我试图了解在使用交叉验证训练模型时应该在哪里发生 SMOTE。我知道所有的预处理步骤都应该针对交叉验证的每一折进行。那么这是否意味着以下两个设置是相同的并且理论上是正确的?
设置 1:使用配方进行预处理,在 trainControl 中进行打击
set.seed(888, sample.kind = "Rounding")
tr_ctrl <- trainControl(summaryFunction = twoClassSummary,
verboseIter = TRUE,
savePredictions = TRUE,
sampling = "smote",
method = "repeatedCV",
number= 2,
repeats = 0,
classProbs = TRUE,
allowParallel = TRUE,
)
cw_smote_recipe <- recipe(husb_beat ~ ., data = nfhs_train) %>%
step_nzv(all_predictors()) %>%
step_naomit(all_predictors()) %>%
step_dummy(all_nominal(), -husb_beat) %>%
step_interact(~starts_with("State"):starts_with("wave"))%>%
step_interact(~starts_with("husb_drink"):starts_with("husb_legal"))
cw_logit1 <- train(cw_smote_recipe, data = nfhs_train,
method = "glm",
family = 'binomial',
metric = "ROC",
trControl = tr_ctrl)
设置 2:使用食谱进行预处理和打击:这是否会在每个 CV 折叠中进行?
set.seed(888, sample.kind = "Rounding")
tr_ctrl <- trainControl(summaryFunction = twoClassSummary,
verboseIter = TRUE,
savePredictions = TRUE,
#sampling = "smote", ## NO LONGER WITHIN TRAINCONTROL
method = "repeatedCV",
number= 2,
repeats = 0,
classProbs = TRUE,
allowParallel = TRUE,
)
smote_recipe <- recipe(husb_beat ~ ., data = nfhs_train) %>%
step_nzv(all_predictors()) %>%
step_naomit(all_predictors()) %>%
step_dummy(all_nominal(), -husb_beat) %>%
step_interact(~starts_with("State"):starts_with("wave"))%>%
step_interact(~starts_with("husb_drink"):starts_with("husb_legal"))%>%
step_smote(husb_beat) ## NEW STEP TO RECIPE
cw_logit2 <- train(smote_recipe, data = nfhs_train,
method = "glm",
family = 'binomial',
metric = "ROC",
trControl = tr_ctrl)
蒂亚!