1、 P220
对该段文字的解决：
得到最大bin 数量后，求最大split 数量。对于无序特征，split = bin 数目/2；对于有序特征，split = bin 数目–1。
其中有读者问到：对于无序特征，split = bin 数目/2这个的由来，解释如下：
1）首先计算numBins：
// 当前的特征数量小于m值，则认为无序
if (numCategories <=maxCategoriesForUnorderedFeature) {//无序时
unorderedFeatures.add(featureIndex)
numBins(featureIndex) = numUnorderedBins(numCategories)
} else {//有序时
numBins(featureIndex) = numCategories
}
根据以上可知，无序时numBins = numUnorderedBins(numCategories)
其中numUnorderedBins函数如下：
/**
* Given the arity of a categorical feature(arity = number of categories),
* return the number of bins for the featureif it is to be treated as an unordered feature.
* There is 1 split for every partitioning ofcategories into 2 disjoint, non-empty sets;
* there are math.pow(2, arity - 1) - 1 suchsplits.
* Each split has 2 corresponding bins.
* 解释：一次划分会有2个bins，好比，切西瓜，一刀下去，分成2块
*/
def numUnorderedBins(arity: Int): Int = 2 * ((1 << arity - 1) - 1)
根据公式：numBins = 2*math.pow(2,arity - 1) – 1
2）根据numBins计算numSplits：
def numSplits(featureIndex: Int): Int = if(isUnordered(featureIndex)) {
numBins(featureIndex) >> 1
} else {
numBins(featureIndex) - 1
}
根据公式：numSplits = numBins/2= math.pow(2, arity - 1) – 1
原文：http://blog.csdn.net/sunbow0/article/details/51211636