Question

我目前正在研究决策树，并且已经实现了自己的决策树。我使用的数据集是电话应用程序的。它具有以下功能名称：

应用名称，类别，等级，评论，大小，安装，类型，价格，内容分级，类型，成功

“成功”基于评分

我的X有：类别，评论，大小，安装，类型，价格，内容分级，类型

我的Y是：成功

目标是构建决策树，并制作一个简短的程序来读取每个节点，然后基于该节点询问用户信息。例如，第一个节点是“评论”，因此程序将提示用户输入评论数。然后读取第二个节点，即Type，然后提示用户输入应用程序的Type等。

我的代码很长，因此出于示例的原因，我将在此处使用代码：https://scikit-learn.org/stable/auto_examples/tree/plot_unveil_tree_structure.html

我的问题是：给定决策树，如何根据用户输入遍历树？

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,))

我正在考虑使用for循环遍历节点，但是我不确定从哪里开始。

根据用户输入遍历决策树

0 个答案: