Cannot run ANOVA to Compare Random Forest Models

Question

I am using tidymodels to fit multiple Random Forest models. I then followed along with this tutorial to compare the model results. The problem is that I get the error: Error in

 UseMethod("anova") : 
  no applicable method for 'anova' applied to an object of class "ranger"

As an example:

set.seed(123)
iris <- iris %>% mutate(
  is_versicolor = ifelse(Species == "versicolor", "versicolor", "not_versicolor")) %>%
  mutate(is_versicolor = factor(is_versicolor, levels = c("versicolor", "not_versicolor")))

iris_split <- initial_split(iris, strata = is_versicolor, prop = 0.8)
iris_train <- training(iris_split)
iris_test  <- testing(iris_split)

rec_normal <- recipe(is_versicolor ~ Petal.Width + Species, data = iris_train)
rec_interaction <- rec_normal %>% 
  step_interact(~ Petal.Width:starts_with("Species"))

iris_model <- rand_forest() %>% set_engine("ranger") %>% set_mode("classification")

# normal workflow
iris_wf <- workflow() %>% 
  add_model(iris_model) %>% 
  add_recipe(rec_normal)

# interaction workflow
iris_wf_interaction <- iris_wf %>% 
  update_recipe(rec_interaction)

# fit models
iris_normal_lf <- last_fit(iris_wf, split = iris_split)
iris_inter_lf <- last_fit(iris_wf_interaction, split = iris_split)

normalmodel <- iris_normal_lf %>% extract_fit_engine()
intermodel  <- iris_inter_lf %>% extract_fit_engine()

anova(normalmodel, intermodel) %>% tidy()

How can I run an ANOVA or ANOVA-type comparison of these models, to see if one is significantly better?

Answer 1

Using a different model pair, and comparing models based on classification accuracy using resamples. Easily extended to other metrics.

library(dplyr)
library(tibble)
library(ggplot2)
library(tidyr)
library(rsample)
library(recipes)
library(parsnip)
library(workflows)
library(tune)
library(yardstick)
library(workflowsets)

set.seed(123)
iris <- iris %>% mutate(
  is_versicolor = ifelse(Species == "versicolor", "versicolor", "not_versicolor")) %>%
  mutate(is_versicolor = factor(is_versicolor, levels = c("versicolor", "not_versicolor")))

iris_split <- initial_split(iris, strata = is_versicolor, prop = 0.8)
iris_train <- training(iris_split)
iris_test  <- testing(iris_split)

# replacing normal and interaction recipes with models
# that give less than 100% accuracy.
rec_normal <- recipe(is_versicolor ~ Sepal.Width, data = iris_train)
rec_alternative <- recipe(is_versicolor ~ Sepal.Length, data = iris_train)

iris_model <- rand_forest() %>% set_engine("ranger") %>% set_mode("classification")

# Create folds
set.seed(234)
iris_folds <- vfold_cv(iris_train)
iris_folds

# Combine models into set
iris_set <-
  workflow_set(
    list(rec_normal, rec_alternative),
    list(iris_model),
    cross = TRUE
  )

doParallel::registerDoParallel()
set.seed(2021)

# fit models
iris_rs <-
  workflow_map(
    iris_set,
    "fit_resamples",
    resamples = iris_folds
  )

# Visualise model performance
autoplot(iris_rs)

# Extract resample accuracies
model_1_rs <- iris_rs[1,][[4]][[1]]$.metrics
model_2_rs <- iris_rs[2,][[4]][[1]]$.metrics
model_acc <- tibble(model_1 = NA, model_2 = NA)
for (i in 1:10) {
  model_acc[i, 1] <- model_1_rs[[i]][[".estimate"]][1]
  model_acc[i, 2] <- model_2_rs[[i]][[".estimate"]][1]
}
model_acc <- model_acc |> pivot_longer(cols = everything(), names_to = "model", values_to = "acc")

# Do ANOVA
aov_results <- aov(acc ~ model, data = model_acc)
summary(aov_results)
ggplot(data = model_acc, aes(fill = model)) +
  geom_density(aes(x = acc, alpha = 0.2)) +
  labs(x = "accuracy")

Giving the p values:

> summary(aov_results)

            Df Sum Sq Mean Sq F value Pr(>F)

model        1 0.0281 0.02813   1.378  0.256

Residuals   18 0.3674 0.02041

Looking at the p values of the model accuracies using a different lens: First visualise the variation:

model_acc |> ggplot(aes(x = model, y = acc)) +
  geom_boxplot() +
  labs(y = 'accuracy')

Then calculate a test statistic:

observed_statistic <- model_acc %>%
  specify(acc ~ model) %>%
  calculate(stat = "diff in means", order = c("model_1", "model_2"))

observed_statistic

Then do a simulation of the distribution:

null_dist_2_sample <- model_acc %>%
  specify(acc ~ model) %>%
  hypothesize(null = "independence") %>%
  generate(reps = 1000, type = "permute") %>%
  calculate(stat = "diff in means" ,order = c("model_1", "model_2"))

and plot:

null_dist_2_sample %>%
  visualize() + 
  shade_p_value(observed_statistic,
                direction = "two-sided") +
  labs(x = "test statistic")

and get the p value:

p_value_2_sample <- null_dist_2_sample %>%
  get_p_value(obs_stat = observed_statistic,
              direction = "two-sided")

p_value_2_sample

# A tibble: 1 × 1
  p_value
    <dbl>
1   0.228

Which is almost the same as the p value from the aov. Note that consistent with the accuracies of the two models being close, the p value is high.

