理解决策树的结构¶
通过对决策树结构的分析，进一步了解特征与预测目标之间的关系。在本例中，我们将展示如何检索：
二叉树结构； 每个节点的深度以及它是否是一片叶子节点； 使用 decision_path方法通过样本到达的节点；使用apply方法的样本所达到的叶子节点； 用来预测样本的规则; 一组样本共享的决策路径。
The binary tree structure has 5 nodes and has the following tree structure:
node=0 test node: go to node 1 if X[:, 3] <= 0.800000011920929 else to node 2.
node=1 leaf node.
node=2 test node: go to node 3 if X[:, 2] <= 4.950000047683716 else to node 4.
node=3 leaf node.
node=4 leaf node.
Rules used to predict sample 0:
decision id node 0 : (X_test[0, 3] (= 2.4) > 0.800000011920929)
decision id node 2 : (X_test[0, 2] (= 5.1) > 4.950000047683716)
The following samples [0, 1] share the node [0 2] in the tree
It is 40.0 % of all nodes.
import numpy as np
from sklearn.model_selection import train_test_split
from sklearn.datasets import load_iris
from sklearn.tree import DecisionTreeClassifier
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)
estimator = DecisionTreeClassifier(max_leaf_nodes=3, random_state=0)
estimator.fit(X_train, y_train)
# The decision estimator has an attribute called tree_ which stores the entire
# tree structure and allows access to low level attributes. The binary tree
# tree_ is 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. NOTE:
# Some of the arrays only apply to either leaves or split nodes, resp. In this
# case the values of nodes of the other type are arbitrary!
#
# Among those arrays, we have:
# - left_child, id of the left child of the node
# - right_child, id of the right child of the node
# - feature, feature used for splitting the node
# - threshold, threshold value at the node
#
# Using those arrays, we can parse the tree structure:
n_nodes = estimator.tree_.node_count
children_left = estimator.tree_.children_left
children_right = estimator.tree_.children_right
feature = estimator.tree_.feature
threshold = estimator.tree_.threshold
# The tree structure can be traversed to compute various properties such
# as the depth of each node and whether or not it is a leaf.
node_depth = np.zeros(shape=n_nodes, dtype=np.int64)
is_leaves = np.zeros(shape=n_nodes, dtype=bool)
stack = [(0, -1)] # seed is the root node id and its parent depth
while len(stack) > 0:
node_id, parent_depth = stack.pop()
node_depth[node_id] = parent_depth + 1
# If we have a test node
if (children_left[node_id] != children_right[node_id]):
stack.append((children_left[node_id], parent_depth + 1))
stack.append((children_right[node_id], parent_depth + 1))
else:
is_leaves[node_id] = True
print("The binary tree structure has %s nodes and has "
"the following tree structure:"
% n_nodes)
for i in range(n_nodes):
if is_leaves[i]:
print("%snode=%s leaf node." % (node_depth[i] * "\t", i))
else:
print("%snode=%s test node: go to node %s if X[:, %s] <= %s else to "
"node %s."
% (node_depth[i] * "\t",
i,
children_left[i],
feature[i],
threshold[i],
children_right[i],
))
print()
# First let's retrieve the decision path of each sample. The decision_path
# method allows to retrieve the node indicator functions. A non zero element of
# indicator matrix at the position (i, j) indicates that the sample i goes
# through the node j.
node_indicator = estimator.decision_path(X_test)
# Similarly, we can also have the leaves ids reached by each sample.
leave_id = estimator.apply(X_test)
# Now, it's possible to get the tests that were used to predict a sample or
# a group of samples. First, let's make it for the sample.
sample_id = 0
node_index = node_indicator.indices[node_indicator.indptr[sample_id]:
node_indicator.indptr[sample_id + 1]]
print('Rules used to predict sample %s: ' % sample_id)
for node_id in node_index:
if leave_id[sample_id] == node_id:
continue
if (X_test[sample_id, feature[node_id]] <= threshold[node_id]):
threshold_sign = "<="
else:
threshold_sign = ">"
print("decision id node %s : (X_test[%s, %s] (= %s) %s %s)"
% (node_id,
sample_id,
feature[node_id],
X_test[sample_id, feature[node_id]],
threshold_sign,
threshold[node_id]))
# For a group of samples, we have the following common node.
sample_ids = [0, 1]
common_nodes = (node_indicator.toarray()[sample_ids].sum(axis=0) ==
len(sample_ids))
common_node_id = np.arange(n_nodes)[common_nodes]
print("\nThe following samples %s share the node %s in the tree"
% (sample_ids, common_node_id))
print("It is %s %% of all nodes." % (100 * len(common_node_id) / n_nodes,))
脚本的总运行时间：(0分0.009秒)
