python实现id3决策树代码连续值

### 实现支持连续值的ID3决策树的Python代码 以下是一个支持连续值的ID3决策树的Python实现。该代码扩展了传统的ID3算法，使其能够处理连续特征。通过寻找最佳分割点来处理连续值。 ```python import numpy as np import pandas as pd from collections import Counter class ID3DecisionTree: def __init__(self, min_samples_split=2, max_depth=10): self.min_samples_split = min_samples_split self.max_depth = max_depth self.tree = None def entropy(self, y): hist = np.bincount(y) ps = hist / len(y) return -np.sum([p * np.log2(p) for p in ps if p > 0]) def best_split(self, X, y): best_gain = -1 split_idx, split_value = None, None _, n_features = X.shape base_entropy = self.entropy(y) for idx in range(n_features): values = X[:, idx] thresholds = np.unique(values) for threshold in thresholds: dy = np.array([1 if v <= threshold else 0 for v in values]) left_indices = np.where(dy == 1)[0] right_indices = np.where(dy == 0)[0] if len(left_indices) == 0 or len(right_indices) == 0: continue p = len(left_indices) / len(y) gain = base_entropy - p * self.entropy(y[left_indices]) - (1 - p) * self.entropy(y[right_indices]) if gain > best_gain: best_gain = gain split_idx = idx split_value = threshold return best_gain, split_idx, split_value def build_tree(self, X, y, depth=0): n_samples, n_features = X.shape if depth >= self.max_depth or n_samples < self.min_samples_split or len(np.unique(y)) == 1: leaf_value = Counter(y).most_common(1)[0][0] return leaf_value best_gain, split_idx, split_value = self.best_split(X, y) if best_gain == -1: leaf_value = Counter(y).most_common(1)[0][0] return leaf_value left_indices = np.where(X[:, split_idx] <= split_value)[0] right_indices = np.where(X[:, split_idx] > split_value)[0] question = "{} <= {}".format(split_idx, split_value) subtree = {question: []} left_tree = self.build_tree(X[left_indices, :], y[left_indices], depth + 1) right_tree = self.build_tree(X[right_indices, :], y[right_indices], depth + 1) if left_tree == right_tree: subtree = left_tree else: subtree[question].append(left_tree) subtree[question].append(right_tree) return subtree def fit(self, X, y): self.tree = self.build_tree(X, y) def predict_sample(self, x, tree): if not isinstance(tree, dict): return tree question = list(tree.keys())[0] feature_idx, comparison_operator, value = self.parse_question(question) if x[feature_idx] <= float(value): answer = tree[question][0] else: answer = tree[question][1] if not isinstance(answer, dict): return answer else: return self.predict_sample(x, answer) def parse_question(self, question): feature_idx, value = question.split(" <= ") return int(feature_idx), "<=", float(value) def predict(self, X): return [self.predict_sample(x, self.tree) for x in X] ``` 此代码中，`best_split` 函数用于寻找最佳分割点以处理连续值。通过计算信息增益[^1]，选择最佳特征和对应的分割值。构建树的过程递归地进行，直到达到最大深度或节点中的样本数不足以继续分裂[^4]。 ###