Dealing with imbalanced classes

PUBLISHED ON JUN 20, 2019 — DEEP LEARNING
In the real world, it is not unusual to encounter datasets which are have imbalanced classes. This post discusses some strategies for dealing with such situations.

Let’s say that we have some classification dataset where items can be categorised as being in one of $N$ classes: $C_1, C_2, \ldots, C_N$. Each example in the dataset has a vector of features, $\boldsymbol{x}$, and a class, $C_k$. For many public datasets, the examples are balanced amongst classes such that $\Pr(C_1) = \Pr(C_2) = \ldots = \Pr(C_N)$. Put another way, if you randomly sample an item from the dataset there is an equal probability of it being in any of the classes.

However, when it comes to solving practical problems, balanced datasets are relatively rare. Instead, we have data that is generally imbalanced: $\Pr(C_1) \ne \Pr(C_2) \ne \ldots \ne \Pr(C_N)$.

In this post we will consider classification models which are both probabilistic and discriminative. That is, after training on a classification dataset, the model is able to estimate $\Pr(C_k|\boldsymbol{x})$—the probability of some input $\boldsymbol{x}$ belonging to some class $C_k$.

One of the great things about having a balanced dataset is that it plays quite nicely with the usual algorithms and objective functions used to train classification models (including neural networks). On the other hand, imbalanced datasets can throw a proverbial spanner into the works by making trivial solutions highly attractive to the learning algorithm. For example, if 99% of the examples belong to a single class, 99% accuracy can be attained by simply predicting that everything is in that class!

Training on an imbalanced dataset

A simple technique for training on an imbalanced dataset is to draw samples in accordance with the distribution of classes. So, whenever you need a training example:

  1. Select a class randomly by sampling from a uniform distribution.
  2. Draw an example from that class.

We’ll refer to the set of examples constructed according to this strategy as the training dataset $A$. Our sampling strategy is specifically designed to be balanced, so ${\Pr}_A(C_k) = \mathrm{constant}$.

An observation that we can make now is that our sampling strategy does not change the distribution of examples within each class since we aren’t doing anything fancy when drawing examples from within a class. This will become important later on.

After training on dataset $A$, we have a model for ${\Pr}_A(C_k|\boldsymbol{x})$. But we’re not finished quite yet…

Evaluating on an imbalanced dataset

Imagine for a moment that we trained a model which classifies children as future millionaires or not. Clearly there is an imbalance between those two classes, and the training data was sampled accordingly. However, there’s a problem when it comes to evaluation: for examples which contain no strong indicators of future financial success, the model is outputting a 50% chance of future millionairism! This is not altogether surprising, since during training either class was equally likely. But even so, clearly a random child is not so likely to be a millionare in practice!

What we want to do is incorporate into our predictions prior knowledge of the likelihood of a random example being in a particular class (following our previous example, the probability of a random child becoming a millionaire).

Let’s call our evaluation dataset $B$. Our prior is ${\Pr}_B(C_k)$, and in general ${\Pr}_B(C_1) \ne {\Pr}_B(C_2) \ne \ldots {\Pr}_B(C_N)$ (classes are imbalanced). The adjusted version of the model which we are hoping to find is ${\Pr}_B(C_k|\boldsymbol{x})$.

Using Bayes’ theorem,

$$ \dfrac{{\Pr}_B(C_k|\boldsymbol{x})} {{\Pr}_A(C_k|\boldsymbol{x})} = \dfrac{{\Pr}_B(\boldsymbol{x}|C_k) {\Pr}_B(C_k) {\Pr}_A(\boldsymbol{x})} {{\Pr}_A(\boldsymbol{x}|C_k) {\Pr}_A(C_k) {\Pr}_B(\boldsymbol{x})} $$