Answer 2

You could compare your random forest models by comparing their accuracies using the aov function. First, you can collect the accuracy with collect_metrics and save them in a data frame to run a model with aov to get the results. Here is a reproducible example:

library(tidymodels)
set.seed(123)
iris <- iris %>% mutate(
  is_versicolor = ifelse(Species == "versicolor", "versicolor", "not_versicolor")) %>%
  mutate(is_versicolor = factor(is_versicolor, levels = c("versicolor", "not_versicolor")))

iris_split <- initial_split(iris, strata = is_versicolor, prop = 0.8)
iris_train <- training(iris_split)
iris_test  <- testing(iris_split)

rec_normal <- recipe(is_versicolor ~ Petal.Width + Species, data = iris_train)
rec_interaction <- rec_normal %>% 
  step_interact(~ Petal.Width:starts_with("Species"))

iris_model <- rand_forest() %>% set_engine("ranger") %>% set_mode("classification")

# normal workflow
iris_wf <- workflow() %>% 
  add_model(iris_model) %>% 
  add_recipe(rec_normal)

# interaction workflow
iris_wf_interaction <- iris_wf %>% 
  update_recipe(rec_interaction)

# fit models
iris_normal_lf <- last_fit(iris_wf, split = iris_split)
iris_inter_lf <- last_fit(iris_wf_interaction, split = iris_split)
#> ! train/test split: preprocessor 1/1: Categorical variables used in `step_interact` should probably be avoided...

normalmodel <- iris_normal_lf %>% extract_fit_engine()
intermodel  <- iris_inter_lf %>% extract_fit_engine()

# Check confusion matrix
iris_normal_lf %>%
  collect_predictions() %>% 
  conf_mat(is_versicolor, .pred_class) 
#>                 Truth
#> Prediction       versicolor not_versicolor
#>   versicolor             10              0
#>   not_versicolor          0             20

iris_inter_lf %>%
  collect_predictions() %>% 
  conf_mat(is_versicolor, .pred_class) 
#>                 Truth
#> Prediction       versicolor not_versicolor
#>   versicolor             10              0
#>   not_versicolor          0             20

# Extract accuracy of models and create dataset
acc_normalmodel <- iris_normal_lf %>% collect_metrics() %>% select(.estimate) %>% slice(1)
acc_intermodel <- iris_normal_lf %>% collect_metrics() %>% select(.estimate) %>% slice(1)
results = data.frame(model = c("normalmodel", "intermodel"),
                     accuracy = c(acc_normalmodel$.estimate, acc_intermodel$.estimate))

# perform ANOVA on the classification accuracy
aov_results <- aov(accuracy ~ model, data = results)
summary(aov_results)
#>             Df   Sum Sq  Mean Sq
#> model        1 4.93e-32 4.93e-32

^{Created on 2022-12-15 with reprex v2.0.2}

As you can see the results doesn't show a p-value, because the degree of freedom is to low (why do I not get a p-value from this anova in r)

---

You could also use the aov on the predictions of both models and compare these performance. Here is a reproducible example:

# Get predictions of both models for not_versicolor
normalmodel_pred<-as.data.frame(normalmodel$predictions)$not_versicolor
intermodel_pred<-as.data.frame(intermodel$predictions)$not_versicolor

summary(aov(normalmodel_pred~intermodel_pred))
#>                  Df Sum Sq Mean Sq F value Pr(>F)    
#> intermodel_pred   1 25.032  25.032    9392 <2e-16 ***
#> Residuals       118  0.314   0.003                   
#> ---
#> Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

^{Created on 2022-12-17 with reprex v2.0.2}

As you can see the p-value is less than 0.05 which suggests that there is a difference between the predictions of the models, which is right if you look at the probabilities of the predictions.

---

More information about ANOVA check this:

• Chapter 7 Understanding ANOVA in R

Answer 3

Just using your code, and adapting Julia Silge's blog on workflowsets:

Predict #TidyTuesday giant pumpkin weights with workflowsets

As ANOVA is not available for ranger, instead generate folds to resample:

set. Seed(234)
iris_folds <- vfold_cv(iris_train)
iris_folds

Combine your recipes into a workflowset:

iris_set <-
  workflow_set(
    list(rec_normal, rec_interaction),
    list(iris_model),
    cross = TRUE
  )

iris_set

Setup parallel processing:

doParallel::registerDoParallel()
set. Seed(2021)

Fit using the folds:

iris_rs <-
  workflow_map(
    iris_set,
    "fit_resamples",
    resamples = iris_folds
  )

autoplot(iris_rs)

This chart would usually address your question of how to compare models.

As "species" is on the righthand side of both recipe formulas, and the response "is_versicolor" is calculated from species, the models are completely accurate.

Finish off the output:

collect_metrics(iris_rs)

final_fit <-
  extract_workflow(iris_rs, "recipe_2_rand_forest") %>%
  fit(iris_train)

There is no tidier for ranger models.

In your code, if you change to:

rec_normal <- recipe(is_versicolor ~ Sepal.Length + Sepal.Width, data = iris_train)
rec_interaction <- recipe(is_versicolor ~ Petal.Width + Petal.Length, data = iris_train)

you can have some fun!

Hope this helps Adam. Just learning the wonderful Tidymodels like you, and look forward to comments. :-)