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我正在使用派对包在 R 中构建 CART 回归树模型，但是当我尝试将模型与测试数据集应用时，我收到错误消息说级别不匹配。

在过去的一周里，我一直在阅读论坛上的帖子，但仍然找不到解决我问题的正确方法。所以我在这里使用我编造的假例子重新发布这个问题。有人可以帮助解释错误消息并提供解决方案吗？

我的训练数据集有大约 1000 条记录，测试数据集有大约 150 条记录。两个数据集中都没有 NA 或空白字段。

我在派对包下使用 ctree 的 CART 模型是：

mytree<- ctree(Rate~Bank+Product+Salary, data=data_train)

数据训练示例：

Rate  Bank  Product  Salary    
1.5    A     aaa     100000
0.6    B     abc      60000
3      C     bac      10000
2.1    D     cba      50000
1.1    E     cca      80000

数据测试示例：

Rate  Bank  Product   Salary
2.0    A     cba       80000
0.5    D     cca      250000
0.8    E     cba      120000
2.1    C     abc       65000

levels(data_train$Bank) : A, B, C, D, E

levels(data_test$Bank): A,D,E,C

我尝试使用以下代码设置为同一级别：

>is.factor(data_test$Bank)

 TRUE 
(Made sure Bank and Products are factors in both datasets)
>levels(data_test$Bank) <-union(levels(data_test$Bank), levels(data_train$Bank))

> levels(data_test$product)<-union(levels(data_test$product),levels(data_train$product))

但是，当我尝试对测试数据集运行预测时，出现以下错误：

> fit1<- predict(mytree,newdata=data_test)

Error in checkData(oldData, RET) : 
  Levels in factors of new data do not match original data

我也尝试了以下方法，但它改变了我的测试数据集的字段......：

级别（data_test$Bank）<-级别（data_train$Bank）

data_test 表已更改：

Rate  Bank(altered)  Bank (original)   
2.0    A              A      
0.5    B              D      
0.8    C              E      
2.1    D              C
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2 回答 2

2

您可以尝试使用可比较的水平重建您的因子，而不是为现有因子分配新的水平。这是一个例子：

# start the party
library(party)

# create training data sample
data_train <- data.frame(Rate = c(1.5, 0.6, 3, 2.1, 1.1),
                         Bank = c("A", "B", "C", "D", "E"),
                         Product = c("aaa", "abc", "bac", "cba", "cca"),
                         Salary = c(100000, 60000, 10000, 50000, 80000))

# create testing data sample
data_test <- data.frame(Rate = c(2.0, 0.5, 0.8, 2.1),
                         Bank = c("A", "D", "E", "C"),
                         Product = c("cba", "cca", "cba", "abc"),
                         Salary = c(80000, 250000, 120000, 65000))

# get the union of levels between train and test for Bank and Product
bank_levels <- union(levels(data_test$Bank), levels(data_train$Bank))
product_levels <- union(levels(data_test$Product), levels(data_train$Product))

# rebuild Bank with union of levels
data_test$Bank <- with(data_test, factor(Bank, levels = bank_levels)) 
data_train$Bank <- with(data_train, factor(Bank, levels = bank_levels)) 

# rebuild Product with union of levels
data_test$Product <- with(data_test, factor(Product, levels = product_levels)) 
data_train$Product <- with(data_train, factor(Product, levels = product_levels)) 

# fit the model
mytree <- ctree(Rate ~ Bank + Product + Salary, data = data_train)

# generate predictions
fit1 <- predict(mytree, newdata = data_test)

> fit1
     Rate
[1,] 1.66
[2,] 1.66
[3,] 1.66
[4,] 1.66
于 2016-05-25T18:14:56.783 回答
0

我正在使用 ctree 的示例，但这基本上是巧妙地使用因子，因此可以用于任何严格依赖因子水平的算法（RandomForest 等）

这一切都是为了了解 R 如何存储和使用因子水平。如果我们在训练数据中使用相同的因子水平（并且以相同的顺序）（是的，即使没有与测试数据合并），我们使用预训练的 ctree 模型进行预测。

实际上，使用 ctree (party) 包进行预测不需要俱乐部训练和测试数据。这是因为当您使用预训练模型时，在运行时生产期间您可能没有那么奢侈的内存和处理器能力。预训练模型减轻了我们在生产环境中基于大量训练数据构建模型的负担。

第 1 步：在构建模型时，您可以将每列的因子级别存储在训练数据中（只要适用）

var_list <- colnames(dtrain)
for(var in var_list)
{
  if(class(dtrain[,var]) == 'character')
  {
    print(var)

    #Fill blanks with "None" to keep the factor levels consistent
    dtrain[dtrain[,var] == '',var] <- 'None'

    col_name_levels <- unique(dtrain[,var])

    #Make sure you have sorted the column levels     
    col_name_levels <- sort(col_name_levels, decreasing = FALSE)

    #Make as factors
    dtrain[,var] <- factor(dtrain[,var], levels = col_name_levels, ordered=TRUE)

    print(levels(dtrain[,var]))

    #This is the trick: Store the exact levels in a CSV which is much easier to load than the whole train data later in prediction phase    
    write.csv(levels(dtrain[,var]), paste0(getwd(),'/Output CSVs/',var,'_levels.csv'), row.names = FALSE)
  }
}


# also store the column names and data types for detecting later
for(col_name in colnames(dtrain))
{
  abc <- data.frame('col_name' = col_name,'class_colname' = paste(class(dtrain[,col_name]), collapse = ' '))

  if(!exists('col_name_type_list'))
  {
    col_name_type_list <- abc
  }else
  {
    col_name_type_list <- rbind(col_name_type_list, abc)
  }
}

#Store for checking later
write.csv(col_name_type_list, filepath, row.names = FALSE)

然后在预测阶段（在生产环境中），只需读取测试数据中每一列的这些级别，丢弃具有新数据的行（ctree 无论如何都无法预测它们），然后使用这些行进行预测。

###############Now in test prediction ###########################


#Read the column list of train data (stored earlier)
col_name_type_list_dtrain <- read.csv( filepath, header = TRUE)


for(i in 1:nrow(col_name_type_list_dtrain))
{
  col_name <- col_name_type_list_dtrain[i,]$col_name
  class_colname <- col_name_type_list_dtrain[i,]$class_colname

  if(class_colname == 'numeric')
  {
    dtest[,col_name] <- as.numeric(dtest[,col_name])
  }

  if(class_colname == 'ordered factor')
  {

    #Now use the column factor levels from train
    remove(col_name_levels)
    col_name_levels <- read.csv( paste0(getwd(),'/Output CSVs/',var,'_levels.csv'), header = TRUE)
    factor_check_flag <- TRUE

    col_name_levels <- as.character(col_name_levels$x)
    print(col_name)
    print('Pre-Existing levels detected')
    print(NROW(col_name_levels))

    #Drop new rows which are not in train; the model cant predict for them
    rows_before_dropping <- nrow(dtest)
    print('Adjusting levels to train......')
    dtest <- dtest[dtest[,col_name] %in% col_name_levels,]
    rows_after_dropping <- nrow(dtest)

    cat('\nDropped Rows for adjusting ',col_name,': ',(rows_before_dropping - rows_after_dropping),'\n')

    #Convert to factors
    dtest[,col_name] <- factor(dtest[,col_name], levels=col_name_levels, ordered=TRUE)

    print(dtest[,col_name])
  }
}
于 2019-04-29T09:39:27.833 回答