"Wrong model type for classification" 在 R-Caret 的回归问题中

Question

我正在尝试使用 R 中 Caret 包中的各种预测算法来解决回归问题，即我的目标变量是连续的。 Caret 认为 classification 是问题的适当 class，当我通过任何回归模型时，我收到一条错误消息，指出 "wrong model type for classification"。为了再现性，让我们看看 Combined Cycle Power Plant Data Set。数据在 CCPP.zip 中。让我们预测功率作为其他变量的函数。功率是一个连续变量。

  library(readxl)
  library(caret)
  power_plant = read_excel("Folds5x2_pp.xlsx")
  apply(power_plant,2, class)   # shows all columns are numeric

  control <- trainControl(method="repeatedcv", number=10, repeats=5)

  my_glm <- train(power_plant[,1:4], power_plant[,5],
           method = "lm",
           preProc = c("center", "scale"),
            trControl = control)

下图是我的截图：

Answer 1

出于某种原因，caret 被 tibbles 弄糊涂了，tibbles 是 read_excel returns 数据框的 tidyverse 变体。通过在将其提供给插入符之前将其转换为简单的数据框，一切正常：

library(readxl)
library(caret)
power_plant = read_excel("Folds5x2_pp.xlsx")
apply(power_plant,2, class)   # shows all columns are numeric

power_plant <- data.frame(power_plant)
control <- trainControl(method="repeatedcv", number=10, repeats=5)

my_glm <- train(power_plant[,1:4], power_plant[,5],
                method = "lm",
                preProc = c("center", "scale"),
                trControl = control)

my_glm

产量：

Linear Regression 

9568 samples
   4 predictor

Pre-processing: centered (4), scaled (4) 
Resampling: Cross-Validated (10 fold, repeated 5 times) 
Summary of sample sizes: 8612, 8612, 8611, 8612, 8612, 8610, ... 
Resampling results:

  RMSE      Rsquared 
  4.556703  0.9287933

Tuning parameter 'intercept' was held constant at a value of TRUE

Answer 2

当我尝试使用公式 = y ~ x 时出现类似的错误，只需省略命名变量并使用 y ~ x 就可以很好地工作。

"Wrong model type for classification" 在 R-Caret 的回归问题中

"Wrong model type for classification" in regression problems in R-Caret

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r-caret