chl*_*chl 51
假设您的意思是计算用于拟合模型的样本的错误率,您可以使用printcp().例如,使用在线示例,
> library(rpart)
> fit <- rpart(Kyphosis ~ Age + Number + Start, data=kyphosis)
> printcp(fit)
Classification tree:
rpart(formula = Kyphosis ~ Age + Number + Start, data = kyphosis)
Variables actually used in tree construction:
[1] Age Start
Root node error: 17/81 = 0.20988
n= 81
CP nsplit rel error xerror xstd
1 0.176471 0 1.00000 1.00000 0.21559
2 0.019608 1 0.82353 0.82353 0.20018
3 0.010000 4 0.76471 0.82353 0.20018
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所述Root node error用于计算的预测性能两种措施,考虑所显示的值时,rel error和xerror列,并且根据复杂参数(第一列):
0.76471 x 0.20988 = 0.1604973(16.0%)是重新替代错误率(即在训练样本上计算的错误率) - 这大致是
class.pred <- table(predict(fit, type="class"), kyphosis$Kyphosis)
1-sum(diag(class.pred))/sum(class.pred)
Run Code Online (Sandbox Code Playgroud)0.82353 X 0.20988 = 0.1728425(17.2%)是交叉验证误差率(使用10倍CV,看到xval在rpart.control();但也参见xpred.rpart()并且plotcp()其依赖于这种测量的).该度量是预测准确性的更客观指标.
请注意,它或多或少与分类准确性一致tree:
> library(tree)
> summary(tree(Kyphosis ~ Age + Number + Start, data=kyphosis))
Classification tree:
tree(formula = Kyphosis ~ Age + Number + Start, data = kyphosis)
Number of terminal nodes: 10
Residual mean deviance: 0.5809 = 41.24 / 71
Misclassification error rate: 0.1235 = 10 / 81
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其中Misclassification error rate从所述训练样本计算的.
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