I know that models such as random forest and boosted trees don't require one-hot encoding for predictor levels, but I don't really get why. If the tree is making a split in the feature space, then isn't there an inherent ordering involved? There must be something I'm missing here.

To add to my confusion I took a problem I was working on and tried using one-hot encoding on a categorical feature versus converting to an integer using xgboost in R. The generalization error using one-hot encoding was marginally better.

Then I took another variable and did the same test, and saw the opposite result.

Can anyone help explain this?

  • $\begingroup$ Short answer: bcs the tree's algorithms do not work with distance and encoding was designed to deal with distance (I think this question was asked long ago, but anyway. Someone can find this useful in the future) $\endgroup$
    – Pithit
    Commented Aug 24, 2020 at 0:08

2 Answers 2


The encoding leads to a question of representation and the way that the algorithms cope with the representation.

Let's consider 3 methods of representing n categorial values of a feature:

  • A single feature with n numeric values.
  • one hot encoding (n Boolean features, exactly one of them must be on)
  • Log n Boolean features,representing the n values.

Note that we can represent the same values in the same methods. The one hot encoding is less efficient, requiring n bits instead of log n bits. More than that, if we are not aware that the n features in the on hot encoding are exclusive, our vc dimension and our hypothesis set are larger.

So, one might wonder why use one hot encoding in the first place?

The problem is that in the single feature representation and the log representation we might use wrong deductions.

In a single feature representation the algorithm might assume order. Usually the encoding is arbitrary and the value 3 is as far for 3 as from 8. However, the algorithm might treat the feature as a numeric feature and come up with rules like "f < 4". Here you might claim that if the algorithm found such a rule, it might be beneficial, even if not intended. While that might be true, small data set, noise and other reason to have a data set that mis represent the underlying distribution might lead to false rules.

Same can happen in logarithmic representation (e.g., having rules like "third bit is on). Here we are likely to get more complex rules, all unintended and sometimes misleading.

So, we should had identical representations, leading to identical results in ideal world. However, in some cases the less efficient representation can lead to worse results while on other cases the badly deduce rules can lead to worse results.

In general, if the values are indeed very distinct in behaviour, the algorithm will probably won't deduce such rule and you will benefit from the more efficient representation. Many times it is hard to analyze it beforehand so what you did, trying both representations, is a good way to choose the proper one.

  • 9
    $\begingroup$ To be honest, I do not really see where you exactly answer the actual question of the post: "Why don't tree ensembles require one-hot-encoding?" . You only speak about algorithms in general and basically what it is the point of one-hot-encoding but nowhere about why "Why don't tree ensembles require one-hot-encoding?". Let me know if I am missing sth. $\endgroup$
    – Outcast
    Commented Sep 4, 2019 at 16:31

xgboost usually performs better after one-hot encoding. Otherwise, it will treat your categorical variables as numerical variables.

But most other tree packages support categorical variables; in other words, they support rules such as: If(Car = Mercedes).

Again, xgboost, unfortunately, does not. Therefore, you must convert your categorical variables to binary variables so that it can do: If(Car[Mercedes] >= 0.5).


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