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[英]Inconsistent “best tune” and “Resampling results across tuning parameters” caret R package
[英]R caret inconsistent results in model tuning
今天,使用插入符號包進行模型調整時,我遇到了一個奇怪的問題:給定調整參數T *的特定組合,如果單獨評估T *或將其作為網格的一部分,則與T *相關的度量(即Cohen K)值就會改變可能的組合。 在下面的實際示例中,將插入符號用於與gbm軟件包進行交互。
# Load libraries and data
library (caret)
data<-read.csv("mydata.csv")
data$target<-as.factor(data$target)
# data are available at https://www.dropbox.com/s/1bglmqd14g840j1/mydata.csv?dl=0
方法1:單獨評估T *
#Define 5-fold cv as validation settings
fitControl <- trainControl(method = "cv",number = 5)
# Define the combination of tuning parameter for this example T*
gbmGrid <- expand.grid(.interaction.depth = 1,
.n.trees = 1000,
.shrinkage = 0.1, .n.minobsinnode=1)
# Fit a gbm with T* as model parameters and K as scoring metric.
set.seed(825)
gbmFit1 <- train(target ~ ., data = data,
method = "gbm",
distribution="adaboost",
trControl = fitControl,
tuneGrid=gbmGrid,
verbose=F,
metric="Kappa")
# The results show that T* is associated with Kappa = 0.47. Remember this result and the confusion matrix.
testPred<-predict(gbmFit1, newdata = data)
confusionMatrix(testPred, data$target)
# output selection
Confusion Matrix and Statistics
Reference
Prediction 0 1
0 832 34
1 0 16
Kappa : 0.4703
步驟2:連同其他調整配置文件一起評估T *
除了考慮調整參數{T}的幾種組合外,這里的一切與過程1相同:
# Notice that the original T* is included in {T}!!
gbmGrid2 <- expand.grid(.interaction.depth = 1,
.n.trees = seq(100,1000,by=100),
.shrinkage = 0.1, .n.minobsinnode=1)
# Fit the gbm
set.seed(825)
gbmFit2 <- train(target ~ ., data = data,
method = "gbm",
distribution="adaboost",
trControl = fitControl,
tuneGrid=gbmGrid2,
verbose=F,
metric="Kappa")
# Caret should pick the model with the highest Kappa.
# Since T* is in {T} I would expect the best model to have K >= 0.47
testPred<-predict(gbmFit2, newdata = data)
confusionMatrix(testPred, data$target)
# output selection
Reference
Prediction 0 1
0 831 47
1 1 3
Kappa : 0.1036
結果與我的預期不一致:{T}中的最佳模型得分K = 0.10。 假設T *的K = 0.47並包含在{T}中,怎么可能? 此外,根據下圖,在步驟2中評估的T *的K現在約為0.01。 關於發生了什么的任何想法? 我想念什么嗎?
我從您的數據和代碼中獲得了一致的重采樣結果。
第一個模型的Kappa = 0.00943
gbmFit1$results
interaction.depth n.trees shrinkage n.minobsinnode Accuracy Kappa AccuracySD
1 1 1000 0.1 1 0.9331022 0.009430576 0.004819004
KappaSD
1 0.0589132
對於n.trees = 1000
,第二個模型具有相同的結果
gbmFit2$results
shrinkage interaction.depth n.minobsinnode n.trees Accuracy Kappa AccuracySD
1 0.1 1 1 100 0.9421803 -0.002075765 0.002422952
2 0.1 1 1 200 0.9387776 -0.008326896 0.002468351
3 0.1 1 1 300 0.9365049 -0.012187900 0.002625886
4 0.1 1 1 400 0.9353749 -0.013950906 0.003077431
5 0.1 1 1 500 0.9353685 -0.013961221 0.003244201
6 0.1 1 1 600 0.9342322 -0.015486214 0.005202656
7 0.1 1 1 700 0.9319658 -0.018574633 0.007033402
8 0.1 1 1 800 0.9319658 -0.018574633 0.007033402
9 0.1 1 1 900 0.9342386 0.010955568 0.003144850
10 0.1 1 1 1000 0.9331022 0.009430576 0.004819004
KappaSD
1 0.004641553
2 0.004654972
3 0.003978702
4 0.004837097
5 0.004878259
6 0.007469843
7 0.009470466
8 0.009470466
9 0.057825336
10 0.058913202
請注意,第二次運行的最佳模型的n.trees = 900
gbmFit2$bestTune
n.trees interaction.depth shrinkage n.minobsinnode
9 900 1 0.1 1
由於train
根據您的指標選擇了“最佳”模型,因此您的第二個預測是使用其他模型(n.trees為900而不是1000)。
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