Question

我是一名精算学生，正在为即将在12月举行的预测分析考试做准备。练习的一部分是使用带有插入符号和xgbTree的boosting建立模型。参见下面的代码，商队数据集来自ISLR包：

server <- function(input, output, session) { 
  df<-reactive({
    df<-iris
    if(input$Petalw==T)
    {
      df<-df[df$Petal.Width==0.2,]
    }
    else{
      df
    }
  }) 
  output$table <- DT::renderDataTable(
    DT::datatable(df(), options = list(searching = FALSE,pageLength = 25))
  )
}
ui <- navbarPage(
  title = 'Select values in two columns based on two inputs respectively',
  fluidRow(
    column(width = 3,
           checkboxInput("Petalw","PetalWithIs0.2",T),
           checkboxInput("PetalL","PetalLengthis1.4",T)
    ),
    column(9,
  tabPanel('Table',       DT::dataTableOutput('table'))
  )
  )
)
shinyApp(ui, server)

expand.grid和trainControl中的定义是由问题指定的，但我不断收到错误消息：

library(caret)
library(ggplot2)
set.seed(1000)
data.Caravan <- read.csv(file = "Caravan.csv")


data.Caravan$Purchase <- factor(data.Caravan$Purchase)
levels(data.Caravan$Purchase) <- c("No", "Yes")


data.Caravan.train <- data.Caravan[1:1000, ]
data.Caravan.test <- data.Caravan[1001:nrow(data.Caravan), ]
grid <- expand.grid(max_depth = c(1:7),
                    nrounds = 500,
                    eta =  c(.01, .05, .01),
                    colsample_bytree = c(.5, .8),
                    gamma = 0,
                    min_child_weight = 1,
                    subsample = .6)

control <- trainControl(method = "cv", 
                        number = 4,
                        classProbs = TRUE,
                        sampling = c("up", "down"))

caravan.boost <- train(formula = Purchase ~ .,
                       data =  data.Caravan.train, 
                       method = "xgbTree", 
                       metric = "Accuracy",
                       trControl = control, 
                       tuneGrid = grid)

如果从trainControl中删除采样方法，则会收到一个新错误，指出“度量标准精度不适用于回归模型”。如果删除“准确性”指标，则会显示错误消息

Error: sampling methods are only implemented for classification problems

最终的问题是，即使目标变量设置为因子变量且classProbs设置为TRUE，插入符号也将问题定义为回归而不是分类。有人可以解释如何告诉插入符号进行分类而不是回归吗？

Answer 1

caret::train没有一个formula参数，而是一个您可以在其中指定公式的form参数。因此，例如，这可行：

caravan.boost <- train(form = Purchase ~ .,
                       data =  data.Caravan.train, 
                       method = "xgbTree", 
                       metric = "Accuracy",
                       trControl = control, 
                       tuneGrid = grid)

#output:
eXtreme Gradient Boosting 

1000 samples
  85 predictor
   2 classes: 'No', 'Yes' 

No pre-processing
Resampling: Cross-Validated (4 fold) 
Summary of sample sizes: 751, 749, 750, 750 
Addtional sampling using up-sampling

Resampling results across tuning parameters:

  eta   max_depth  colsample_bytree  Accuracy   Kappa     
  0.01  1          0.5               0.7020495  0.10170007
  0.01  1          0.8               0.7100335  0.09732773
  0.01  2          0.5               0.7730581  0.12361444
  0.01  2          0.8               0.7690620  0.11293561
  0.01  3          0.5               0.8330506  0.14461709
  0.01  3          0.8               0.8290146  0.06908344
  0.01  4          0.5               0.8659949  0.07396586
  0.01  4          0.8               0.8749790  0.07451637
  0.01  5          0.5               0.8949792  0.07599005
  0.01  5          0.8               0.8949792  0.07525191
  0.01  6          0.5               0.9079873  0.09766492
  0.01  6          0.8               0.9099793  0.10420720
  0.01  7          0.5               0.9169833  0.11769151
  0.01  7          0.8               0.9119753  0.10873268
  0.05  1          0.5               0.7640699  0.08281792
  0.05  1          0.8               0.7700580  0.09201503
  0.05  2          0.5               0.8709909  0.09034807
  0.05  2          0.8               0.8739990  0.10440898
  0.05  3          0.5               0.9039792  0.12166348
  0.05  3          0.8               0.9089832  0.11850402
  0.05  4          0.5               0.9149793  0.11602447
  0.05  4          0.8               0.9119713  0.11207786
  0.05  5          0.5               0.9139633  0.11853793
  0.05  5          0.8               0.9159754  0.11968085
  0.05  6          0.5               0.9219794  0.11744643
  0.05  6          0.8               0.9199794  0.12803204
  0.05  7          0.5               0.9179873  0.08701058
  0.05  7          0.8               0.9179793  0.10702619

Tuning parameter 'nrounds' was held constant at a value of 500
Tuning parameter 'gamma' was held constant
 at a value of 0
Tuning parameter 'min_child_weight' was held constant at a value of 1
Tuning
 parameter 'subsample' was held constant at a value of 0.6
Accuracy was used to select the optimal model using the largest value.
The final values used for the model were nrounds = 500, max_depth = 6, eta = 0.05, gamma =
 0, colsample_bytree = 0.5, min_child_weight = 1 and subsample = 0.6.

您还可以使用非公式界面，在其中分别指定x和y：

caravan.boost <- train(x = data.Caravan.train[,-ncol(data.Caravan.train)],
                       y =  data.Caravan.train$Purchase, 
                       method = "xgbTree", 
                       metric = "Accuracy",
                       trControl = control, 
                       tuneGrid = grid)

请注意，当x中存在因子变量时，由于大多数算法的公式接口都调用model.matrix，因此这两种规范方法并不总是产生相同的结果。

要获取数据：

library(ISLR)
data(Caravan)

脱字号包错误-分类v回归

1 个答案: