# Quick Start Guide: Optimal Policy Trees with Numeric Treatment

This is an R version of the corresponding OptimalTrees quick start guide.

In this example we will give a demonstration of how to use Optimal Policy Trees with numeric treatment options. For this example, we will the auto-mpg dataset, where the task is usually to predict a car's fuel efficiency (in miles-per-gallon, or MPG) based on other characteristics of the car. To apply a prescriptive lens to this case, we will instead treat the amount of acceleration as a treatment that can be controlled, and try to find the value that optimizes the MPG for a given car.

Note: this case is not intended to serve as a practical application of policy trees, but rather to serve as an illustration of the training and evaluation process. For a real-world case study using similar techniques, see the grocery pricing case study.

First we load in the data and drop 6 rows with missing values:

df <- na.omit(read.csv("auto-mpg.csv", na.strings = "?"))

  mpg cylinders displacement horsepower weight acceleration model.year origin
1  18         8          307        130   3504         12.0         70      1
2  15         8          350        165   3693         11.5         70      1
3  18         8          318        150   3436         11.0         70      1
4  16         8          304        150   3433         12.0         70      1
5  17         8          302        140   3449         10.5         70      1
6  15         8          429        198   4341         10.0         70      1
car.name
1 chevrolet chevelle malibu
2         buick skylark 320
3        plymouth satellite
4             amc rebel sst
5               ford torino
6          ford galaxie 500
[ reached 'max' / getOption("max.print") -- omitted 386 rows ]

Policy trees are trained using a features matrix/dataframe X as usual and a rewards matrix that has one column for each potential treatment that contains the outcome for each sample under that treatment.

There are two ways to get this rewards matrix:

• in rare cases, the problem may have full information about the outcome associated with each treatment for each sample
• more commonly, we have observational data, and use this partial data to train models to estimate the outcome associated with each treatment

Refer to the documentation on data preparation for more information on the data format.

In this case, the dataset is observational, and so we will use RewardEstimation to estimate our rewards matrix.

## Reward Estimation

Warning

Please refer to the Reward Estimation documentation for a detailed description on how to perform reward estimation for numeric treatments properly. For simplicity and to keep the focus on Optimal Policy Trees, this quick start guide does not cover tuning the reward estimation parameters, but in real problems this tuning is an important step.

First, we split into training and testing:

X <- subset(df, select = -c(mpg, acceleration, car.name))
treatments <- df$acceleration outcomes <- df$mpg

split <- iai::split_data("policy_maximize", X, treatments, outcomes,
seed = 123, train_proportion = 0.5)
train_X <- split$train$X
train_treatments <- split$train$treatments
train_outcomes <- split$train$outcomes
test_X <- split$test$X
test_treatments <- split$test$treatments
test_outcomes <- split$test$outcomes


Note that we have used a training/test split of 60%/40%, so that we save more data for testing to ensure high-quality reward estimation on the test set.

The treatment, acceleration, is a numeric value, so we follow the process for estimating rewards with numeric treatments. We will consider prescribing acceleration values from 11 to 20, in steps of 3:

treatment_candidates <- seq(11, 20, 3)


The outcome is continuous, so we use a numeric_regression_reward_estimator to estimate the MPG under our candidate acceleration values using a random forest model:

reward_lnr = iai::numeric_regression_reward_estimator(
propensity_estimator = iai::random_forest_regressor(),
outcome_estimator = iai::random_forest_regressor(),
reward_estimator = "doubly_robust",
propensity_min_value = 0.1,
random_seed = 1,
)
train_rewards <- iai::fit_predict(reward_lnr, train_X, train_treatments,
train_outcomes, treatment_candidates)
train_rewards$rewards   11 14 17 20 1 15.61183 14.324114 14.98341 15.88650 2 17.89016 16.706663 15.34452 16.37598 3 15.21088 11.313360 11.79762 16.05733 4 13.55233 11.037557 11.79762 16.05733 5 14.02919 11.451840 11.65667 16.34512 6 15.04257 15.661554 14.97618 16.30873 7 15.03297 13.564238 13.49279 16.25742 8 13.27892 15.067589 18.12993 17.50950 9 22.60668 22.551235 21.33069 18.97877 10 22.12423 17.811266 18.46614 20.78981 11 24.25679 23.646085 20.63887 19.28444 12 27.32696 27.999439 32.29332 24.77390 13 23.34823 23.554402 25.16990 25.22115 14 12.82628 -2.076632 13.48320 15.39619 15 14.03892 9.442752 12.84378 16.25809 [ reached 'max' / getOption("max.print") -- omitted 181 rows ] train_rewards$score$propensity  $17
[1] 0.1205333

