Recently I encountered an interesting paper from the facebook research team that outlines a method for using decision trees (specifically boosted trees) to create transformed data to be used as input to a final logistic regression. I thought this was really cool and wanted to try and recreate the method in R. I’ll let the paper authors explain the process for how they went about this.

We treat each individual tree as a categorical feature that takes as value the index of the leaf an instance ends up falling in. We use 1-of-K coding of this type of features. For example, consider the boosted tree model in Figure 1 with 2 subtrees, where the first subtree has 3 leafs and the second 2 leafs. If an instance ends up in leaf 2 in the first subtree and leaf 1 in second subtree, the overall input to the linear classifier will be the binary vector [0, 1, 0, 1, 0], where the first 3 entries correspond to the leaves of the first subtree and last 2 to those of the second subtree.

Essentially each tree represents a dummy variable and the terminal node that each observation ends up in represents a different level of that variable. Here is a picture from the paper diagramming the process: I’m by no means an expert on using decision rules as input to regression models but I have not heard of this exact way to transform raw data to use as input to a different model but if there has been other research on this topic please feel free to share, I’d love to read more about it.

Their paper shows that this method was more accurate than either the Gradient Boosting Machine (GBM) or the Logistic Regression alone but they didn’t go into any detail on how they traversed the tree so I’ll try to focus on that since it was not the easiest thing to code up. I’ll be using the Lending Club Loan Data found on Kaggle as my test set for no other reason than it has many samples and many different features. It’s actually a little too big, the steps in this script do take a while so if you want to recreate the results you’ve been warned. I’ll be doing some small data cleansing which you can see in the source code for this post but I won’t be showing it here to save space. The dataset is a mix of categorical and numeric values so this should be a good use case for using the trees to find relationships between the underlying variables and the loan status. I assume, and hope to test, if the data was just numeric variables then the trees wouldn’t improve much over just using smoothing splines.

### Modeling Strategy

Although the paper showed that this method was more accurate on their data I’d still like to test its accuracy in comparison to other methods. To do this we need to plan our modeling strategy. We have two models to fit to the data; A GBM to transform the raw data to the leaf positions and a Logistic Regression on the transformed data. We should use different datasets to train these two models so that we don’t overfit the Logistic Regression. We will also need a third validation dataset to test the accuracy on data that the models haven’t seen so we have an unbiased estimate of how well the models perform. I’m going to do a 40-40-20 split on the loan data. Here is a totally sweet diagram to show what I mean: ### Train a Gradient Boosting Model

Let’s get started training the GBM. While using the `xgboost` or `h2o` package would be faster than the `gbm` package I am more familiar with the output of the `gbm` function and know that it contains the tree and split information we need. Traversing each tree to find the leaf positions took a lot of thought and trial and error so I figured the time I saved on tuning the model wouldn’t make up for the extra time it took me to code it up.

``````set.seed(21091)
set_id <- sample(c(1, 2, 3), size = nrow(clean_data), replace = T, prob = c(.4, .4, .2))

training_data_1 <- clean_data[set_id == 1, ]
training_data_2 <- clean_data[set_id == 2, ]
validation_data <- clean_data[set_id == 3, ]

gbm_model <- gbm(formula = bad_loan ~ .,
data = training_data_1,
interaction.depth = 2,
distribution = "bernoulli",
cv.folds = 5,
shrinkage = .1,
n.trees = 800)
`````` ``````# Refit the model to the appropriate number of trees
gbm_model <- gbm(formula = bad_loan ~ .,
data = training_data_1,
interaction.depth = 2,
distribution = "bernoulli",
shrinkage = .1,
n.trees = 290,
keep.data = F)
``````

### Traverse the Trees

Now that we have the fitted model we need to send the second training set through the GBM to find the leaf positions of each observation. To do this I wrote a helper function called `traverse_trees`. It may be tough to follow along but here is the basic structure:

1. For each tree find the split decisions
2. For each split decision generate a logical index of the underlying data for the left and right path
3. Combine any multi-layer splits.
4. Use the logical indexes to generate the dummy variables for each terminal node

There are certainly ways to make this function better. I rely on the depth of the GBM to always be two so I don’t do any recursive partitioning to make it more robust. It’s all embarrassingly parallel so could easily be made faster. But for a quick exploratory analysis I’m cool with it.

