但是用pred <- predict(svmmodel4, svm_test)就行了。

支持支持，学习了

我又用《机器学习与R语言》的数据letterdata试了一下，但还是必须有响应变量，没有的话会出现Error in eval(expr, envir, enclos) : object 'letter' not found

letter=read.csv(“letter-recognition.data",sep=",",header=FALSE)
colnames(letter)=c("letter","x-box","y-box","width","high","onpix","x-bar","y-bar",
                   "x2bar","y2bar","xybar","x2ybr","xy2br","x-ege","xegvy","y-ege",
                   "yegvx")

library(caret)
set.seed(2014)
intrain=createDataPartition(y=letter$letter,p=0.8,list = FALSE)
letter_train=letter[intrain,]
letter_test=letter[-intrain,]

#kernlab包下的ksvm算法
library(kernlab)
letter_classifer=ksvm(letter~.,data=letter_train,kernel="vanilladot")
#模型评估
letter_pred=predict(letter_classifer,letter_test[,-1])
agreement= letter_pred==letter_test$letter#返回的逻辑值，预测与实际是否相等
table(agreement)
prop.table(table(agreement))
是这个数据集吧，我这里没问题。

那就奇怪了，我做的怎么不行呢？

请问In .local(x, ...) : Variable(s) `' constant. Cannot scale data.是什么意思呢？就算有这个提醒也还能继续操作吗？