With some minor rearranging for clarity,

$$ {\Pr}_B(C_k|\boldsymbol{x}) = {\Pr}_A(C_k|\boldsymbol{x}) \cdot \dfrac{{\Pr}_B(C_k)} {{\Pr}_A(C_k)} \cdot \dfrac{{\Pr}_B(\boldsymbol{x}|C_k)} {{\Pr}_A(\boldsymbol{x}|C_k)} \cdot \dfrac{{\Pr}_A(\boldsymbol{x})} {{\Pr}_B(\boldsymbol{x})} $$

Remember how we observed before that our sampling strategy did not change the distribution of examples within each class? Well, we can write that formally as ${\Pr}_B(\boldsymbol{x}|C_k) = {\Pr}_A(\boldsymbol{x}|C_k)$. That is,

$$ \dfrac{{\Pr}_B(\boldsymbol{x}|C_k)} {{\Pr}_A(\boldsymbol{x}|C_k)} = 1 $$

So:

$$ {\Pr}_B(C_k|\boldsymbol{x}) = {\Pr}_A(C_k|\boldsymbol{x}) \cdot \dfrac{{\Pr}_B(C_k)} {{\Pr}_A(C_k)} \cdot \dfrac{{\Pr}_A(\boldsymbol{x})} {{\Pr}_B(\boldsymbol{x})} $$

Furthermore, ${\Pr}_A(\boldsymbol{x})$ and ${\Pr}_B(\boldsymbol{x})$ are constant with respect to the class, $C_k$.

So:

$$ {\Pr}_B(C_k|\boldsymbol{x}) \propto {\Pr}_A(C_k|\boldsymbol{x}) \dfrac{{\Pr}_B(C_k)} {{\Pr}_A(C_k)} $$

Now we know that ${\Pr}_B(C_k|\boldsymbol{x})$ is a probability distribution and therefore the probabilities of all classes must sum to one ($\sum_k{{\Pr}_B(C_k|\boldsymbol{x})}=1$). So, by normalising the probability distribution we get:

$$ {\Pr}_B(C_k|\boldsymbol{x}) = \dfrac{ {\Pr}_A(C_k|\boldsymbol{x}) \dfrac{{\Pr}_B(C_k)} {{\Pr}_A(C_k)} }{ \sum_i{ {\Pr}_A(C_i|\boldsymbol{x}) \dfrac{{\Pr}_B(C_i)} {{\Pr}_A(C_i)} } } $$

This formula works regardless of how examples from dataset $A$ are distributed amongst classes. However, since we know that the training data was resampled such that the classes are uniformly distributed, we can make one last simplification by cancelling the constant dataset $A$ class probability terms:

$$ {\Pr}_B(C_k|\boldsymbol{x}) = \dfrac{ {\Pr}_A(C_k|\boldsymbol{x}) {\Pr}_B(C_k) }{ \sum_i{ {\Pr}_A(C_i|\boldsymbol{x}) {\Pr}_B(C_i) } } $$

Bingo bango, we’re done.

An example

Let’s run through a quick example based on the millionaire detector described earlier on.

We’ll say that 5% of people go on to become millionaires:

$${\Pr}_B(\mathrm{millionaire}) = 0.05$$

Now we run our classification model on a kid called Billy and get a 70% probability of the child becoming a millionaire:

$${\Pr}_A(\mathrm{millionaire}|\mathrm{billy}) = 0.7$$

How good is that for the kid? Let’s find out.

Recall that the classifier is assumed to have been trained on a balanced dataset, so we calculate the adjusted probability (based on our prior) as follows:

$$ {\Pr}_B(\mathrm{millionaire}|\mathrm{billy}) = \dfrac{ 0.7 \times 0.05 }{ (0.7 \times 0.05) + (0.3 \times 0.95) } \approx 0.11 $$

So little Billy shouldn’t get their hopes up too high, they only have an ~11% probability of becoming a millionaire. Too bad.

An important caveat

It’s assumed that the model’s outputs are well-calibrated. If the predicted probabilities aren’t meaningful, the Bayesian evaluation-time strategy described here probably won’t work too well.


This article was motivated by a Reddit post about common machine learning interview questions.