``````traverse_trees <- function(boosted_model, predict_data){

n_trees <- boosted_model\$n.trees
category_splits <- boosted_model\$c.splits
col_types <- vapply(predict_data, class, character(1))
matrix_stuff <- c()

for(i in seq_len(n_trees)){
#extract the tree
tree <- pretty.gbm.tree(boosted_model, i.tree = i)

# add one to each var (the gbm.object is 0 indexed, gross)
tree\$SplitVar <- tree\$SplitVar + 1

# Get info about the non-terminal nodes into a list
split_info <- tree %>%
mutate(node = row.names(.)) %>%
filter(SplitVar != 0) %>%
apply(1, function(x) {
temp_list <- list(node = unname(x["node"]),
split_var = as.numeric(unname(x["SplitVar"])),
split_num = as.numeric(unname(x["SplitCodePred"])),
left_child = unname(x["LeftNode"]),
right_child = unname(x["RightNode"]),
missing_child = unname(x["MissingNode"]))
return(temp_list)
})

# function to get logical indicators for the observations at each split
traverse <- function(tree_info, category_splits, obs_data, col_types){

split_var <- names(obs_data)[tree_info\$split_var]
# check for missing values in the data
missing_flag <- any(is.na(obs_data[[split_var]]))

if(col_types[tree_info\$split_var] == "factor") {
# find the direction the tree puts each value
level_directions <- category_splits[[tree_info\$split_num + 1]]
var_levels <- levels(obs_data[[split_var]])
# assign the levels to the left and right splits
left_levels <- var_levels[level_directions == -1]
right_levels <- var_levels[level_directions == 1]
# turn each into a list
left_list <- list(lgl = obs_data[[split_var]] %in% left_levels,
parent = tree_info\$node)
right_list <- list(lgl = obs_data[[split_var]] %in% right_levels,
parent = tree_info\$node)
} else {
# I actually don't know if it's <= & > or < & >=
left_list <- list(lgl = obs_data[[split_var]] <= tree_info\$split_num,
parent = tree_info\$node)
right_list <- list(lgl = obs_data[[split_var]] > tree_info\$split_num,
parent = tree_info\$node)
}

# if the tree uses missing values then just add on an extra list
if(missing_flag) {
miss_list <- list(lgl = is.na(obs_data[[split_var]]),
parent = tree_info\$node)
split_list <- list(left_list, right_list, miss_list)
names(split_list) <- c(tree_info\$left_child, tree_info\$right_child, tree_info\$missing_child)
} else {
split_list <- list(left_list, right_list)
names(split_list) <- c(tree_info\$left_child, tree_info\$right_child)
}

return(split_list)
}

# for each split node find the children
terminal_info <- map(split_info, traverse, category_splits = category_splits, obs_data = predict_data, col_types = col_types) %>% flatten

terminal_nodes <- row.names(tree)[tree\$SplitVar == 0]
splits <- names(terminal_info)
# for each terminal node find out if it has a parent
# if so then include the parent logical vector
full <- map_if(terminal_info, ~ .x\$parent %in% splits, function(x, tree_info) {
x[["lgl"]] <- x[["lgl"]] & tree_info[[x\$parent]][["lgl"]]
return(x)
}, terminal_info)

output <- full[terminal_nodes] %>% # only keep the terminal nodes
# if there are no missing values then remove that terminal node
map(extract2, "lgl") %>%
unlist

obs <- matrix(output * 1, nrow = nrow(predict_data), byrow = F)
stopifnot(all(rowSums(obs) == 1))
matrix_stuff <- cbind(matrix_stuff, obs)
}

