尝试在 R 中创建 ROC 曲线时出现预测函数错误答案

【问题标题】：Getting error with prediction function when trying to create an ROC curve in R尝试在 R 中创建 ROC 曲线时出现预测函数错误
【发布时间】：2016-10-01 00:56:58
【问题描述】：

在对 UCI 存储库中提供的印度肝病患者数据集执行分类后尝试实施 ROC 曲线。收到错误。以下是 R 中的代码，后跟错误，然后是数据集头部的 dput。

代码

library(ROCR)
library(ggplot2)
Data<-read.csv("C:/Users/Dell/Desktop/Codes and Datasets/ilpd.csv")
nrow(Data)  
set.seed(9850)
gp<-runif(nrow(Data))
Data<-Data[order(gp),]
idx <- createDataPartition(y = Data$Class, p = 0.7, list = FALSE)
train<-Data[idx,]
test<-Data[-idx,]
ncol(train)
ncol(test)

#svm here
svmmodel<-svm(Class~.,train,
            kernel="sigmoid")
prob<-predict(svmmodel,test,type="response")
#plot(prob)

pred <- prediction(prob, test$Class) #Getting error at this line

错误

Error in prediction(prob, test$Class) : Format of predictions is invalid.

数据集

structure(list(age = c(55L, 48L, 14L, 17L, 40L, 37L), gender = c(0L, 
0L, 0L, 0L, 1L, 0L), TB = c(0.9, 2.4, 0.9, 0.9, 0.9, 0.7), DB = c(0.2, 
1.1, 0.3, 0.2, 0.3, 0.2), Alkphos = c(116L, 554L, 310L, 224L, 
293L, 235L), SGPT = c(36L, 141L, 21L, 36L, 232L, 96L), sgot = c(16L, 
73L, 16L, 45L, 245L, 54L), TP = c(6.2, 7.5, 8.1, 6.9, 6.8, 9.5
), ALB = c(3.2, 3.6, 4.2, 4.2, 3.1, 4.9), AG = c(1, 0.9, 1, 1.55, 
0.8, 1), Class = structure(c(2L, 1L, 2L, 1L, 1L, 1L), .Label = c("One", 
"Two"), class = "factor")), .Names = c("age", "gender", "TB", 
"DB", "Alkphos", "SGPT", "sgot", "TP", "ALB", "AG", "Class"), row.names =    c(216L, 
405L, 316L, 103L, 20L, 268L), class = "data.frame")

我可以获得 ROC 曲线，但现在我对它的解释感到困惑。需要进一步的帮助。

【问题讨论】：

标签： r prediction curve roc

【解决方案1】：

问题在于prob 是因子，而prediction() 正在寻找数字向量。

SVM 自然输出类预测，但你可以覆盖它并使用

#svm here, note the probability=TRUE
svmmodel<-svm(Class~.,train,
              kernel="sigmoid", probability = TRUE)

## In the ?predict.svm you can see probability = TRUE is needed to output probability
## type="repsonse" and type = "prob" would do nothing. 
pred.output <-predict(svmmodel,test,probability = TRUE)

## It outputs the probabilities as an attribute, so you need to go in an grab them
prob <- attr(pred.output, "probabilities")[,2]


pred <- prediction(prob, test$Class) #Now this works

【讨论】：