从KNN函数中获取AUC值

Question

我正在使用class包来使用KNN算法。我也使用ROCR包来计算AUC值。

knn_one<-knn(train, test, train$Digit, k=1)

计算另一种方法的AUC值，例如分类树，我使用了以下一系列命令：

treeTrain_Pred<-predict(Tree_Train, test , type = "prob")[,2]
Pred<-prediction(treeTrain_Pred, test$Digit)
Perf<-performance(Pred, "auc")
Perf@y.values[[1]]

然而，当我尝试

时

knn_one = predict(knn_one, test, type="prob")[,2]

我收到以下错误：

Error in UseMethod("predict") : 
no applicable method for 'predict' applied to an object of class "factor"

如何修复此问题并获取KNN功能的AUC值？

Answer 1

对于knn模型没有预测方法，而是在单个调用中训练和接收预测。声纳数据示例：

library(mlbench)
data(Sonar)   

创建数据分区：

set.seed(1)
tr_ind <- sample(1:nrow(Sonar), 150)
train <- Sonar[tr_ind,]
test <- Sonar[-tr_ind,]

mod <- class::knn(cl = train$Class,
                  test = test[,1:60],
                  train = train[,1:60],
                  k = 5,
                  prob = TRUE)

现在预测的概率在：

attributes(mod)$prob

library(pROC)

roc(test$Class, attributes(mod)$prob)
#output
Call:
roc.default(response = test$Class, predictor = attributes(mod)$prob)

Data: attributes(mod)$prob in 30 controls (test$Class M) < 28 cases (test$Class R).
Area under the curve: 0.4667

plot(roc(test$Class, attributes(mod)$prob),
     print.thres = T,
     print.auc=T)

让我们试试k = 4

mod <- class::knn(cl = train$Class,
                  test = test[,1:60],
                  train = train[,1:60],
                  k = 4,
                  prob = TRUE)

plot(roc(test$Class, attributes(mod)$prob),
     print.thres = T,
     print.auc = T,
     print.auc.y = 0.2)