Question

library(rpart)
train <- data.frame(ClaimID = c(1,2,3,4,5,6,7,8,9,10),
                    RearEnd = c(TRUE, TRUE, TRUE, FALSE, FALSE, FALSE, FALSE, TRUE, TRUE, FALSE),
                    Whiplash = c(TRUE, TRUE, TRUE, TRUE, TRUE, FALSE, FALSE, FALSE, FALSE, TRUE),
                    Activity = factor(c("active", "very active", "very active", "inactive", "very inactive", "inactive", "very inactive", "active", "active", "very active"),
                                      levels=c("very inactive", "inactive", "active", "very active"),
                                      ordered=TRUE),
                    Fraud = c(FALSE, TRUE, TRUE, FALSE, FALSE, TRUE, TRUE, FALSE, FALSE, TRUE))
mytree <- rpart(Fraud ~ RearEnd + Whiplash + Activity, data = train, method = "class", minsplit = 2, minbucket = 1, cp=-1)
prp(mytree, type = 4, extra = 101, leaf.round = 0, fallen.leaves = TRUE, 
    varlen = 0, tweak = 1.2)

enter image description here

然后使用printcp我可以看到交叉验证结果

> printcp(mytree)

Classification tree:
rpart(formula = Fraud ~ RearEnd + Whiplash + Activity, data = train, 
    method = "class", minsplit = 2, minbucket = 1, cp = -1)

Variables actually used in tree construction:
[1] Activity RearEnd  Whiplash

Root node error: 5/10 = 0.5

n= 10 

    CP nsplit rel error xerror xstd
1  0.6      0       1.0    2.0  0.0
2  0.2      1       0.4    0.4  0.3
3 -1.0      3       0.0    0.4  0.3

所以根节点错误是0.5，从我的理解是错误分类错误。但是我在计算敏感性（真阳性的比例）和特异性（真阴性的比例）方面遇到了麻烦。如何根据rpart输出计算出这些？

（以上示例来自http://gormanalysis.com/decision-trees-in-r-using-rpart/）

Answer 1

您可以使用caret包来执行此操作：

数据：

library(rpart)
train <- data.frame(ClaimID = c(1,2,3,4,5,6,7,8,9,10),
                    RearEnd = c(TRUE, TRUE, TRUE, FALSE, FALSE, FALSE, FALSE, TRUE, TRUE, FALSE),
                    Whiplash = c(TRUE, TRUE, TRUE, TRUE, TRUE, FALSE, FALSE, FALSE, FALSE, TRUE),
                    Activity = factor(c("active", "very active", "very active", "inactive", "very inactive", "inactive", "very inactive", "active", "active", "very active"),
                                      levels=c("very inactive", "inactive", "active", "very active"),
                                      ordered=TRUE),
                    Fraud = c(FALSE, TRUE, TRUE, FALSE, FALSE, TRUE, TRUE, FALSE, FALSE, TRUE))
mytree <- rpart(Fraud ~ RearEnd + Whiplash + Activity, data = train, method = "class", minsplit = 2, minbucket = 1, cp=-1)

解决方案

library(caret)

#calculate predictions
preds <- predict(mytree, train)

#calculate sensitivity
> sensitivity(factor(preds[,2]), factor(as.numeric(train$Fraud)))
[1] 1

#calculate specificity
> specificity(factor(preds[,2]), factor(as.numeric(train$Fraud)))
[1] 1

sensitivity和specificity都将预测作为第一个参数，观察值（响应变量，即train$Fraud）作为第二个参数。

根据文件记录，预测和观察值都需要作为具有相同水平的因素提供给函数。

在这种情况下，特异性和敏感性都是1，因为预测是100％准确的。

Answer 2

根节点错误是树根处的错误分类错误。因此在添加任何节点之前发生错误分类错误。不是最终树的错误分类错误。

R：如何计算rpart树的灵敏度和特异性

2 个答案: