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## Introduction

This blog posts will use several packages from the {tidymodels} collection of packages, namely {recipes}, {rsample} and {parsnip} to train a random forest the tidy way. I will also use {mlrMBO} to tune the hyper-parameters of the random forest.

## Set up

library("tidyverse")
library("tidymodels")
library("parsnip")
library("brotools")
library("mlbench")

Load the data, included in the {mlrbench} package:

data("BostonHousing2")

I will train a random forest to predict the housing price, which is the cmedv column:

head(BostonHousing2)
##         town tract      lon     lat medv cmedv    crim zn indus chas   nox
## 1     Nahant  2011 -70.9550 42.2550 24.0  24.0 0.00632 18  2.31    0 0.538
## 2 Swampscott  2021 -70.9500 42.2875 21.6  21.6 0.02731  0  7.07    0 0.469
## 3 Swampscott  2022 -70.9360 42.2830 34.7  34.7 0.02729  0  7.07    0 0.469
## 4 Marblehead  2031 -70.9280 42.2930 33.4  33.4 0.03237  0  2.18    0 0.458
## 5 Marblehead  2032 -70.9220 42.2980 36.2  36.2 0.06905  0  2.18    0 0.458
## 6 Marblehead  2033 -70.9165 42.3040 28.7  28.7 0.02985  0  2.18    0 0.458
##      rm  age    dis rad tax ptratio      b lstat
## 1 6.575 65.2 4.0900   1 296    15.3 396.90  4.98
## 2 6.421 78.9 4.9671   2 242    17.8 396.90  9.14
## 3 7.185 61.1 4.9671   2 242    17.8 392.83  4.03
## 4 6.998 45.8 6.0622   3 222    18.7 394.63  2.94
## 5 7.147 54.2 6.0622   3 222    18.7 396.90  5.33
## 6 6.430 58.7 6.0622   3 222    18.7 394.12  5.21

Only keep relevant columns:

boston <- BostonHousing2 %>%
select(-medv, -town, -lon, -lat) %>%
rename(price = cmedv)

I remove town, lat and lon because the information contained in the column tract is enough.

To train and evaluate the model’s performance, I split the data in two. One data set, which I call the training set, will be further split into two down below. I won’t touch the second data set, the test set, until the very end.

train_test_split <- initial_split(boston, prop = 0.9)

housing_train <- training(train_test_split)

housing_test <- testing(train_test_split)

I want to train a random forest to predict price of houses, but random forests have so-called hyperparameters, which are parameters that cannot be estimated, or learned, from the data. Instead, these parameters have to be chosen by the analyst. In order to choose them, you can use values from the literature that seemed to have worked well (like is done in Macro-econometrics) or you can further split the train set into two, create a grid of hyperparameter, train the model on one part of the data for all values of the grid, and compare the predictions of the models on the second part of the data. You then stick with the model that performed the best, for example, the model with lowest RMSE. The thing is, you can’t estimate the true value of the RMSE with only one value. It’s like if you wanted to estimate the height of the population by drawing one single observation from the population. You need a bit more observations. To approach the true value of the RMSE for a give set of hyperparameters, instead of doing one split, I’ll do 30. I then compute the average RMSE, which implies training 30 models for each combination of the values of the hyperparameters I am interested in.

First, let’s split the training data again, using the mc_cv() function from {rsample} package. This function implements Monte Carlo cross-validation:

validation_data <- mc_cv(housing_train, prop = 0.9, times = 30)

What does validation_data look like?

validation_data
## # # Monte Carlo cross-validation (0.9/0.1) with 30 resamples
## # A tibble: 30 x 2
##    splits       id
##
##  1  Resample01
##  2  Resample02
##  3  Resample03
##  4  Resample04
##  5  Resample05
##  6  Resample06
##  7  Resample07
##  8  Resample08
##  9  Resample09
## 10  Resample10
## # ... with 20 more rows

Let’s look further down:

validation_data$splits[[1]] ## <411/45/456> The first value is the number of rows of the first set, the second value of the second, and the third was the original amount of values in the training data, before splitting again. How should we call these two new data sets? The author of {rsample}, Max Kuhn, talks about the analysis and the assessment sets: Now, in order to continue I need pre-process the data. I will do this in three steps. The first and the second step are used to center and scale the numeric variables and the third step converts character and factor variables to dummy variables. This is needed because I will train a random forest, which cannot handle factor variables directly. Let’s define a recipe to do that, and start by pre-processing the testing set. I write a wrapper function around the recipe, because I will need to apply this recipe to various data sets: simple_recipe <- function(dataset){ recipe(price ~ ., data = dataset) %>% step_center(all_numeric()) %>% step_scale(all_numeric()) %>% step_dummy(all_nominal()) } Once the recipe is defined, I can use the prep() function, which estimates the parameters from the data which are needed to process the data. For example, for centering, prep() estimates the mean which will then be subtracted from the variables. With bake() the estimates are then applied on the data: testing_rec <- prep(simple_recipe(housing_test), testing = housing_test) test_data <- bake(testing_rec, newdata = housing_test) It is important to split the data before using prep() and bake(), because if not, you will use observations from the test set in the prep() step, and thus introduce knowledge from the test set into the training data. This is called data leakage, and must be avoided. This is why it is necessary to first split the training data into an analysis and an assessment set, and then also pre-process these sets separately. However, the validation_data object cannot now be used with recipe(), because it is not a dataframe. No worries, I simply need to write a function that extracts the analysis and assessment sets from the validation_data object, applies the pre-processing, trains the model, and returns the RMSE. This will be a big function, at the center of the analysis. But before that, let’s run a simple linear regression, as a benchmark. For the linear regression, I will not use any CV, so let’s pre-process the training set: trainlm_rec <- prep(simple_recipe(housing_train), testing = housing_train) trainlm_data <- bake(trainlm_rec, newdata = housing_train) linreg_model <- lm(price ~ ., data = trainlm_data) broom::augment(linreg_model, newdata = test_data) %>% rmse(price, .fitted) ## # A tibble: 1 x 3 ## .metric .estimator .estimate ## ## 1 rmse standard 0.469 broom::augment() adds the predictions to the test_data in a new column, .fitted. I won’t use this trick with the random forest, because there is no augment() method for random forests from the {ranger} which I’ll use. I’ll add the predictions to the data myself. Ok, now let’s go back to the random forest and write the big function: my_rf <- function(mtry, trees, split, id){ analysis_set <- analysis(split) analysis_prep <- prep(simple_recipe(analysis_set), training = analysis_set) analysis_processed <- bake(analysis_prep, newdata = analysis_set) model <- rand_forest(mtry = mtry, trees = trees) %>% set_engine("ranger", importance = 'impurity') %>% fit(price ~ ., data = analysis_processed) assessment_set <- assessment(split) assessment_prep <- prep(simple_recipe(assessment_set), testing = assessment_set) assessment_processed <- bake(assessment_prep, newdata = assessment_set) tibble::tibble("id" = id, "truth" = assessment_processed$price,
"prediction" = unlist(predict(model, new_data = assessment_processed)))
}

The rand_forest() function is available from the {parsnip} package. This package provides an unified interface to a lot of other machine learning packages. This means that instead of having to learn the syntax of range() and randomForest() and, and… you can simply use the rand_forest() function and change the engine argument to the one you want (ranger, randomForest, etc).

Let’s try this function:

results_example <- map2_df(.x = validation_data$splits, .y = validation_data$id,
~my_rf(mtry = 3, trees = 200, split = .x, id = .y))
## # A tibble: 6 x 3
##   id           truth prediction
##
## 1 Resample01  0.0235    -0.104
## 2 Resample01 -0.135     -0.0906
## 3 Resample01 -0.378     -0.158
## 4 Resample01 -0.232      0.0623
## 5 Resample01 -0.0859     0.0173
## 6 Resample01  0.169      0.303

I can now compute the RMSE when mtry = 3 and trees = 200:

results_example %>%
group_by(id) %>%
rmse(truth, prediction) %>%
summarise(mean_rmse = mean(.estimate)) %>%
pull
## [1] 0.4319164

The random forest has already lower RMSE than the linear regression. The goal now is to lower this RMSE by tuning the mtry and trees hyperparameters. For this, I will use Bayesian Optimization methods implemented in the {mlrMBO} package.

## Bayesian hyperparameter optimization

I will re-use the code from above, and define a function that does everything from pre-processing to returning the metric I want to minimize by tuning the hyperparameters, the RMSE:

tuning <- function(param, validation_data){

mtry <- param[1]
trees <- param[2]

results <- purrr::map2_df(.x = validation_data$splits, .y = validation_data$id,
~my_rf(mtry = mtry, trees = trees, split = .x, id = .y))

results %>%
group_by(id) %>%
rmse(truth, prediction) %>%
summarise(mean_rmse = mean(.estimate)) %>%
pull
}

This is exactly the code from before, but it now returns the RMSE. Let’s try the function with the values from before:

tuning(c(3, 200), validation_data)
## [1] 0.4330951

Let’s also plot the value of RMSE for mtry = 3 and trees from 200 to 300. This takes some time, because I need to evaluate this costly function 100 times. If evaluating the function was cheap, I could have made a 3D plot by varying values of mtry too, but then again if evaluating the function was cheap, I would run an exhaustive grid search to find the hyperparameters instead of using Bayesian optimization.

plot_points <- crossing("mtry" = 3, "trees" = seq(200, 300))

plot_data <- plot_points %>%
mutate(value = map_dbl(seq(200, 300), ~tuning(c(3, .), validation_data)))
plot_data %>%
ggplot(aes(y = value, x = trees)) +
geom_line(colour = "#82518c") +
theme_blog() +
ggtitle("RMSE for mtry = 3")

For mtry = 3 the minimum seems to lie around 255. The function to minimize is not smooth at all.

I now follow the code that can be found in the arxiv paper to run the optimization. I think I got the gist of the paper, but I did not understand everything yet. For now, I am still experimenting with the library at the moment, but from what I understand, a simpler model, called the surrogate model, is used to look for promising points and to evaluate the value of the function at these points. This seems somewhat similar (in spirit) to the Indirect Inference method as described in Gourieroux, Monfort, Renault.

Let’s first load the package and create the function to optimize:

library("mlrMBO")
fn <- makeSingleObjectiveFunction(name = "tuning",
fn = tuning,
par.set = makeParamSet(makeIntegerParam("x1", lower = 3, upper = 8),
makeIntegerParam("x2", lower = 50, upper = 500)))

This function is based on the function I defined before. The parameters to optimize are also defined as are their bounds. I will look for mtry between the values of 3 and 8, and trees between 50 and 500.

Now comes the part I didn’t quite get.

# Create initial random Latin Hypercube Design of 10 points
library(lhs)# for randomLHS
des <- generateDesign(n = 5L * 2L, getParamSet(fn), fun = randomLHS)

I think this means that these 10 points are the points used to start the whole process. I did not understand why they have to be sampled from a hypercube, but ok. Then I choose the surrogate model, a random forest too, and predict the standard error. Here also, I did not quite get why the standard error can be an option.

# Specify kriging model with standard error estimation
surrogate <- makeLearner("regr.ranger", predict.type = "se", keep.inbag = TRUE)

Here I define some options:

# Set general controls
ctrl <- makeMBOControl()
ctrl <- setMBOControlTermination(ctrl, iters = 10L)
ctrl <- setMBOControlInfill(ctrl, crit = makeMBOInfillCritEI())

And this is the optimization part:

# Start optimization
result <- mbo(fn, des, surrogate, ctrl, more.args = list("validation_data" = validation_data))
result
## Recommended parameters:
## x1=6; x2=381
## Objective: y = 0.393
##
## Optimization path
## 10 + 10 entries in total, displaying last 10 (or less):
##    x1  x2         y dob eol error.message exec.time            ei
## 11  6 370 0.3943479   1  NA               8.913 -3.134568e-05
## 12  6 362 0.3950402   2  NA               8.844 -2.987934e-05
## 13  6 373 0.3939587   3  NA               8.939 -2.259674e-05
## 14  6 394 0.3962875   4  NA               9.342 -7.427682e-06
## 15  6 368 0.3944954   5  NA               8.760 -4.121337e-06
## 16  6 378 0.3938796   6  NA               8.949 -4.503591e-07
## 17  6 381 0.3934176   7  NA               9.109 -1.141853e-06
## 18  6 380 0.3948077   8  NA               9.026 -4.718394e-08
## 19  6 381 0.3932636   9  NA               9.022 -9.801395e-08
## 20  6 383 0.3953004  10  NA               9.184 -1.579619e-09
##    error.model train.time prop.type propose.time           se      mean
## 11              0.014 infill_ei        0.449 0.0010924600 0.3955131
## 12              0.012 infill_ei        0.458 0.0007415920 0.3948705
## 13              0.012 infill_ei        0.460 0.0006116756 0.3947185
## 14              0.012 infill_ei        0.729 0.0003104694 0.3943572
## 15              0.023 infill_ei        0.444 0.0003446061 0.3945085
## 16              0.013 infill_ei        0.458 0.0002381887 0.3944642
## 17              0.013 infill_ei        0.492 0.0002106454 0.3943200
## 18              0.013 infill_ei        0.516 0.0002093524 0.3940764
## 19              0.014 infill_ei        0.756 0.0002481260 0.3941597
## 20              0.013 infill_ei        0.483 0.0001687982 0.3939285

So the recommended parameters are 6 for mtry and 381 for trees. The value of the RMSE is lower than before, and equals 0.393. Let’s now train the random forest on the training data with this values. First, I pre-process the training data:

training_rec <- prep(simple_recipe(housing_train), testing = housing_train)

train_data <- bake(training_rec, newdata = housing_train)

Let’s now train our final model and predict the prices:

final_model <- rand_forest(mtry = 6, trees = 381) %>%
set_engine("ranger", importance = 'impurity') %>%
fit(price ~ ., data = train_data)

price_predict <- predict(final_model, new_data = select(test_data, -price))

Let’s transform the data back and compare the predicted prices to the true ones visually:

cbind(price_predict * sd(housing_train$price) + mean(housing_train$price),
housing_test$price) ## .pred housing_test$price
## 1  26.66387               24.0
## 2  18.05651               18.9
## 3  17.73167               15.0
## 4  19.58751               18.2
## 5  15.83327               13.9
## 6  15.23911               13.2
## 7  20.71109               20.0
## 8  24.57467               25.0
## 9  23.30817               21.4
## 10 22.49698               22.9
## 11 22.80554               21.4
## 12 19.12121               18.6
## 13 19.63354               19.3
## 14 18.33941               20.4
## 15 17.03028               14.4
## 16 45.58864               50.0
## 17 28.26440               32.0
## 18 28.48881               30.5
## 19 28.18277               31.1
## 20 45.12203               48.5
## 21 45.13364               50.0
## 22 20.95897               24.4
## 23 32.70919               31.6
## 24 18.88725               18.5
## 25 23.80696               22.0
## 26 24.05506               23.2
## 27 33.69617               37.3
## 28 26.93756               27.9
## 29 22.71415               23.9
## 30 24.88508               27.1
## 31 24.63913               22.5
## 32 30.73453               33.1
## 33 23.35807               22.1
## 34 21.89201               23.1
## 35 23.22168               23.1
## 36 20.15691               19.3
## 37 21.12050               22.7
## 38 12.88927               13.1
## 39 19.25563               10.4
## 40 11.32007               10.5
## 41 16.30650               13.1
## 42 14.14549                8.3
## 43 11.08126                5.0
## 44 14.14806               13.4
## 45 20.56828               17.8
## 46 14.43175               14.4
## 47 20.00864               19.5
## 48 13.46410               12.0
## 49 20.81910               21.8
## 50 21.72144               21.2

Let’s now compute the RMSE:

tibble::tibble("truth" = test_data\$price,
"prediction" = unlist(price_predict)) %>%
rmse(truth, prediction)
## # A tibble: 1 x 3
##   .metric .estimator .estimate
##
## 1 rmse    standard       0.327

Very nice.

Hope you enjoyed! If you found this blog post useful, you might want to follow me on twitter for blog post updates and buy me an espresso.

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