例如，假设我有一个受访者，我问他/她是否患有某种疾病。从那里，我问他/她的父亲是否也患过这种疾病。如果对后一个问题的回答是肯定的，那么我会问父亲现在是否已经治愈。如果父亲没有患过这种疾病，那么这个问题就不适用。
我可以在R中或其他地方创建这样一个“决策树”吗？
这里有一些可用的数据，其中0表示“否”，1表示“是”：
person_disease <- c(rep(1, 10), rep(0, 20))father_disease <- c(rep(1, 7), rep(0,18), rep(1,5))father_cured <- c( rep(0, 4), rep(1,3), rep(NA,18),rep(1,5) )##df <- data.frame(person_disease, father_disease, father_cured)
回答：
你可以使用data.tree包来实现这一点。有很多方法可以达到你想要的效果。例如：
person_disease <- c(rep(1, 10), rep(0, 20))father_disease <- c(rep(1, 7), rep(0,18), rep(1,5))father_cured <- c( rep(0, 4), rep(1,3), rep(NA,18),rep(1,5) )df <- data.frame(person_disease, father_disease, father_cured)library(data.tree)#这里，树是“手动”构建的#然而，根据你的数据和需求，你可能希望直接从数据中生成树#许多这样的例子可以在vignettes中找到，参见browseVignettes("data.tree")disease <- Node$new("Disease", data = df)father_disease_yes <- disease$AddChild("Father Disease Yes", label = "Father Disease", edge = "yes", condition = function(df) df[df$person_disease == 1,])father_cured_yes <- father_disease_yes$AddChild("Father Cured Yes", label = "Father Cured", edge = "yes", condition = function(df) df[df$father_cured == 1,])father_disease_no <- disease$AddChild("Father Disease No", label = "Father Disease", edge = "no", condition = function(df) df[df$person_disease == 0,])#数据过滤（前序遍历）#另一种方法是递归执行disease$Do(function(node) { for (child in node$children) { child$data <- child$condition(node$data) }})print(disease, total = function(node) nrow(node$data))#绘图#(有更多可用的选项，参见?plot.Node)SetEdgeStyle(disease, fontname = "helvetica", arrowhead = "none", label = function(node) paste0(node$edge, "\n", "total = ", nrow(node$data)))SetNodeStyle(disease, fontname = "helvetica", label = function(node) node$label)plot(disease)
