library(tidymodels)
regr-tree02
statlearning
trees
tidymodels
mtcars
string
Aufgabe
Berechnen Sie einfaches Prognosemodell auf Basis eines Entscheidungsbaums!
Modellformel: am ~ .
(Datensatz mtcars
)
Berichten Sie die Modellgüte (ROC-AUC).
Hinweise:
- Tunen Sie alle Parameter (die der Engine anbietet).
- Führen Sie eine \(v=2\)-fache Kreuzvalidierung durch (weil die Stichprobe so klein ist).
- Beachten Sie die üblichen Hinweise.
Lösung
Setup
library(tidymodels)
data(mtcars)
library(tictoc) # Zeitmessung
Für Klassifikation verlangt Tidymodels eine nominale AV, keine numerische:
<-
mtcars %>%
mtcars mutate(am = factor(am))
Daten teilen
<- initial_split(mtcars)
d_split <- training(d_split)
d_train <- testing(d_split) d_test
Modell(e)
<-
mod_tree decision_tree(mode = "classification",
cost_complexity = tune(),
tree_depth = tune(),
min_n = tune())
Rezept(e)
<-
rec1 recipe(am ~ ., data = d_train)
Resampling
<- vfold_cv(d_train, v = 2) rsmpl
Workflow
<-
wf1 workflow() %>%
add_recipe(rec1) %>%
add_model(mod_tree)
Tuning/Fitting
<-
fit1 tune_grid(object = wf1,
metrics = metric_set(roc_auc),
resamples = rsmpl)
Bester Kandidat
autoplot(fit1)
show_best(fit1)
# A tibble: 5 × 9
cost_complexity tree_depth min_n .metric .estimator mean n std_err
<dbl> <int> <int> <chr> <chr> <dbl> <int> <dbl>
1 1.07e- 8 4 2 roc_auc binary 0.764 2 0.0139
2 9.57e- 4 10 10 roc_auc binary 0.764 2 0.0139
3 1.25e-10 14 20 roc_auc binary 0.764 2 0.0139
4 6.83e- 2 11 17 roc_auc binary 0.764 2 0.0139
5 3.01e- 5 3 7 roc_auc binary 0.764 2 0.0139
# ℹ 1 more variable: .config <chr>
Finalisieren
<-
wf1_finalized %>%
wf1 finalize_workflow(select_best(fit1))
Last Fit
<-
final_fit last_fit(object = wf1_finalized, d_split)
collect_metrics(final_fit)
# A tibble: 3 × 4
.metric .estimator .estimate .config
<chr> <chr> <dbl> <chr>
1 accuracy binary 0.75 Preprocessor1_Model1
2 roc_auc binary 0.75 Preprocessor1_Model1
3 brier_class binary 0.25 Preprocessor1_Model1
Categories:
- statlearning
- trees
- tidymodels
- string