Breiman的随机森林是否使用信息增益或Gini指数？

我想知道Breiman的随机森林（R randomForest包中的随机森林）是用作分割标准（属性选择标准）还是信息增益或基尼系数？我试图在http://www.stat.berkeley.edu/~breiman/RandomForests/cc_home.htm以及R中randomForest包的文档中找到它。但是我发现的唯一发现是，可以将Gini索引用于可变重要性计算。

A. Liaw在R中的randomForest软件包是原始代码的一部分，原始代码是c代码（已翻译）和一些剩余的fortran代码以及R包装程序代码的混合。为了确定跨断点和mtry变量的总体最佳分割，代码使用类似于gini-gain的评分函数：

$GiniGain(N,X)=Gini(N)-\frac{\lvert N_{1} \rvert }{\lvert N \rvert }Gini(N_{1})-\frac{\lvert N_{2} \rvert }{\lvert N \rvert }Gini(N_{2})$

其中是给定的特征，N是要在其上进行拆分的节点，N 1和N 2是通过拆分N创建的两个子节点。| 。| 是节点中元素的数量。$X$$N$$N_{1}$$N_{2}$$N$$\lvert . \rvert$

并且，其中K是节点中类别的数量$Gini(N)=1-\sum_{k=1}^{K}p_{k}^2$$K$

但是，应用的评分功能并不完全相同，而是等效的计算效率更高的版本。和| N | 对于所有比较的分割都是常数，因此省略。$Gini(N)$

还让我们检查零件是否在node（1）中的患病率平方和计算为$\frac{\lvert N_{2} \rvert }{\lvert N \rvert }Gini(N_{2}) \propto |N_2| Gini(N_{2}) = |N_2| (1-\sum_{k=1}^{K}p_{k}^2 ) = |N_2| \sum \frac{nclass_{2,k}^2}{|N_2|^2}$

$nclass_{1,k}$$|N_2|$

$1-$

$|N_1| \sum_{k=1}^{K}p_{1,k}^2 + |N_2| \sum_{k=1}^{K}p_{2,k}^2 = |N_1|\sum_{k=1}^{K}\frac{nclass_{1,k}^2}{|N_1|^2} + |N_2|\sum_{k=1}^{K}\frac{nclass_{2,k}^2}{|N_2|^2}$ $= \sum_{k=1}^{K}\frac{nclass_{2,k}^2}{1} |N_1|^{-1} + \sum_{k=1}^{K}\frac{nclass_{2,k}^2}{1} |N_1|^{-2}$ $= nominator_1/denominator_1 + nominator_2/denominator_2$

The implementation also allows for classwise up/down weighting of samples. Also very important when the implementation update this modified gini-gain, moving a single sample from one node to the other is very efficient. The sample can be substracted from nominators/denominators of one node and added to the others. I wrote a prototype-RF some months ago, ignorantly recomputing from scratch gini-gain for every break-point and that was slower :)

If several splits scores are best, a random winner is picked.

This answer was based on inspecting source file "randomForest.x.x.tar.gz/src/classTree.c" line 209-250