不一致的“最佳调整”和“跨调整参数重新采样结果”插入符号 R 包

Question

我正在尝试使用带有调谐网格的 Caret 创建模型

svmGrid

然后再次使用此网格的子集：

svmGrid

问题是我得到了不同的“最佳调谐”和“跨调谐参数的重新采样结果”，尽管为第一个调谐网格选择的 C 参数值也出现在第二个调谐网格中。

在对采样参数使用不同选项以及在 trainControl() 中使用不同的 summaryFunction 选项时，我也会遇到这些差异

不用说，由于每次都选择不同的最佳模型，它会影响测试集上的预测结果。

有人知道为什么会这样吗？

可重现的数据集：

library(caret)
library(doMC)
registerDoMC(cores = 8)

set.seed(2969)
imbal_train <- twoClassSim(100, intercept = -20, linearVars = 20)
imbal_test  <- twoClassSim(100, intercept = -20, linearVars = 20)
table(imbal_train$Class)

使用第一个曲调网格运行

svmGrid <-  expand.grid(C = c(0.0001,0.001,0.01,0.1,1,10,20,30,40,50,100))

up_fitControl = trainControl(method = "cv", number = 10 , savePredictions = TRUE, allowParallel = TRUE, sampling = "up", seeds = NA)


set.seed(5627)
up_inside <- train(Class ~ ., data = imbal_train,
                   method = "svmLinear",
                   trControl = up_fitControl,
                   tuneGrid = svmGrid,
                   scale = FALSE)

up_inside

首次运行输出：

> up_inside
Support Vector Machines with Linear Kernel 

100 samples
 25 predictors
  2 classes: 'Class1', 'Class2' 

No pre-processing
Resampling: Cross-Validated (10 fold) 
Summary of sample sizes: 90, 91, 90, 90, 89, 90, ... 
Addtional sampling using up-sampling

Resampling results across tuning parameters:

  C      Accuracy   Kappa         Accuracy SD  Kappa SD 
  1e-04  0.7734343   0.252201364  0.1227632    0.3198165
  1e-03  0.8225253   0.396439198  0.1245455    0.3626456
  1e-02  0.7595960   0.116150973  0.1431780    0.3046825
  1e-01  0.7686869   0.051430454  0.1167093    0.2712062
  1e+00  0.7695960  -0.004261294  0.1162279    0.2190151
  1e+01  0.7093939   0.111852756  0.2030250    0.3810059
  2e+01  0.7195960   0.040458804  0.1932690    0.2580560
  3e+01  0.7195960   0.040458804  0.1932690    0.2580560
  4e+01  0.7195960   0.040458804  0.1932690    0.2580560
  5e+01  0.7195960   0.040458804  0.1932690    0.2580560
  1e+02  0.7195960   0.040458804  0.1932690    0.2580560

Accuracy was used to select the optimal model using  the largest value.
The final value used for the model was C = 0.001. 

使用第二个曲调网格运行

svmGrid <-  expand.grid(C = c(0.0001,0.001,0.01,0.1,1,10,20,30,40,50))

up_fitControl = trainControl(method = "cv", number = 10 , savePredictions = TRUE, allowParallel = TRUE, sampling = "up", seeds = NA)


set.seed(5627)
up_inside <- train(Class ~ ., data = imbal_train,
                   method = "svmLinear",
                   trControl = up_fitControl,
                   tuneGrid = svmGrid,
                   scale = FALSE)

up_inside

第二次运行输出：

> up_inside
Support Vector Machines with Linear Kernel 

100 samples
 25 predictors
  2 classes: 'Class1', 'Class2' 

No pre-processing
Resampling: Cross-Validated (10 fold) 
Summary of sample sizes: 90, 91, 90, 90, 89, 90, ... 
Addtional sampling using up-sampling

Resampling results across tuning parameters:

