R中的Maptree决策树图:显示每个节点的分布
我使用包树生成了一个决策树。 我正在用maptree绘制它。我的代码如下所示:R中的Maptree决策树图:显示每个节点的分布,r,tree,classification,mean,median,R,Tree,Classification,Mean,Median,我使用包树生成了一个决策树。 我正在用maptree绘制它。我的代码如下所示: library(tree) library(maptree) treeTree = tree(fmla,data = df, control = tree.control(n,mincut = 50)) prune.tree(treeTree, eps = 10^(-3) ) pdf( paste(fName, varName,"tree", "tree.pdf", sep
library(tree)
library(maptree)
treeTree = tree(fmla,data = df, control = tree.control(n,mincut = 50))
prune.tree(treeTree, eps = 10^(-3) )
pdf( paste(fName, varName,"tree", "tree.pdf", sep = "_") )
missc = deviance(treeTree, detail=TRUE)
text(treeTree, splits = TRUE, label = "yval", all = FALSE,
pretty = NULL, digits = getOption("digits") - 3,adj =
par("adj"), xpd = TRUE)
title(main=varName)
dev.off()
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NO NO
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NO NO
不幸的是,这似乎绘制了每个端点节点的中间值。所以我得到一棵树,看起来像这样:
library(tree)
library(maptree)
treeTree = tree(fmla,data = df, control = tree.control(n,mincut = 50))
prune.tree(treeTree, eps = 10^(-3) )
pdf( paste(fName, varName,"tree", "tree.pdf", sep = "_") )
missc = deviance(treeTree, detail=TRUE)
text(treeTree, splits = TRUE, label = "yval", all = FALSE,
pretty = NULL, digits = getOption("digits") - 3,adj =
par("adj"), xpd = TRUE)
title(main=varName)
dev.off()
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NO NO
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NO NO
也就是说,有很多分裂,但我们总是以第二名结束
这是因为对于我的结果变量(代表患病患者的风险),中位风险始终为0
我想展示每个节点的平均值,或者更好的是,每个选项的结果分数
我已经阅读了maptree手册,但不知道该怎么做