fly6516
1d2138e0d2
- 实现了 C4.5 决策树算法，包括信息熵、信息增益、树的构建和预测等功能 - 添加了实验脚本，使用 Iris 和 Wine Quality 数据集进行性能比较- 生成了实验报告，总结了 C4.5 算法与 Logistic 回归的性能差异
88 lines
3.7 KiB
Python
88 lines
3.7 KiB
Python
from collections import Counter
|
|
|
|
import numpy as np
|
|
from sklearn.base import BaseEstimator, ClassifierMixin # 导入BaseEstimator和ClassifierMixin
|
|
|
|
class C45DecisionTree(BaseEstimator, ClassifierMixin): # 继承BaseEstimator和ClassifierMixin
|
|
def __init__(self):
|
|
self.tree = None # 初始化决策树结构为空
|
|
|
|
def entropy(self, y):
|
|
"""
|
|
计算信息熵 (Entropy)
|
|
:param y: 标签列表
|
|
:return: 信息熵值
|
|
"""
|
|
counts = np.bincount(y) # 统计每个类别的数量
|
|
probabilities = counts / len(y) # 计算每个类别的概率
|
|
return -np.sum([p * np.log2(p) for p in probabilities if p > 0]) # 使用信息熵公式计算熵值
|
|
|
|
def information_gain(self, X, y, feature_index):
|
|
"""
|
|
计算信息增益 (Information Gain)
|
|
:param X: 特征矩阵
|
|
:param y: 标签列表
|
|
:param feature_index: 当前特征索引
|
|
:return: 信息增益值
|
|
"""
|
|
total_entropy = self.entropy(y) # 计算总熵
|
|
values, counts = np.unique(X[:, feature_index], return_counts=True) # 获取当前特征的唯一值及其数量
|
|
weighted_entropy = sum((counts[i] / len(y)) * self.entropy(y[X[:, feature_index] == value])
|
|
for i, value in enumerate(values)) # 计算加权熵
|
|
return total_entropy - weighted_entropy # 信息增益等于总熵减去加权熵
|
|
|
|
def fit(self, X, y):
|
|
"""
|
|
构建决策树
|
|
:param X: 特征矩阵
|
|
:param y: 标签列表
|
|
"""
|
|
self.tree = self._build_tree(X, y) # 调用递归函数构建决策树
|
|
return self # 返回自身以符合scikit-learn的接口规范
|
|
|
|
def _build_tree(self, X, y):
|
|
"""
|
|
递归构建决策树
|
|
:param X: 特征矩阵
|
|
:param y: 标签列表
|
|
:return: 决策树节点
|
|
"""
|
|
if len(np.unique(y)) == 1: # 如果所有样本属于同一类别，则返回该类别
|
|
return y[0]
|
|
if X.shape[1] == 0: # 如果没有剩余特征，则返回多数类别
|
|
return Counter(y).most_common(1)[0][0]
|
|
|
|
# 选择信息增益最大的特征
|
|
best_feature = np.argmax([self.information_gain(X, y, i) for i in range(X.shape[1])])
|
|
values = np.unique(X[:, best_feature]) # 获取当前特征的唯一值
|
|
tree = {best_feature: {}} # 构建树节点，使用字典表示
|
|
for value in values:
|
|
sub_X = X[X[:, best_feature] == value] # 划分子集，获取当前特征值对应的样本
|
|
sub_y = y[X[:, best_feature] == value] # 获取对应的标签
|
|
tree[best_feature][value] = self._build_tree(sub_X, sub_y) # 递归构建子树
|
|
return tree
|
|
|
|
def predict_sample(self, tree, x):
|
|
"""
|
|
预测单个样本
|
|
:param tree: 决策树
|
|
:param x: 单个样本
|
|
:return: 预测类别
|
|
"""
|
|
if not isinstance(tree, dict): # 如果是叶子节点，直接返回类别
|
|
return tree
|
|
feature = list(tree.keys())[0] # 获取当前节点的特征
|
|
value = x[feature] # 获取样本在该特征上的值
|
|
subtree = tree[feature].get(value) # 获取对应的子树
|
|
if subtree is None: # 如果子树不存在，返回多数类别
|
|
return Counter(x).most_common(1)[0][0]
|
|
return self.predict_sample(subtree, x) # 递归预测
|
|
|
|
def predict(self, X):
|
|
"""
|
|
预测多个样本
|
|
:param X: 特征矩阵
|
|
:return: 预测类别列表
|
|
"""
|
|
predictions = [self.predict_sample(self.tree, x) for x in X] # 对每个样本调用predict_sample方法
|
|
return np.array(predictions, dtype=int) # 确保返回的预测结果是整数类型 |