从ctree对象中提取预测器

Question

我已经检查了binary tree类方法和如何从ctree函数中提取树结构？ (这有助于理解S4对象结构和槽)，但还不清楚如何获得ctree对象的最终预测器。对于rpart，我会使用类似于

 extract_preds <- function( tt ){
   leaves <- tt$frame$var == '<leaf>'
   as.character( unique( tt$frame$var[ leaves==F ] ) )
 }

是否有类似的快捷方式可用，或者我是否必须编写递归函数来遍历ctree对象并提取预测器？那，还是打印输出的正则节？谢谢。

更新:使用baydoganm下面的代码。仍然需要弄清楚如何通过递归正确地更新res：

 library(party)

 ctree_preds <- function(tr,vnames){    
    res <- character(0)
    traverse <- function(treenode,vnames,res){
    if(treenode$terminal){
        return(res)
    } else {
        res <- c(res,vnames[treenode$psplit$variableID])
        traverse(treenode$left , vnames, res )
        traverse(treenode$right, vnames, res )
        }
    }
    traverse(tr,vnames,res)
    return(unique(res))
 }

 airq <- subset(airquality, !is.na(Ozone))
 airct <- ctree(Ozone ~ ., data = airq,
                         controls = ctree_control(maxsurrogate = 3))
 plot(airct)

 ctree_preds(airct@tree,names(airq)[-1])

Answer 1

下面是我实现的从ctree对象遍历树的脚本。我在party包中使用了相同的示例，即airct数据集。

require(party)
data(airquality)

traverse <- function(treenode){
    if(treenode$terminal){
        bas=paste("Current node is terminal node with",treenode$nodeID,'prediction',treenode$prediction)
        print(bas)
        return(0)
    } else {
        bas=paste("Current node",treenode$nodeID,"Split var. ID:",treenode$psplit$variableName,"split value:",treenode$psplit$splitpoint,'prediction',treenode$prediction)
        print(bas)
}
traverse(treenode$left)
traverse(treenode$right)
}

airq <- subset(airquality, !is.na(Ozone))
airct <- ctree(Ozone ~ ., data = airq,
controls = ctree_control(maxsurrogate = 3))
plot(airct)

traverse(airct@tree)

这个函数，traverse，只是以深度一阶的方式遍历树.您可以通过更改递归部分来更改遍历的顺序。

此外，如果您想返回其他节点特征，我建议检查ctree对象的结构。

编辑:次要的代码修订。

Answer 2

麦美塔 R包的ctree2sas()函数将已安装的ctree模型转换为SAS代码。它可以很容易地适应其他语言，并且通常对对象的内部结构有指导意义。

Answer 3

split <- 
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split <- split[order(split)]