R中的C树：在非二叉树中以列表的形式获取所有叶节点拆分变量

Question

我们当前项目中使用的决策树使用条件推理（C Tree）算法。我可以使用下面的代码提取二进制c-tree的split变量：

    #develop ctree decision tree
    prod_discount_data_ctree <- ctree(Discount~Prod, data=prod_discount_data, controls = ctree_control(minsplit=30))
    plot(prod_discount_data_ctree)

    #extract the left and right terminal node split rule
    lvls <- levels(prod_discount_data_ctree@tree$psplit$splitpoint)

    #left leaf node split variable
    left.df = lvls[prod_discount_data_ctree@tree$psplit$splitpoint == 1]

    #right leaf node split variable
    right.df = lvls[prod_discount_data_ctree@tree$psplit$splitpoint == 0]

如果树只有一个节点（深度= 1），它可以分成2个叶子节点，则可以正常工作。但是如果树有一个节点（节点1）分成多个节点（节点2,5），这些节点进一步分裂成叶节点（节点2 {3,4}节点5 {6,7}），我该如何穿越更深层次并获取叶节点拆分变量？ 根据这个例子，我希望以4个列表的形式为节点3,4,6,7分割变量。

Answer 1

我尝试了所有可能的选项，最后找到了一种在C树内遍历的方法，并获得每个叶节点的拆分变量。如果有人想在将来推荐，请粘贴代码段。

    if (nrow(SubBrandright_total) > 200) {

      sec_discount_data <- subset(SubBrandright_total, select=c(Discount,Sector))
      sec_discount_data_ctree <- ctree(Discount~Sector, data=sec_discount_data, controls = ctree_control(minsplit=30))

      sec_lvls_r <- levels(sec_discount_data_ctree@tree$psplit$splitpoint)
      #Testing if the node is terminal [TRUE] or not [FALSE]
      #print(sec_discount_data_ctree@tree$terminal)
      #print(sec_discount_data_ctree@tree$left$terminal)
      #print(sec_discount_data_ctree@tree$left$left$terminal)
      #print(sec_discount_data_ctree@tree$left$right$terminal)

      sec_left_left.df = sec_lvls_r[sec_discount_data_ctree@tree$left$psplit$splitpoint == 1]
      sec_left.df = sec_lvls_r[sec_discount_data_ctree@tree$psplit$splitpoint == 1]

      #Using setdiff to get right leaf node from Node minus left leaf node
      sec_left_right.df = setdiff(sec_left.df,sec_left_left.df)

      print("Sector Segmentation")
      print(sec_left_left.df)
      print(sec_left_right.df)

      sec_right.df = sec_lvls_r[sec_discount_data_ctree@tree$psplit$splitpoint == 0]
      sec_right_right.df = sec_lvls_r[sec_discount_data_ctree@tree$right$psplit$splitpoint == 0]

      #Using setdiff to get left leaf node from Node minus right leaf node
      sec_right_left.df = setdiff(sec_right.df,sec_right_right.df)
      print(sec_right_left.df)
      print(sec_right_right.df)

}