks everybody.


Your instinct is correct -- this is still a binary problem. The feature vectors $x$ and labels $y$ have just been "compressed" in your representation. Consider some feature vector $x$ that has associated label $y = (0.25, 0.75)$. This is the exact same as having $$ (X,Y)= \left(\begin{bmatrix} x \\ x \\ x \\ x \\ \end{bmatrix}, \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \\ 0 & 1 \\ \end{bmatrix} \right) $$ as parts of your feature and label matrix.

Of course the order isn't imporant, so you could also write $$ Y= \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \\ 0 & 1 \\ \end{bmatrix} ,$$ or any other ordering, for your labels of a particular $(x,y)$.

If you un-compress your data using this method, it's exactly the same as the ordinary binary case.

Note that $(x,y)$ are just stand-ins for any tuple of feature vectors and labels. There might be another feature vector $z \neq x$ which also has label $y = [0.25, 0.75]$. If we de-compress this and append it to the previous result, we have $$ (X,Y)= \left(\begin{bmatrix} x \\ x \\ x \\ x \\ z \\ z \\ z \\ z \\ \end{bmatrix}, \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \\ 0 & 1 \\ 0 & 1 \\ 0 & 1 \\ 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \right) .$$

  • $\begingroup$ Won't the classifier then predict the more likely thing to be the class when presented with $x$? The last step of a classifier based on probabilities is typically to argmax, and steps before this point are analogous to a regressor that tries to match real numbers in [0,1] (modulo some log-prob softmaxer as the last step). $\endgroup$ Feb 14 '18 at 15:01
  • $\begingroup$ @pvlkmrv That's a very roundabout way of saying that, in a vacuum, $\mathbb{P}(y=1|x) > \mathbb{P}(y=0|x)$, which is what we already know by inspection. But if we take all other feature and label vectors into account, including ones such as $z \neq x$ associated to $y = [0.25, 0.75]$, it's not necessarily true that the model will produce that estimate, because the labels, the features and the model all influence the predicted probabilities. $\endgroup$
    – Sycorax
    Feb 14 '18 at 15:47
  • 1
    $\begingroup$ I like this explanation-- It confirms how I was trying to think about it. Thank you. The nature of this problem is supplying varying "shades" of binary classification in order to inform the classifier of confidence factors, which is what I'm trying to encapsulate, and you've explained it nicely. $\endgroup$
    – oxmpbeta
    Feb 14 '18 at 21:52
  • $\begingroup$ @oxmpbeta I'm glad you like it. If you've found this helpful, please consider upvoting and/or accepting this answer. $\endgroup$
    – Sycorax
    Feb 14 '18 at 22:03

Since you only have two outputs A and B and 1 - A = B, you can discard one of them because you can always reconstruct what the other should be from a single answer.

A classifier considers classes to be completely disjoint, that no pair is any more similar than any other, so it's not really quite what you want. You're correct that some classifiers output probabilities or log probabilities, but these are really confidences (How likely is it the classification is A?), and you want to be 100% confident the answer is [0.5 0.5] not have completely split confidence, which is what would happen if you trained as you outline.

If you want there to be a relationship between outputs that are "closer", then I would train a univariate regressor to come up with that one value. Since your answers are discrete, round it to the nearest value from the set of possibilities.


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