Let $R^2$ be the R-squared defined for regression models. Given a vector x, the target, and a vector, y, the prediction of the same length, then we can compute $R$.

However, there is also the adjusted $R^2$ (see formula below) which adds $n$ and $p$ where $n$ is the number of rows and $p$ the number of features in the modelling data.

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However, even though we can compute $R^2$ for a tree-based model, I think $p$ is not defined in the case of a tree model. Hence adjusted $R^2$ doesn't make sense for a tree-based model.

Is this correct?

  • $\begingroup$ $R^2$ is in fact quite problematic for nonlinear models. Search this Stack for the details, but the gist is that the predictions and residuals are not orthogonal in nonlinear models. Anyway, I don’t see a way to make $p$ make sense, but the point of adjusted $R^2$ is to do inference. If you’re using a tree (I’m really thinking RF), your goal is to predict, not to do inference. $\endgroup$
    – Dave
    Mar 25 '20 at 5:38

It is often said on this site that R-squared is problematic for nonlinear models, search ... , but I cannot find a canonic thread. A published paper, with simulations, is this, and they also point to information measures like AIC as better alternatives. From an answer in this R-help thread

There is a good reason that an nls model fit in R does not provide r-squared - r-squared doesn't make sense for a general nls model.

One way of thinking of r-squared is as a comparison of the residual sum of squares for the fitted model to the residual sum of squares for a trivial model that consists of a constant only. You cannot guarantee that this is a comparison of nested models when dealing with an nls model. If the models aren't nested this comparison is not terribly meaningful.

So the answer is that you probably don't want to do this in the first place.

-- Douglas Bates

(This would apply equally to the adjusted version.) But that answer also points to a solution: R-squared is a kind of comparison to a simple, default (and nested) model, and if you can find some other simple model for comparison, you can use that. But, for a regression tree, the constant only model corresponds to a default tree with only one node, the root, so R-squared do make sense in this case.

But your question about the meaning of $p$ (the number of parameters) for a tree model is still interesting! In models like splines, gams and many others, we use a measure called the effective number of parameters or effective number of degrees of freedom. See this list of questions with answers. But can this be defined for a very non-linear model like trees (or forests?) The book The Elements of Statistical Learning by Hastie, Tibshirani & Friedman offers a discussion as Exercise 9.5 (page 336, second Edition.) What is proposed there is to define the degrees of freedom of the fit as $\DeclareMathOperator{\C}{\mathbb{Cov}} \sum_i \C(y_i, \hat{y}_i)/\sigma^2$. Doing the exercise is left for you. It should involve some simulation.

Some related posts: Decision Tree Quality Metric, computing AIC or BIC for nonlinear regression models


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