Understanding the decision tree structure

The decision tree structure can be analysed to gain further insight on the relation between the features and the target to predict. In this example, we show how to retrieve:

  • the binary tree structure;

  • the depth of each node and whether or not it’s a leaf;

  • the nodes that were reached by a sample using the decision_path method;

  • the leaf that was reached by a sample using the apply method;

  • the rules that were used to predict a sample;

  • the decision path shared by a group of samples.

import numpy as np
from matplotlib import pyplot as plt

from sklearn import tree
from sklearn.datasets import load_iris
from sklearn.model_selection import train_test_split
from sklearn.tree import DecisionTreeClassifier

Train tree classifier

First, we fit a DecisionTreeClassifier using the load_iris dataset.

iris = load_iris()
X = iris.data
y = iris.target
X_train, X_test, y_train, y_test = train_test_split(X, y, random_state=0)

clf = DecisionTreeClassifier(max_leaf_nodes=3, random_state=0)
clf.fit(X_train, y_train)
DecisionTreeClassifier(max_leaf_nodes=3, random_state=0)
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Tree structure

The decision classifier has an attribute called tree_ which allows access to low level attributes such as node_count, the total number of nodes, and max_depth, the maximal depth of the tree. The tree_.compute_node_depths() method computes the depth of each node in the tree. tree_ also stores the entire binary tree structure, represented as a number of parallel arrays. The i-th element of each array holds information about the node i. Node 0 is the tree’s root. Some of the arrays only apply to either leaves or split nodes. In this case the values of the nodes of the other type is arbitrary. For example, the arrays feature and threshold only apply to split nodes. The values for leaf nodes in these arrays are therefore arbitrary.

Among these arrays, we have:

  • children_left[i]: id of the left child of node i or -1 if leaf node

  • children_right[i]: id of the right child of node i or -1 if leaf node

  • feature[i]: feature used for splitting node i

  • threshold[i]: threshold value at node i

  • n_node_samples[i]: the number of training samples reaching node i

  • impurity[i]: the impurity at node i

  • weighted_n_node_samples[i]: the weighted number of training samples reaching node i

  • value[i, j, k]: the summary of the training samples that reached node i for class j and output k.

Using the arrays, we can traverse the tree structure to compute various properties. Below, we will compute the depth of each node and whether or not it is a leaf.

n_nodes = clf.tree_.node_count
children_left = clf.tree_.children_left
children_right = clf.tree_.children_right
feature = clf.tree_.feature
threshold = clf.tree_.threshold
values = clf.tree_.value

node_depth = np.zeros(shape=n_nodes, dtype=np.int64)
is_leaves = np.zeros(shape=n_nodes, dtype=bool)
stack = [(0, 0)]  # start with the root node id (0) and its depth (0)
while len(stack) > 0:
    # `pop` ensures each node is only visited once
    node_id, depth = stack.pop()
    node_depth[node_id] = depth

    # If the left and right child of a node is not the same we have a split
    # node
    is_split_node = children_left[node_id] != children_right[node_id]
    # If a split node, append left and right children and depth to `stack`
    # so we can loop through them
    if is_split_node:
        stack.append((children_left[node_id], depth + 1))
        stack.append((children_right[node_id], depth + 1))
        is_leaves[node_id] = True

    "The binary tree structure has {n} nodes and has "
    "the following tree structure:\n".format(n=n_nodes)
for i in range(n_nodes):
    if is_leaves[i]:
            "{space}node={node} is a leaf node with value={value}.".format(
                space=node_depth[i] * "\t", node=i, value=values[i]
            "{space}node={node} is a split node with value={value}: "
            "go to node {left} if X[:, {feature}] <= {threshold} "
            "else to node {right}.".format(
                space=node_depth[i] * "\t",
The binary tree structure has 5 nodes and has the following tree structure:

node=0 is a split node with value=[[0.33035714 0.30357143 0.36607143]]: go to node 1 if X[:, 3] <= 0.800000011920929 else to node 2.
        node=1 is a leaf node with value=[[1. 0. 0.]].
        node=2 is a split node with value=[[0.         0.45333333 0.54666667]]: go to node 3 if X[:, 2] <= 4.950000047683716 else to node 4.
                node=3 is a leaf node with value=[[0.         0.91666667 0.08333333]].
                node=4 is a leaf node with value=[[0.         0.02564103 0.97435897]].

What is the values array used here?

The tree_.value array is a 3D array of shape [n_nodes, n_classes, n_outputs] which provides the count of samples reaching a node for each class and for each output. Each node has a value array which is the number of weighted samples reaching this node for each output and class.

For example, in the above tree built on the iris dataset, the root node has value = [37, 34, 41], indicating there are 37 samples of class 0, 34 samples of class 1, and 41 samples of class 2 at the root node. Traversing the tree, the samples are split and as a result, the value array reaching each node changes. The left child of the root node has value = [37, 0, 0] because all 37 samples in the left child node are from class 0.

Note: In this example, n_outputs=1, but the tree classifier can also handle multi-output problems. The value array at each node would just be a 2D array instead.

We can compare the above output to the plot of the decision tree.

plot unveil tree structure

Decision path

We can also retrieve the decision path of samples of interest. The decision_path method outputs an indicator matrix that allows us to retrieve the nodes the samples of interest traverse through. A non zero element in the indicator matrix at position (i, j) indicates that the sample i goes through the node j. Or, for one sample i, the positions of the non zero elements in row i of the indicator matrix designate the ids of the nodes that sample goes through.

The leaf ids reached by samples of interest can be obtained with the apply method. This returns an array of the node ids of the leaves reached by each sample of interest. Using the leaf ids and the decision_path we can obtain the splitting conditions that were used to predict a sample or a group of samples. First, let’s do it for one sample. Note that node_index is a sparse matrix.

node_indicator = clf.decision_path(X_test)
leaf_id = clf.apply(X_test)

sample_id = 0
# obtain ids of the nodes `sample_id` goes through, i.e., row `sample_id`
node_index = node_indicator.indices[
    node_indicator.indptr[sample_id] : node_indicator.indptr[sample_id + 1]

print("Rules used to predict sample {id}:\n".format(id=sample_id))
for node_id in node_index:
    # continue to the next node if it is a leaf node
    if leaf_id[sample_id] == node_id:

    # check if value of the split feature for sample 0 is below threshold
    if X_test[sample_id, feature[node_id]] <= threshold[node_id]:
        threshold_sign = "<="
        threshold_sign = ">"

        "decision node {node} : (X_test[{sample}, {feature}] = {value}) "
        "{inequality} {threshold})".format(
            value=X_test[sample_id, feature[node_id]],
Rules used to predict sample 0:

decision node 0 : (X_test[0, 3] = 2.4) > 0.800000011920929)
decision node 2 : (X_test[0, 2] = 5.1) > 4.950000047683716)

For a group of samples, we can determine the common nodes the samples go through.

sample_ids = [0, 1]
# boolean array indicating the nodes both samples go through
common_nodes = node_indicator.toarray()[sample_ids].sum(axis=0) == len(sample_ids)
# obtain node ids using position in array
common_node_id = np.arange(n_nodes)[common_nodes]

    "\nThe following samples {samples} share the node(s) {nodes} in the tree.".format(
        samples=sample_ids, nodes=common_node_id
print("This is {prop}% of all nodes.".format(prop=100 * len(common_node_id) / n_nodes))
The following samples [0, 1] share the node(s) [0 2] in the tree.
This is 40.0% of all nodes.

Total running time of the script: (0 minutes 0.076 seconds)

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