  C      Accuracy   Kappa         Accuracy SD  Kappa SD 
  1e-04  0.8125253   0.392165694  0.13043060   0.3694786
  1e-03  0.8114141   0.375569633  0.12291273   0.3549978
  1e-02  0.7995960   0.205413345  0.06734882   0.2662161
  1e-01  0.7495960   0.017139266  0.09742161   0.2270128
  1e+00  0.7695960  -0.004261294  0.11622791   0.2190151
  1e+01  0.7093939   0.111852756  0.20302503   0.3810059
  2e+01  0.7195960   0.040458804  0.19326904   0.2580560
  3e+01  0.7195960   0.040458804  0.19326904   0.2580560
  4e+01  0.7195960   0.040458804  0.19326904   0.2580560
  5e+01  0.7195960   0.040458804  0.19326904   0.2580560

Accuracy was used to select the optimal model using  the largest value.
The final value used for the model was C = 1e-04.

Answer 1

如果您不在caret 中提供种子，它会为您选择种子。由于网格的长度不同，因此折叠的种子会略有不同。

下面，我粘贴了示例，我刚刚重命名了您的第二个模型，以便更容易获得比较的输出：

> up_inside$control$seeds[[1]]
 [1] 825016 802597 128276 935565 324036 188187 284067  58853 923008 995461  60759
> up_inside2$control$seeds[[1]]
 [1] 825016 802597 128276 935565 324036 188187 284067  58853 923008 995461
> up_inside$control$seeds[[2]]
 [1] 966837 256990 592077 291736 615683 390075 967327 349693  73789 155739 916233
# See how the first seed here is the same as the last seed of the first model
> up_inside2$control$seeds[[2]]
 [1]  60759 966837 256990 592077 291736 615683 390075 967327 349693  73789

如果您现在继续设置自己的种子，您将获得相同的输出：

# Seeds for your first train
myseeds <- list(c(1:10,1000), c(11:20,2000), c(21:30, 3000),c(31:40, 4000),c(41:50, 5000),
                c(51:60, 6000),c(61:70, 7000),c(71:80, 8000),c(81:90, 9000),c(91:100, 10000), c(343))
# Seeds for your second train
myseeds2 <- list(c(1:10), c(11:20), c(21:30),c(31:40),c(41:50),c(51:60),
                 c(61:70),c(71:80),c(81:90),c(91:100), c(343))

> up_inside
Support Vector Machines with Linear Kernel 

100 samples
 25 predictor
  2 classes: 'Class1', 'Class2' 

No pre-processing
Resampling: Cross-Validated (10 fold) 
Summary of sample sizes: 90, 91, 90, 90, 89, 90, ... 
Addtional sampling using up-sampling

Resampling results across tuning parameters:

  C      Accuracy   Kappa      
  1e-04  0.7714141  0.239823027
  1e-03  0.7914141  0.332834590
  1e-02  0.7695960  0.207000745
  1e-01  0.7786869  0.103957926
  1e+00  0.7795960  0.006849817
  1e+01  0.7093939  0.111852756
  2e+01  0.7195960  0.040458804
  3e+01  0.7195960  0.040458804
  4e+01  0.7195960  0.040458804
  5e+01  0.7195960  0.040458804
  1e+02  0.7195960  0.040458804

Accuracy was used to select the optimal model using  the largest value.
The final value used for the model was C = 0.001. 
> up_inside2
Support Vector Machines with Linear Kernel 

100 samples
 25 predictor
  2 classes: 'Class1', 'Class2' 

No pre-processing
Resampling: Cross-Validated (10 fold) 
Summary of sample sizes: 90, 91, 90, 90, 89, 90, ... 
Addtional sampling using up-sampling

Resampling results across tuning parameters:

  C      Accuracy   Kappa      
  1e-04  0.7714141  0.239823027
  1e-03  0.7914141  0.332834590
  1e-02  0.7695960  0.207000745
  1e-01  0.7786869  0.103957926
  1e+00  0.7795960  0.006849817
  1e+01  0.7093939  0.111852756
  2e+01  0.7195960  0.040458804
  3e+01  0.7195960  0.040458804
  4e+01  0.7195960  0.040458804
  5e+01  0.7195960  0.040458804

Accuracy was used to select the optimal model using  the largest value.
The final value used for the model was C = 0.001.