如何从插入符:: train对象绘制最终的c50决策树模型（库C50）

Question

我使用caret库中的train函数训练了Decision Tree模型：

gr = expand.grid(trials = c(1, 10, 20), model = c("tree", "rules"), winnow = c(TRUE, FALSE))
dt = train(y ~ ., data = train, method = "C5.0", trControl = trainControl(method = 'cv', number = 10), tuneGrid = gr)

现在，我想为最终模型绘制Decision Tree。但是此命令不起作用：

plot(dt$finalModel)

Error in data.frame(eval(parse(text = paste(obj$call)[xspot])), eval(parse(text = paste(obj$call)[yspot])),  : 
  arguments imply differing number of rows: 4160, 208, 0

有人已经在这里问过这个问题：topic

建议的解决方案是使用拟合的train对象中的bestTune手动定义相关的c5.0模型。然后正常绘制c5.0模型：

c5model = C5.0(x = x, y = y, trials = dt$bestTune$trials, rules = dt$bestTune$model == "rules", control = C5.0Control(winnow = dt$bestTune$winnow))
plot(c5model)

我试图这样做。是的，它可以绘制c5.0模型，BUT来自train对象的预测概率和手动重新创建的c5.0模型不匹配。我猜这是因为在手动重新创建的[[c5.0模型中，我们没有实现10倍交叉验证。

所以，我的问题是：是否可以从

caret :: train

对象中提取最终的c5.0模型并绘制该Decision Tree？

Answer 1

预测的概率应该相同，请参见下文：

library(MASS)
library(caret)
library(C50)
library(partykit)

traindata = Pima.tr
testdata = Pima.te

gr = expand.grid(trials = c(1, 2), 
model = c("tree"), winnow = c(TRUE, FALSE))

dt = train(x = traindata[,-ncol(testdata)], y = traindata[,ncol(testdata)], 
method = "C5.0",trControl = trainControl(method = 'cv', number=3),tuneGrid=gr)

c5model = C5.0.default(x = traindata[,-ncol(testdata)], y = traindata[,ncol(testdata)], 
trials = dt$bestTune$trials, rules = dt$bestTune$model == "rules", 
control = C5.0Control(winnow = dt$bestTune$winnow))

all.equal(predict(c5model,testdata[,-ncol(testdata)],type="prob"),
predict(dt$finalModel,testdata[,-ncol(testdata)],type="prob"))
[1] TRUE

所以我建议您仔细检查预测是否相同。
您看到从插入符号绘制最终模型的错误来自于$ call下存储的内容，这很奇怪，我们可以用一个可以进行绘制的调用来代替它：
plot(c5model)

finalMod = dt$finalModel
finalMod$call = c5model$call
plot(finalMod)

或者您可以像使用培训结果那样重写它，但是您可以看到它与表达式有点复杂（或者至少我不太满意）：
newcall = substitute(C5.0.default(x = X, y = Y, trials = ntrials, rules = RULES, control = C5.0Control(winnow = WINNOW)),
list(
X = quote(traindata[, -ncol(traindata)]),
Y = quote(traindata[, ncol(traindata)]),
RULES = dt$bestTune$model == "rules",
ntrials = dt$bestTune$trials,
WINNOW = dt$bestTune$winnow)
)

finalMod = dt$finalModel
finalMod$call = newcall