绘制支持向量机的ROC

问题描述

我尝试按照示例https://rpubs.com/JanpuHou/359286绘制svm的ROC，但是我在最后一行代码中始终遇到错误：这是数据集的开头头（数据）

growth LogSales Age    LogTA CoAge CoAge2 Reg DigMkt
1     No 15.87283  45 15.32751     8     64   0      1
2    Yes 16.05044  44 15.27176     7     49   0      1
3    Yes 15.36307  32 15.20180     3      9   1      0
4    Yes 15.09644  31 14.97866     2      4   1      0
5    Yes 16.90655  59 16.58810    11    121   1      0
6    Yes 16.45457  58 15.95558    10    100   1      0

我的代码：

split = sample.split(data,SplitRatio = 0.70)
training = subset(data,split==T)
testing = subset(data,split==F)

###Making growth last to allow for variable importnce


###Fitting model
svm_Lin = svm(growth~.,data = training,kernel = "linear",cost =1,scale = T,probability = TRUE)

##Prediction
pred = predict(svm_Lin,testing)
table(predict = pred,truth = testing$growth)
confusionMatrix(table(pred,testing$growth))
###ROC Curve
library(ROCR)
p<- predict(svm_Lin,testing,type="decision")
pr<-prediction(p,testing$growth)
pref <- performance(pr,"tpr","fpr")
plot(pref)

当我运行以下行：pr<-prediction(p,testing$growth)时，我收到以下错误消息

Error: Format of predictions is invalid. It Couldn't be coerced to a list.

感谢您提供任何解决方法的帮助。

解决方法

我建议采用下一种方法。您遇到的主要问题是，来自svm的预测属于类型因子，因此ROCR函数无法对其进行比较。我将对您的问题进行一些修改。您拥有二进制数据，因此可以将目标变量作为两个级别的因数使用。然后，在ROCR部分中，您必须将因子转换为数值。这样，您的代码就会起作用。

此外，来自caTools包的采样方法正在产生NA。因此，我使用rsample包添加了类似的方法。这里是代码。

library(ROCR)
library(e1071)
library(rsample)
#Data
data <- structure(list(growth = c("Yes","Yes","No","No"),LogSales = c(15.36307,15.36307,16.05044,16.45457,16.90655,15.87283,15.87283),Age = c(32L,32L,44L,58L,59L,45L,45L),LogTA = c(15.2018,15.2018,15.27176,15.95558,16.5881,15.32751,15.32751),CoAge = c(3L,3L,7L,10L,11L,8L,8L),CoAge2 = c(9L,9L,49L,100L,121L,64L,64L),Reg = c(1L,1L,0L,0L),DigMkt = c(0L,1L
    )),row.names = c("3","3.1","2","6","5","2.1","2.2","6.1","2.3","5.1","1","1.1","5.2","6.2","5.3","5.4","2.4","2.5","1.2","1.3"),class = "data.frame")

现在，我们格式化目标变量：

#Format objective var to have a factor
data$growth[data$growth=='No']<-0
data$growth[data$growth=='Yes']<-1
data$growth <- factor(data$growth,levels = c(0,1),labels = c(0,1))

rsample中的拆分方法：

#Split
split <- initial_split(data,prop = 0.7,strata = 'growth')
#Create training and test set
training <- training(split)
testing <- testing(split)

我们拟合了模型：

###Fitting model
svm_Lin = svm(growth~.,data = training,kernel = "linear",cost =1,scale = T,probability = TRUE,type="C-classification")

我们对测试集进行预测：

###Predict for ROC Curve
testing$p <- predict(svm_Lin,testing,type="response")

现在，我们格式化输出变量并准备使用ROCR函数：

由于因子从1开始，数字1的类的值为2，数字0的类的值为1。可以通过将其变为数字并减去1来转换为0-1。

#Format variables
testing$growth <- as.numeric(testing$growth)-1
testing$p <- as.numeric(testing$p)-1

最后，我们建立ROC曲线：

#Build ROCR scheme
pr<-prediction(testing$p,testing$growth)
pref <- performance(pr,"tpr","fpr")
plot(pref)

输出：

r r roc svm svm