“ svr”的参数调整

时间:2019-02-26 15:12:36

标签: r svm

当我尝试使用“ svr”作为函数的第一个参数时,如何使用带有支持向量REGRESSION而不是分类的调整函数,它不起作用,并且找不到单个调整回归的示例。这就是我使用e1071软件包执行的代码:

tuneSVR <- tune(svm, 
                train.x = train_[, -which(names(train) %in% c("Identifier", "Sales"))],
                train.y = train$Sales,
                data = train,
                ranges = list(epsilon = seq(0,1,0.1), cost = 2^(seq(0.5,8,.5))))

是不是因为我的问题是回归问题?这些代码行要花很长时间才能正常执行吗?以及如何计算svr的R平方?

1 个答案:

答案 0 :(得分:1)

e1071::svm()中,可以从响应变量中自动推断出问题类型,但是可以使用type参数将其覆盖。 tune()函数及其用于svm的包装的行为类似:

library( e1071 )

# Option 1: using the wrapper
tuneSVR1 <- tune.svm( mtcars[,c("drat","wt")], mtcars$mpg )

# Option 2: using the base function
tuneSVR2 <- tune( svm, mpg~drat+wt, data=mtcars )

# In both cases, the methods correctly infer that the task is regression
tuneSVR2$best.model

# Call:
# best.tune(method = svm, train.x = mpg ~ drat + wt, data = mtcars)
# 
# Parameters:
#    SVM-Type:  eps-regression 
#  SVM-Kernel:  radial 
#        cost:  1 
#       gamma:  0.5 
#     epsilon:  0.1 
# 
# Number of Support Vectors:  28

Since R-squared between two vectors is just the square of their Pearson correlation,您可以直接将其计算为

ypred <- predict( tuneSVR2$best.model, mtcars[,c("drat","wt")] )
cor( ypred, mtcars$mpg )^2
# [1] 0.8325807