$20 [1] 0.200617$11
[1] 0.2871446

$14 [1] 0.1653518 train_rewards$score$outcome  $17
[1] 0.8727428

$20 [1] 0.3228448$11
[1] 0.6603072

$14 [1] 0.8352894 We can see that the R2 of the internal outcome models for each candidate treatment are between 0.32 and 0.87, whereas the propensity models have R2 between 0.12 and 0.29. The non-zero propensity scores tell us that there is a small amount of treatment assignment bias in this data (keeping in mind that we arbitrarily selected a feature to use as the treatment column). It seems we can predict the outcomes reasonably well, so we have a good base for the reward estimates. ## Optimal Policy Trees Now that we have a complete rewards matrix, we can train a tree to learn an optimal prescription policy that maximizes MPG. We will use a grid_search to fit an optimal_tree_policy_maximizer (note that if we were trying to minimize the outcomes, we would use optimal_tree_policy_minimizer): grid <- iai::grid_search( iai::optimal_tree_policy_maximizer( random_seed = 1, minbucket = 15, ), max_depth = 4:5, ) iai::fit(grid, train_X, train_rewards$rewards)
iai::get_learner(grid)

Optimal Trees Visualization

The resulting tree recommends different accelerations based on the characteristics of the car. The intensity of the color in each leaf shows the difference in quality between the best and second-best acceleration values.

We can see that both extremes of our candidate acceleration values are prescribed by the tree. Older cars with lower horsepower receive the lowest acceleration, whereas the highest acceleration is best for many different groups of cars.

We can make treatment prescriptions using predict:

prescriptions <- iai::predict(grid, train_X)


The prescriptions are always returned as strings matching the column names of the input rewards matrix. In our case the treatments are numeric values, and if we want them in numeric form to use later we can convert them to numeric treatments using convert_treatments_to_numeric:

iai::convert_treatments_to_numeric(prescriptions)

 [1] 20 20 20 20 20 20 20 20 20 20 11 11 20 20 20 11 20 20 11 20 20 20 11 20 11
[26] 11 11 11 11 11 11 11 20 20 20 20 20 20 20 20 11 11 20 20 11 20 20 20 20 20
[51] 20 20 11 11 20 20 11 11 20 20
[ reached getOption("max.print") -- omitted 136 entries ]

If we want more information about the relative performance of treatments for these points, we can predict the full treatment ranking with predict_treatment_rank:

rank <- iai::predict_treatment_rank(grid, train_X)

       [,1] [,2] [,3] [,4]
[1,] "20" "11" "17" "14"
[2,] "20" "11" "17" "14"
[3,] "20" "11" "17" "14"
[4,] "20" "11" "17" "14"
[5,] "20" "11" "17" "14"
[6,] "20" "11" "17" "14"
[7,] "20" "11" "17" "14"
[8,] "20" "11" "17" "14"
[9,] "20" "11" "17" "14"
[10,] "20" "11" "17" "14"
[11,] "11" "17" "14" "20"
[12,] "11" "17" "14" "20"
[13,] "20" "11" "17" "14"
[14,] "20" "11" "17" "14"
[15,] "20" "11" "17" "14"
[ reached getOption("max.print") -- omitted 181 rows ]

For each point in the data, this gives the treatments in order of effectiveness. As before, this are returned as strings, but we can convert the treatments to numeric values with convert_treatments_to_numeric:

iai::convert_treatments_to_numeric(rank)

       [,1] [,2] [,3] [,4]
[1,]   20   11   17   14
[2,]   20   11   17   14
[3,]   20   11   17   14
[4,]   20   11   17   14
[5,]   20   11   17   14
[6,]   20   11   17   14
[7,]   20   11   17   14
[8,]   20   11   17   14
[9,]   20   11   17   14
[10,]   20   11   17   14
[11,]   11   17   14   20
[12,]   11   17   14   20
[13,]   20   11   17   14
[14,]   20   11   17   14
[15,]   20   11   17   14
[ reached getOption("max.print") -- omitted 181 rows ]

To quantify the difference in performance behind the treatment rankings, we can use predict_treatment_outcome to extract the estimated quality of each treatment for each point:

iai::predict_treatment_outcome(grid, train_X)

