Question

我想在R中的Caret包中使用相互作用项进行强大的线性回归，但我得到以下错误：

train.default（x，y，weights = w，...）出错：停止进入另外：警告信息：在nominalTrainWorkflow中（x = x，y = y，wts = weights，info = trainInfo，：重新采样中缺少值绩效衡量标准。

在我的代码下面：

mod <- train(
Pac ~ clearSkyPOA + clearSkyPOA*TotalCover+Temp2,
data = training,
method = "rlm",
metric = "RMSE",
preProc= c("center","scale","BoxCox"),
trControl =  trainControl(method="cv", number = 5),
na.action =na.omit)

如果我删除了互动字词：＆＃39; clearSkyPOA * TotalCover＆＃39;，它按预期工作。例如，使用代码：

mod <- train(
    Pac ~ clearSkyPOA + TotalCover+Temp2,
    data = training,
    method = "rlm",
    metric = "RMSE",
    preProc= c("center","scale","BoxCox"),
    trControl =  trainControl(method="cv", number = 5),
    na.action=na.omit
  )

我获得了以下结果：

Robust Linear Model 

4363 samples
   3 predictor

Pre-processing: centered (3), scaled (3), Box-Cox transformation (2) 
Resampling: Cross-Validated (5 fold) 
Summary of sample sizes: 3490, 3490, 3491, 3491, 3490 
Resampling results across tuning parameters:

  intercept  psi           RMSE      Rsquared 
  FALSE      psi.huber     291.3261  0.7501889
  FALSE      psi.hampel    291.3261  0.7501889
  FALSE      psi.bisquare  291.3470  0.7499932
   TRUE      psi.huber     115.0178  0.7488397
   TRUE      psi.hampel    114.2018  0.7500523
   TRUE      psi.bisquare  115.4231  0.7483018

RMSE was used to select the optimal model using  the smallest value.
The final values used for the model were intercept = TRUE and psi = psi.hampel.

我错过了什么吗？以下是来自dput（training）的20个样本的结果：

structure(list(Pac = c(3.42857142857143, 38.25, 120.916666666667, 
258, 367.166666666667, 269.083333333333, 233.75, 112.416666666667, 
21.9166666666667, 0.2, 1.5, 12.4166666666667, 134.916666666667, 
104.333333333333, 394.583333333333, 342.5, 303.333333333333, 
151.5, 42.0833333333333, 4.83333333333333), clearSkyPOA = c(63.0465796511235, 
230.023517163135, 472.935466225438, 646.271261971453, 739.926063392829, 
751.872076941902, 681.91937141018, 531.40317803238, 306.020562749019, 
120.318359249055, 68.2689523552881, 229.800769386719, 473.162397232603, 
647.082096293271, 741.364282016807, 753.955817698295, 684.656233771643, 
534.787114500355, 309.953073794329, 114.55351678131), TotalCover = c(0.602923, 
0.5798824, 0.5095124, 0.3896642, 0.2744389, 0.232004, 0.3052016, 
0.4355463, 0.5392107, 0.5571411, 0.4599758, 0.4555472, 0.4434351, 
0.41583, 0.3704268, 0.306295, 0.2271317, 0.1551105, 0.1170307, 
0.1307881), Temp = c(13.72545, 13.91255, 14.04348, 14.06298, 
13.98118, 13.82455, 13.61805, 13.3806, 13.12966, 12.87026, 12.37558, 
12.76012, 13.12112, 13.37877, 13.5505, 13.67806, 13.7903, 13.86462, 
13.86556, 13.76468), Temp2 = c(188.3879777025, 193.5590475025, 
197.2193305104, 197.7674064804, 195.4733941924, 191.1181827025, 
185.4512858025, 179.04045636, 172.3879717156, 165.6435924676, 
153.1549803364, 162.8206624144, 172.1637900544, 178.9914867129, 
183.61605025, 187.0893253636, 190.17237409, 192.2276877444, 192.2537541136, 
189.4664155024)), .Names = c("Pac", "clearSkyPOA", "TotalCover", 
"Temp", "Temp2"), row.names = c(NA, 20L), class = "data.frame")

R中的Caret封装的鲁棒线性回归

0 个答案: