3
$\begingroup$

I have some data (14000 examples, with 1000 features, very sparse). My goal is to predict 50 binary values for each example, and there usually are multiple positive values. I want to train a logistic regression and a feedforward neural network, but to do so I need a smooth loss function - meaning a function that has a continuous derivative. What function's optimum values are the optimum values of this metric? What if the metric was e.g. 0.3*precision + 0.5*recall + 0.2*accuracy? Or F1 score?

My current solution is to use binary crossentropy, and then select such a threshold for the probability that maximizes the given metric. That feels very "hacky" though.

$\endgroup$
  • 1
    $\begingroup$ If you weight accuracy more than recall it seems to me that you are giving more weight to precision rather than recall. Therefore you do not want false positives. Thus I think the cost weight of a positive sample should be increased. There are ways to make cross-entropy cost sensitive I guess: eg stats.stackexchange.com/questions/68940/… $\endgroup$ – Simone Apr 24 '17 at 17:11
  • $\begingroup$ It's not hacky, that's the correct thing to do. Think of it this way, if instead of needing a model, someone told you the true conditional class probabilities $P(y \mid x)$, then you would know everything there is to know about the situation, and your job would necessarily be to set a threshold. $\endgroup$ – Matthew Drury Apr 25 '17 at 4:37
4
+50
$\begingroup$

Your procedure isn't hacky, its the correct thing to do.

As a thought experiment, imagine a situation where an oracle told you the true class probabilities

$$P(y \mid X)$$

If you actually knew this function, you would have complete knowledge about the situation.

Now suppose you need to make a hard classification, i.e., you actually need to take each possible value of $y$ and assign it to either the $1$ class or the $0$ class. Further, you would like to do so in order to minimize some cost function which depends on your choices.

Suppose you use a rule $R$ that does not arise from setting some threshold on the conditional probabilities, i.e. the direction of class assignment somewhere disagrees with the direction of the class probabilities. Then there would be two data points $(x_1, y_1)$ and $(x_2, y_2)$ with

$$ P(y_1 \mid x_1) < P(y_2 \mid x_2)$$

yet

$$ R(x_1) = 1, R(x_2) = 0 $$

Then, interpreting "average" as the behaviour I would observe across repeatedly sampled datasets including $x_1$ and $x_2$, the average accuracy of this rule is:

$$\frac{P(y_1 \mid x_1) + (1 - P(y_2 \mid x_2))}{2} = \frac{P(y_1 \mid x_1) - P(y_2 \mid x_2) + 1}{2}$$

While the average accuracy of the rule that makes the opposite assignment is:

$$ \frac{P(y_2 \mid x_2) - P(y_1 \mid x_1) + 1}{2} $$

Because $ P(y_1 \mid x_1) < P(y_2 \mid x_2) $, the average accuracy of the second rule is always larger. You should be able to make similar arguments with other metrics for hard classification, precision and recall for example.

So the direction of class assignment must agree with the directionality of the conditional class probabilities, otherwise a better rule is available (at least on average).

Note: It is important in the above argument that the evaluation metric be a function of only the true $y$ and classified $y$ alone. If, instead, it is a function of $x$ as well (say customers who own a home are more valuable to us than those without a home, where home ownership is a feature in our prediction scheme) then it may pay to set a different threshold for different subsets of our population. The argument above still tells us that our classification rule should be a monotonic function of the class probabilities within each subpopulation.

$\endgroup$
2
$\begingroup$

There are a few ways to optimize trade-offs between precision and recall:

  1. As Mattew said, you can modify the classification threshold: lower threshold (eg 0.1 on logistic regression) yields to high recall and lower precision. Viceversa, higher threshold (eg 0.9) yields to high precision and lower recall. I guess that this is what you are doing at the moment;
  2. You fix the classification threshold to the default value: ie 0.5 for logistic regression. Then you modify the cost of a positive instance in the loss function. You might want to use something like the weighted cross entropy: $-[weight \cdot y \cdot \log(p) + (1 - y) \log(1 - p)]$. Higher $weight$ for the positive class yields to higher precision and lower recall;
  3. You fix the threshold and the loss and you change the training data set. If you oversample the positive class you will classify with higher precision and lower recall.
$\endgroup$
1
$\begingroup$

A loss function needs to have a non-zero gradient almost everywhere in your domain. As you say, or imply, this is not the case for either accuracy or recall.

You would need to convert your binary targets into reals somehow, eg use sigmoid as your output, which is a real, and the goal is to minimize the L1 or L2 distance of your output from your targets, expressed as reals, ie 0, or 1.

It's not impossible to train with a loss function that involves discrete values, but it tends to need some kind of MCMC, variational, lots of complicated maths. Making your loss function differentiable will be significantly easier.

$\endgroup$
  • $\begingroup$ Yes, I use sigmoid outputs and gradient descent to optimize some loss (it could be MSE, MAE, crossentropy and so on), all in keras. My question is which loss fits my metric and why? $\endgroup$ – Hristo Buyukliev Apr 1 '17 at 21:40
1
$\begingroup$

Usually you wouldn't directly build a custom loss function. Separate between estimating probabilities and making decisions. Build a classifier estimating probabilities by minimizing logloss, probably using a softmax in the output layer. Then make decisions based on the probabilities which minimize your own custom expected loss.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.