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I understand the concepts of L1 and L2 loss (i.e. L1 loss will force some parameter coefficients to zero while L2 will only make them approach zero). What do these do when implemented in XGBoost? Does L1 loss prune the tree more significantly than L2 loss?

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    $\begingroup$ Your question seems to be about L1/L2 regularization, not about L1/L2 loss functions. If that's the case, then see datascience.stackexchange.com/q/57255/55122 $\endgroup$
    – Ben Reiniger
    Apr 21, 2020 at 17:03
  • $\begingroup$ @BenReiniger You are correct, thank you for the link! $\endgroup$
    – Kyle
    Apr 22, 2020 at 15:43

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