链表回归树在r中的表转换

Question

我使用了来自此链接 ..It的CHAID包，它给了我一个可以绘制的chaid对象。我想要一个决策表，其中包含列中的每个决策规则，而不是一个决策树。.But我不知道如何访问这个chaid object..Kindly中的节点和路径。我遵循了此链接给出的程序

我不能在这里发布我的数据，因为它太long.So了，我正在发布一个代码，它使用带有chaid的示例数据集来执行任务。

复制自chaid:的帮助手册

library("CHAID")

  ### fit tree to subsample
  set.seed(290875)
  USvoteS <- USvote[sample(1:nrow(USvote), 1000),]

  ctrl <- chaid_control(minsplit = 200, minprob = 0.1)
  chaidUS <- chaid(vote3 ~ ., data = USvoteS, control = ctrl)

  print(chaidUS)
  plot(chaidUS)

输出：

Model formula:
vote3 ~ gender + ager + empstat + educr + marstat

Fitted party:
[1] root
|   [2] marstat in married
|   |   [3] educr <HS, HS, >HS: Gore (n = 311, err = 49.5%)
|   |   [4] educr in College, Post Coll: Bush (n = 249, err = 35.3%)
|   [5] marstat in widowed, divorced, never married
|   |   [6] gender in male: Gore (n = 159, err = 47.8%)
|   |   [7] gender in female
|   |   |   [8] ager in 18-24, 25-34, 35-44, 45-54: Gore (n = 127, err = 22.0%)
|   |   |   [9] ager in 55-64, 65+: Gore (n = 115, err = 40.9%)

Number of inner nodes:    4
Number of terminal nodes: 5

所以我的问题是如何用列中的每个决策规则(分支/路径)来获取决策表中的树数据，我不知道如何从这个chaid对象访问不同的树路径。

Answer 1

CHAID包使用派对包 (递归分区)树结构。您可以通过使用各方节点来遍历树--节点可以是终端，也可以有一个节点列表，其中包含有关决策规则(split)和拟合数据的信息。

下面的代码遍历树并创建决策表。它是为演示目的而编写的，并且只在一个示例树上进行测试。

tree2table <- function(party_tree) {

  df_list <- list()
  var_names <-  attr( party_tree$terms, "term.labels")
  var_levels <- lapply( party_tree$data, levels)

  walk_the_tree <- function(node, rule_branch = NULL) {
    # depth-first walk on partynode structure (recursive function)
    # decision rules are extracted for every branch
    if(missing(rule_branch)) {
      rule_branch <- setNames(data.frame(t(replicate(length(var_names), NA))), var_names)
      rule_branch <- cbind(rule_branch, nodeId = NA)
      rule_branch <- cbind(rule_branch, predict = NA)
    }
    if(is.terminal(node)) {
      rule_branch[["nodeId"]] <- node$id
      rule_branch[["predict"]] <- predict_party(party_tree, node$id) 
      df_list[[as.character(node$id)]] <<- rule_branch
    } else {
      for(i in 1:length(node)) {
        rule_branch1 <- rule_branch
        val1 <- decision_rule(node,i)
        rule_branch1[[names(val1)[1]]] <- val1
        walk_the_tree(node[i], rule_branch1)
      }
    }
  }

  decision_rule <- function(node, i) {
    # returns split decision rule in data.frame with variable name an values
    var_name <- var_names[node$split$varid[[1]]]
    values_vec <- var_levels[[var_name]][ node$split$index == i]
    values_txt <- paste(values_vec, collapse = ", ")
    return( setNames(values_txt, var_name))
  }
  # compile data frame list
  walk_the_tree(party_tree$node)
  # merge all dataframes
  res_table <- Reduce(rbind, df_list)
  return(res_table)
}

使用CHAID树对象调用函数：

table1 <- tree2table(chaidUS)

结果应该是这样的：

gender   ager                       empstat   educr              marstat                          nodeId   predict  
-------- -------------------------- --------- ------------------ -------------------------------- -------- ---------
NA       NA                         NA        <HS, HS, >HS       married                          3        Gore     
NA       NA                         NA        College, Post Coll married                          4        Bush     
male     NA                         NA        NA                 widowed, divorced, never married 6        Gore     
female   18-24, 25-34, 35-44, 45-54 NA        NA                 widowed, divorced, never married 8        Gore     
female   55-64, 65+                 NA        NA                 widowed, divorced, never married 9        Gore

Answer 2

首先，感谢这次精彩的表演。为了得到预测的类概率，从我这边做一点修改，而不是predict_party(party_tree，节点$id)，尝试predict_party(party_tree，节点$id，type = 'prob')。另外，要获得特定的类概率，使用predict_party(party_tree，节点$id，type = 'prob')1或predict_party(party_tree，节点$id，type = 'prob')2。