Question

我正在阅读Luis Torgo使用R教科书进行数据挖掘的第2章中的代码，也可以在这里找到： http://www.dcc.fc.up.pt/~ltorgo/DataMiningWithR/code2.html。

我想准确理解bestScores函数输出中的分数。我认为它们是NMSE（归一化均方误差），但我认为归一化意味着这些分数介于0和1之间。表面上看，分数越低越好，但我想确定。请注意，experimentalComparison函数运行大约需要1-2分钟。

if (require(rpart)==F) install.packages("rpart"); require(rpart)
if (require(DMwR)==F) install.packages("DMwR"); require(DMwR)

data(algae)
algae <- algae[-manyNAs(algae), ]
clean.algae <- knnImputation(algae, k = 10)

lm.a1 <- lm(a1 ~ .,data=clean.algae[,1:12])
rt.a1 <- rpart(a1 ~ .,data=algae[,1:12])

final.lm <- step(lm.a1)

lm.predictions.a1 <- predict(final.lm,clean.algae)
rt.predictions.a1 <- predict(rt.a1,algae)

cv.rpart <- function(form,train,test,...) {
  m <- rpartXse(form,train,...)
  p <- predict(m,test)
  mse <- mean((p-resp(form,test))^2)
  c(nmse=mse/mean((mean(resp(form,train))-resp(form,test))^2))
}
cv.lm <- function(form,train,test,...) {
  m <- lm(form,train,...)
  p <- predict(m,test)
  p <- ifelse(p < 0,0,p)
  mse <- mean((p-resp(form,test))^2)
  c(nmse=mse/mean((mean(resp(form,train))-resp(form,test))^2))
}

res <- experimentalComparison(
  c(dataset(a1 ~ .,clean.algae[,1:12],'a1')),
  c(variants('cv.lm'), 
    # 3 tree models each with a different 
    # complexity
    variants('cv.rpart',se=c(0,0.5,1))),
  # 3 times 10-fold cross-validation
  # 1234 is seed 
  cvSettings(3,10,1234))

getVariant('cv.rpart.v1',res)

DSs <- sapply(names(clean.algae)[12:18],
              function(x,names.attrs) { 
                f <- as.formula(paste(x,"~ ."))
                # dataset is a class of objects that represent all necessary
                # information on a predictive task
                # dataset(formula, data, name)
                dataset(f,clean.algae[,c(names.attrs,x)],x) 
              },
              names(clean.algae)[1:11])

res.all <- experimentalComparison(
  DSs,
  c(variants('cv.lm'),
    variants('cv.rpart',se=c(0,0.5,1))
  ),
  cvSettings(5,10,1234))

bestScores(res.all)

这是输出：

> bestScores(res.all)
$a1
          system   score
nmse cv.rpart.v1 0.64231

$a2
          system score
nmse cv.rpart.v3     1

$a3
          system score
nmse cv.rpart.v2     1

$a4
          system score
nmse cv.rpart.v2     1

$a5
       system     score
nmse cv.lm.v1 0.9316803

$a6
       system     score
nmse cv.lm.v1 0.9359697

$a7
          system    score
nmse cv.rpart.v3 1.029505

Answer 1

基于this网站，NMSE可以大于1.数字越小，模型在空间和时间上表现越好。

从DMwR包中的bestScores函数中了解分数

1 个答案: