我正在尝试用分类树概率绘制 ROC 曲线。但是,当我绘制曲线时,它不存在。我正在尝试绘制 ROC 曲线,然后从曲线下的区域中找到 AUC 值。有谁知道如何解决这一问题?如果可以的话,谢谢。二元列 Risk 代表风险错误分类,我认为这是我的标签。我应该在代码中的不同点应用 ROC 曲线方程吗?
这是数据框:
library(ROCR)
data(Risk.table)
pred = prediction(Risk.table$Predicted.prob, Risk.table2$Risk)
perf = performance(pred, measure="tpr", x.measure="fpr")
perf
plot(perf)
Predicted.prob Actual.prob predicted actual Risk
1 0.5384615 0.4615385 G8 V4 0
2 0.1212121 0.8787879 V4 V4 1
3 0.5384615 0.4615385 G8 G8 1
4 0.9000000 0.1000000 G8 G8 1
5 0.1212121 0.8787879 V4 V4 1
6 0.1212121 0.8787879 V4 V4 1
7 0.9000000 0.1000000 G8 G8 1
8 0.5384615 0.4615385 G8 V4 0
9 0.5384615 0.4615385 G8 V4 0
10 0.1212121 0.8787879 V4 G8 0
11 0.1212121 0.8787879 V4 V4 1
12 0.9000000 0.1000000 G8 V4 0
13 0.9000000 0.1000000 G8 V4 0
14 0.1212121 0.8787879 G8 V4 1
15 0.9000000 0.1000000 G8 G8 1
16 0.5384615 0.4615385 G8 V4 0
17 0.9000000 0.1000000 G8 V4 0
18 0.1212121 0.8787879 V4 V4 1
19 0.5384615 0.4615385 G8 V4 0
20 0.1212121 0.8787879 V4 V4 1
21 0.9000000 0.1000000 G8 G8 1
22 0.5384615 0.4615385 G8 V4 0
23 0.9000000 0.1000000 G8 V4 0
24 0.1212121 0.8787879 V4 V4 1
这是此代码输出的 ROC 曲线,但缺少该曲线:
我又试了一次,这条ROC曲线是错误的
我使用以下代码构建了上述数据框:
包含所有数据的初始数据帧称为 shuffle.cross.validation2
#Split data 70:30 after shuffling the data frame
index<-1:nrow(LDA.scores1)
trainindex.LDA3=sample(index, trunc(length(index)*0.70),replace=FALSE)
LDA.70.trainset3<-shuffle.cross.validation2[trainindex.LDA3,]
LDA.30.testset3<-shuffle.cross.validation2[-trainindex.LDA3,]
使用包 rpart() 运行分类树
tree.split3<-rpart(Family~., data=LDA.70.trainset3, method="class")
tree.split3
summary(tree.split3)
print(tree.split3)
plot(tree.split3)
text(tree.split3,use.n=T,digits=0)
printcp(tree.split3)
tree.split3
预测预测数据和实际数据
res3=predict(tree.split3,newdata=LDA.30.testset3)
res4=as.data.frame(res3)
使用 NA 创建两列(实际和预测分类率)
res4$predicted<-NA
res4$actual<-NA
for (i in 1:length(res4$G8)){
if(res4$R2[i]>res4$V4[i]) {
res4$predicted[i]<-"G8"
}
else {
res4$predicted[i]<-"V4"
}
print(i)
}
res4
res4$actual<-LDA.30.testset3$Family
res4
Risk.table$Risk<-NA
Risk.table
创建二元预测列
for (i in 1:length(Risk.table$Risk)){
if(Risk.table$predicted[i]==res4$actual[i]) {
Risk.table$Risk[i]<-1
}
else {
Risk.table$Risk[i]<-0
}
print(i)
}
为上述 V4 和 G8 两个族创建预测概率和实际概率
#Confusion Matrix
cm=table(res4$actual, res4$predicted)
names(dimnames(cm))=c("actual", "predicted")
朴素贝叶斯
index<-1:nrow(significant.lda.Wilks2)
trainindex.LDA.help1=sample(index, trunc(length(index)*0.70), replace=FALSE)
sig.train=significant.lda.Wilks2[trainindex.LDA.help1,]
sig.test=significant.lda.Wilks2[-trainindex.LDA.help1,]
library(klaR)
nbmodel<-NaiveBayes(Family~., data=sig.train)
prediction<-predict(nbmodel, sig.test)
NB<-as.data.frame(prediction)
colnames(NB)<-c("Actual", "Predicted.prob", "acual.prob")
NB$actual2 = NA
NB$actual2[NB$Actual=="G8"] = 1
NB$actual2[NB$Actual=="V4"] = 0
NB2<-as.data.frame(NB)
plot(fit.perf, col="red"); #Naive Bayes
plot(perf, col="blue", add=T); #Classification Tree
abline(0,1,col="green")
使用 caret 包的原始朴素贝叶斯代码
library(caret)
library(e1071)
train_control<-trainControl(method="repeatedcv", number=10, repeats=3)
model<-train(Matriline~., data=LDA.scores, trControl=train_control, method="nb")
predictions <- predict(model, LDA.scores[,2:13])
confusionMatrix(predictions,LDA.scores$Family)
结果
Confusion Matrix and Statistics
Reference
Prediction V4 G8
V4 25 2
G8 5 48
Accuracy : 0.9125
95% CI : (0.828, 0.9641)
No Information Rate : 0.625
P-Value [Acc > NIR] : 4.918e-09
Kappa : 0.8095
Mcnemar's Test P-Value : 0.4497
Sensitivity : 0.8333
Specificity : 0.9600
Pos Pred Value : 0.9259
Neg Pred Value : 0.9057
Prevalence : 0.3750
Detection Rate : 0.3125
Detection Prevalence : 0.3375
Balanced Accuracy : 0.8967
'Positive' Class : V4