Question

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。

Answer 1

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", "Yes", "Yes", "Yes", 
"Yes", "Yes", "Yes", "Yes", "Yes", "No", "No", "Yes", "Yes", 
"Yes", "Yes", "Yes", "Yes", "No", "No"), LogSales = c(15.36307, 
15.36307, 16.05044, 16.45457, 16.90655, 16.05044, 16.05044, 16.45457, 
16.05044, 16.90655, 15.87283, 15.87283, 16.90655, 16.45457, 16.90655, 
16.90655, 16.05044, 16.05044, 15.87283, 15.87283), Age = c(32L, 
32L, 44L, 58L, 59L, 44L, 44L, 58L, 44L, 59L, 45L, 45L, 59L, 58L, 
59L, 59L, 44L, 44L, 45L, 45L), LogTA = c(15.2018, 15.2018, 15.27176, 
15.95558, 16.5881, 15.27176, 15.27176, 15.95558, 15.27176, 16.5881, 
15.32751, 15.32751, 16.5881, 15.95558, 16.5881, 16.5881, 15.27176, 
15.27176, 15.32751, 15.32751), CoAge = c(3L, 3L, 7L, 10L, 11L, 
7L, 7L, 10L, 7L, 11L, 8L, 8L, 11L, 10L, 11L, 11L, 7L, 7L, 8L, 
8L), CoAge2 = c(9L, 9L, 49L, 100L, 121L, 49L, 49L, 100L, 49L, 
121L, 64L, 64L, 121L, 100L, 121L, 121L, 49L, 49L, 64L, 64L), 
    Reg = c(1L, 1L, 0L, 1L, 1L, 0L, 0L, 1L, 0L, 1L, 0L, 0L, 1L, 
    1L, 1L, 1L, 0L, 0L, 0L, 0L), DigMkt = c(0L, 0L, 1L, 0L, 0L, 
    1L, 1L, 0L, 1L, 0L, 1L, 1L, 0L, 0L, 0L, 0L, 1L, 1L, 1L, 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：

绘制支持向量机的ROC

1 个答案: