注意:以下解决方案仅适用于 Scala。我没有找到用 Python 做的方法。
假设您只想像示例中那样直观地表示树,也许一种选择是调整 Spark GitHub 上 Node.scala 代码中的 subtreeToString 方法,以包含每个节点拆分的概率,例如以下sn-p:
def subtreeToString(rootNode: Node, indentFactor: Int = 0): String = {
def splitToString(split: Split, left: Boolean): String = {
split.featureType match {
case Continuous => if (left) {
s"(feature ${split.feature} <= ${split.threshold})"
} else {
s"(feature ${split.feature} > ${split.threshold})"
}
case Categorical => if (left) {
s"(feature ${split.feature} in ${split.categories.mkString("{", ",", "}")})"
} else {
s"(feature ${split.feature} not in ${split.categories.mkString("{", ",", "}")})"
}
}
}
val prefix: String = " " * indentFactor
if (rootNode.isLeaf) {
prefix + s"Predict: ${rootNode.predict.predict} \n"
} else {
val prob = rootNode.predict.prob*100D
prefix + s"If ${splitToString(rootNode.split.get, left = true)} " + f"(Prob: $prob%04.2f %%)" + "\n" +
subtreeToString(rootNode.leftNode.get, indentFactor + 1) +
prefix + s"Else ${splitToString(rootNode.split.get, left = false)} " + f"(Prob: ${100-prob}%04.2f %%)" + "\n" +
subtreeToString(rootNode.rightNode.get, indentFactor + 1)
}
}
我在Iris dataset 上运行的模型上对其进行了测试,得到了以下结果:
scala> println(subtreeToString(model.topNode))
If (feature 2 <= -0.762712) (Prob: 35.35 %)
Predict: 1.0
Else (feature 2 > -0.762712) (Prob: 64.65 %)
If (feature 3 <= 0.333333) (Prob: 52.24 %)
If (feature 0 <= -0.666667) (Prob: 92.11 %)
Predict: 3.0
Else (feature 0 > -0.666667) (Prob: 7.89 %)
If (feature 2 <= 0.322034) (Prob: 94.59 %)
Predict: 2.0
Else (feature 2 > 0.322034) (Prob: 5.41 %)
If (feature 3 <= 0.166667) (Prob: 50.00 %)
Predict: 3.0
Else (feature 3 > 0.166667) (Prob: 50.00 %)
Predict: 2.0
Else (feature 3 > 0.333333) (Prob: 47.76 %)
Predict: 3.0
可以使用类似的方法来创建包含此信息的树结构。主要区别是将打印的信息(split.feature、split.threshold、predict.prob 等)存储为 val 并使用它们来构建结构。