How is xgboost quality calculated?

Question

Could someone explain how the Quality column in the xgboost R package is calculated in the xgb.model.dt.tree function?

In the documentation it

Answer 1

OK, I think I've worked it out. The value for reg_lambda is not 0 by default as given in the documentation, but is actually 1 (from param.h)

Also, it appears that the factor of a half is not applied when calculating the gain, so the Quality column is double what you would expect. Lastly, I also don't think gamma (also called min_split_loss) is applied to this calculation either (from update_hitmaker-inl.hpp)

Instead, gamma is used to determine whether to invoke pruning, but is not reflected in the gain calculation itself, as the documentation suggests.

If you apply these changes, you do indeed get 4000.53 as the Quality for node 0 of tree 0, as in the original question. I'll raise this as an issue to the xgboost guys, so the documentation can be changed accordingly.