         11       14       17       20
1  16.95464 15.73563 16.44407 18.33659
2  16.95464 15.73563 16.44407 18.33659
3  16.95464 15.73563 16.44407 18.33659
4  16.95464 15.73563 16.44407 18.33659
5  16.95464 15.73563 16.44407 18.33659
6  16.95464 15.73563 16.44407 18.33659
7  16.95464 15.73563 16.44407 18.33659
8  16.95464 15.73563 16.44407 18.33659
9  16.95464 15.73563 16.44407 18.33659
10 16.95464 15.73563 16.44407 18.33659
11 27.01673 24.65935 26.18041 23.08957
12 27.01673 24.65935 26.18041 23.08957
13 16.95464 15.73563 16.44407 18.33659
14 16.95464 15.73563 16.44407 18.33659
15 16.95464 15.73563 16.44407 18.33659
[ reached 'max' / getOption("max.print") -- omitted 181 rows ]

## Evaluating Optimal Policy Trees

It is critical for a fair evaluation that we do not evaluate the quality of the policy using rewards from our existing reward estimator trained on the training set. This is to avoid any information from the training set leaking through to the out-of-sample evaluation.

Instead, what we need to do is to estimate a new set of rewards using only the test set, and evaluate the policy against these rewards:

test_rewards <- iai::fit_predict(reward_lnr, test_X, test_treatments,
test_outcomes, treatment_candidates)
test_rewards$rewards   11 14 17 20 1 18.87610 17.348683 15.442553 13.02367 2 18.54885 17.177284 16.064991 13.10194 3 16.90637 17.195855 16.408816 13.26537 4 14.70057 12.306050 11.470901 12.95174 5 15.65078 13.336453 11.553937 12.99407 6 15.09902 16.819106 13.569338 12.67351 7 24.33651 23.789861 23.599930 22.83901 8 26.44638 27.176601 25.299428 24.02699 9 23.16950 23.695650 30.648817 23.61563 10 23.68355 21.405816 24.498837 20.54564 11 31.47751 25.039407 22.125471 20.36907 12 18.56323 21.176058 21.268225 21.27784 13 14.18914 4.528854 14.285364 17.85454 14 14.04011 13.358946 -7.535757 -18.70225 15 24.73810 26.717745 27.413140 27.73378 [ reached 'max' / getOption("max.print") -- omitted 181 rows ] test_rewards$score$propensity  $17
[1] 0.1580786

$20 [1] 0.2672843$11
[1] 0.3930468

$14 [1] 0.217413 test_rewards$score$outcome  $17
[1] 0.7164027

$20 [1] 0.7470992$11
[1] 0.5085259

$14 [1] 0.8895272 We see the internal models of our test reward estimator have scores that are similar to those on the test set. As with the training set, this gives us confidence that the estimated rewards are a fair reflection of reality, and will serve as a good basis for evaluation. We can now evaluate the quality using these new estimated rewards. First, we will calculate the average predicted MPG under the treatments prescribed by the tree for the test set. To do this, we use predict_outcomes which uses the model to make prescriptions and looks up the predicted outcomes under these prescriptions: policy_outcomes <- iai::predict_outcomes(grid, test_X, test_rewards$rewards)

 [1]  13.02367  13.10194  13.26537  12.95174  12.99407  12.67351  22.83901
[8]  26.44638  23.16950  23.68355  20.36907  18.56323  17.85454 -18.70225
[15]  24.73810  19.84227  18.41289  18.02786  17.85454  12.66219  12.66219
[22]  12.99407  18.75970  17.48188  26.20989  28.31080  23.51552  25.02196
[29]  25.42858  17.85454  14.89811  12.74661  17.85454  14.02707  15.65969
[36]  15.65969  26.84839  28.07555  27.50332  15.65969  12.88554  14.38841
[43]  14.84884  14.84884  12.88554  15.65969  19.30518  20.20801  14.84884
[50]  14.84884  19.30518  24.62845  27.34884  22.65649  25.94121  20.78518
[57]  15.65969  30.14560  28.72838  22.97014
[ reached getOption("max.print") -- omitted 136 entries ]

We can then get the average estimated MPG under our treatments:

mean(policy_outcomes)

[1] 23.98876

We can compare this number to the average estimated MPG under a baseline policy that assigns the same treatment to all observations regardless of their features. Let us compare to a policy that assigns 17 everywhere:

mean(test_rewards\$rewards[,"17"])

[1] 23.1136

We can see that the personalization offered by the tree policy indeed improves upon a baseline where each observation receives the same treatment.