return(matrix_stuff)
}

tree_output <- traverse_trees(gbm_model, training_data_2)
``````

To recap what we just did here is how this process works for the first tree. We can use the `pretty.gbm.tree` function from the `gbm` package to view the splits in the first tree:

SplitVar SplitCodePred LeftNode RightNode MissingNode ErrorReduction Weight Prediction
0 36 1.8000000 1 5 6 4268.57 177368 -0.0004912
1 38 0.0000000 2 3 4 614.49 172512 -0.0376891
2 -1 -0.0680307 -1 -1 -1 0.00 149788 -0.0680307
3 -1 0.2032145 -1 -1 -1 0.00 19216 0.2032145
4 -1 -0.0617460 -1 -1 -1 0.00 3508 -0.0617460
5 -1 1.3209838 -1 -1 -1 0.00 4856 1.3209838
6 -1 -0.0004912 -1 -1 -1 0.00 177368 -0.0004912

The tree is 0 indexed so we want to look at the 37th and 39th variable in our data:

0 0.00 Jun-2014
0 0.00 Jan-2016
1 269.29 Nov-2012
0 0.00 Jan-2015
0 0.00 Jan-2015

This tree uses the `recoveries` variable to make its first split and for small amounts of that (less than 1.8) it checks the month of `last_pymnt_d`. There are some missing observations in that variable so we have 4 nodes for this tree

``````##      [,1] [,2] [,3] [,4]
## [1,]    1    0    0    0
## [2,]    1    0    0    0
## [3,]    0    0    0    1
## [4,]    1    0    0    0
## [5,]    1    0    0    0
``````

### Train a Logistic Regression Model

Now we can get on to our logistic regression. I’ll use the `cv.glmnet` function from the `glmnet` package to fit a Penalized Logistic Regression using 5-fold cross validation.

``````Y <- training_data_2\$bad_loan

ridge_regression <- cv.glmnet(x = tree_output, y = Y, nfolds = 5, alpha = 0)
`````` ### Model Performance

``````val_output <- traverse_trees(gbm_model, validation_data)

gbm_preds <- predict(gbm_model, newdata = validation_data, type = "response", n.trees = 290)
boosted_logistic_preds <- predict(ridge_regression, newx = val_output, type = "link", s = "lambda.1se")[, 1]
boosted_logistic_preds <- 1 / (1 + exp(-boosted_logistic_preds))

perf_gbm <- c("MSE" = mean((val_response - gbm_preds)^2), "AUC" = glmnet::auc(val_response, gbm_preds))
perf_boost_ridge <- c("MSE" = mean((val_response - boosted_logistic_preds)^2), "AUC" = glmnet::auc(val_response, boosted_logistic_preds))

rbind(perf_gbm, perf_boost_ridge)
``````
``````##                         MSE       AUC
## perf_gbm         0.01619194 0.9725718
## perf_boost_ridge 0.24403963 0.9672341
``````

Wow, the GBM alone performs much better by Mean Square Error (MSE) than the boosted tree + ridge regression model. Let’s look at how the models are calibrated. The fact that the boosted + ridge model performed so poorly by MSE but still has a relatively high AUC implies, at least to me, that the penalized regression was too restrictive. The model separates the bad from good very well (as evidenced by a high AUC) but is poorly calibrated. Perhaps if we used an actual logistic regression and not ridge regression we would see better results? And I know this is using the validation set to influence your decisions but I gots to know. Unfortunately with a dataset of this size (the training output used to fit the ridge regression was 354852 rows by 1104 columns) R’s base `glm` won’t work. We can use the `h2o` package to fit one though. h2o is a pretty incredible resource to fit large models by using distributed computing on a JVM, if you haven’t heard of them I really encourage you to check out their resources.

``````library(h2o)
# initialize large local cluster
localH2O <- h2o.init(min_mem_size = "1g", max_mem_size = "4g")
# load the tree output into the h2o environment
logistic_h2o_df <- as.h2o(data.frame(tree_output))
# fit the regression model
logistic_h2o_model <- h2o.glm(x = seq(1, ncol(tree_output)),
training_frame = logistic_h2o_df,
lambda = 0,
family = "binomial")
# get validation data into h2o
logisitc_val_df <- as.h2o(data.frame(val_output))
h2o_preds <- predict(logistic_h2o_model, newdata = logisitc_val_df)
h2o_preds <- as.data.frame(h2o_preds)
``````
``````##                         MSE       AUC
## perf_gbm         0.01619194 0.9725718
## perf_boost_ridge 0.24403963 0.9672341
## perf_h2o         0.01226116 0.9800423
`````` And there it is, using a full Logistic Regression Model we get improved results over the GBM alone by MSE, AUC, and the calibration of the model.

This was a fun exercise for me that allowed me to work out my R muscles and learn a new modeling strategy. I hope you enjoyed it too and if you have any questions or comments please leave a comment or reach out on twitter. As always you can view the source code for this post on github