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在指定要在 caret::train 中使用的配方后，我正在尝试预测新样本。我对此有几个问题，因为我在插入符号/食谱文档中找不到。

  1. 我应该使用 predict() 还是 predict.train()？有什么不同？
  2. 在使用 predict 之前，我应该先用准备好的配方烘焙测试数据吗？在 train() 中直接使用 preProcess 时，建议不要预处理新数据，因为 train 对象会自动执行此操作。使用食谱时也一样吗？

下面是一个可重现的示例，说明了我的过程以及使用 predict 与 predict.train 时的预测差异

library(recipes)
library(caret)
# Data ----
data("credit_data")

credit_train <- credit_data[1:3500,]
credit_test <- credit_data[-(1:3500),]

# Set up recipe ----

set.seed(0)
Rec.Obj = recipe(Status ~ ., data = credit_train) %>%
    step_knnimpute(all_predictors()) %>% 
    step_center(all_numeric())%>%
    step_scale(all_numeric())

# Control parameters ----
set.seed(0)
TC = trainControl("cv",number = 10, savePredictions = "final", classProbs = TRUE, returnResamp = "final")


set.seed(0)
Model.Output = train(Rec.Obj,
                     credit_train,
                     trControl = TC,
                     tuneLength = 1,
                     metric = "Accuracy",
                     method = "glm")

# Preped recipe ----
set.seed(0)
prep.rec <- 
    prep(Rec.Obj, newdata = credit_train)

# Baked data for observation ----
set.seed(0)
bake.train <- bake(prep.rec, new_data = credit_train)
bake.test <- bake(prep.rec, new_data = credit_test)

# investigation of prediction methods ----

# no application of recipe to newdata
set.seed(0)
predict.norm = predict(Model.Output, credit_test, type = "raw")
predict.train = predict.train(Model.Output, credit_test,  type = "raw")

identical(predict.norm,predict.train)
# evaluates to FALSE

# Apply recipe to new data (bake.test)
predict.norm.baked = predict(Model.Output, bake.test, type = "raw")
predict.train.baked = predict.train(Model.Output, bake.test, type = "raw")

identical(predict.norm.baked, predict.train.baked)
# evaluates to FALSE

# Comparison of both predict() funcs
identical(predict.norm, predict.norm.baked)
# evaluates to FALSE
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1 回答 1

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配方嵌入到train对象中。答案不同有两个原因：

  1. 由于您正在为配方（内部Model.Output）提供要重新处理的已处理数据。你不应该提供predict()烘焙数据；只需使用predict()并给它原始测试集..

  2. 让 S3 做它的事：predict.train用于 x/y 接口和predict.train.recipe配方接口。只需使用predict()就会做适当的事情。

于 2019-03-20T21:44:51.470 回答