python代码利用ID3算法实现对葡萄酒数据集的分类

好的，以下是利用ID3算法实现对葡萄酒数据集的分类的Python代码：

import pandas as pd
import numpy as np
from math import log2

# 计算信息熵
def calc_entropy(data):
    n = len(data)
    label_counts = {}
    for feat_vec in data:
        current_label = feat_vec[-1]
        if current_label not in label_counts.keys():
            label_counts[current_label] = 0
        label_counts[current_label] += 1
    entropy = 0.0
    for key in label_counts:
        prob = float(label_counts[key]) / n
        entropy += -prob * log2(prob)
    return entropy

# 划分数据集
def split_data(data, axis, value):
    ret_data = []
    for feat_vec in data:
        if feat_vec[axis] == value:
            reduced_feat_vec = feat_vec[:axis]
            reduced_feat_vec.extend(feat_vec[axis+1:])
            ret_data.append(reduced_feat_vec)
    return ret_data

# 选择最佳划分特征
def choose_best_feature_to_split(data):
    num_features = len(data[0]) - 1
    base_entropy = calc_entropy(data)
    best_info_gain = 0.0
    best_feature = -1
    for i in range(num_features):
        feat_list = [example[i] for example in data]
        unique_vals = set(feat_list)
        new_entropy = 0.0
        for value in unique_vals:
            sub_data = split_data(data, i, value)
            prob = len(sub_data) / float(len(data))
            new_entropy += prob * calc_entropy(sub_data)
        info_gain = base_entropy - new_entropy
        if info_gain > best_info_gain:
            best_info_gain = info_gain
            best_feature = i
    return best_feature

# 定义叶子节点
def majority_cnt(class_list):
    class_count = {}
    for vote in class_list:
        if vote not in class_count.keys():
            class_count[vote] = 0
        class_count[vote] += 1
    sorted_class_count = sorted(class_count.items(), key=lambda x: x[1], reverse=True)
    return sorted_class_count[0][0]

# 创建决策树
def create_tree(data, labels):
    class_list = [example[-1] for example in data]
    if class_list.count(class_list[0]) == len(class_list):
        return class_list[0]
    if len(data[0]) == 1:
        return majority_cnt(class_list)
    best_feat = choose_best_feature_to_split(data)
    best_feat_label = labels[best_feat]
    my_tree = {best_feat_label: {}}
    del(labels[best_feat])
    feat_values = [example[best_feat] for example in data]
    unique_vals = set(feat_values)
    for value in unique_vals:
        sub_labels = labels[:]
        my_tree[best_feat_label][value] = create_tree(split_data(data, best_feat, value), sub_labels)
    return my_tree

# 测试决策树
def classify(input_tree, feat_labels, test_vec):
    first_str = list(input_tree.keys())[0]
    second_dict = input_tree[first_str]
    feat_index = feat_labels.index(first_str)
    for key in second_dict.keys():
        if test_vec[feat_index] == key:
            if type(second_dict[key]).__name__ == 'dict':
                class_label = classify(second_dict[key], feat_labels, test_vec)
            else:
                class_label = second_dict[key]
    return class_label

# 加载数据集
def load_dataset():
    df = pd.read_csv('wine.csv')
    data = np.array(df)
    labels = df.columns.tolist()
    return data, labels

# 主函数
if __name__ == '__main__':
    data, labels = load_dataset()
    my_tree = create_tree(data.tolist(), labels)
    print(my_tree)
    test_vec = [1, 13.05, 3.86, 2.32, 22.5, 85, 1.65, 1.59, 0.61, 2.87, 580]
    class_label = classify(my_tree, labels, test_vec)
    print(class_label)

在这个例子中，我们使用了葡萄酒数据集。数据集包含13个特征和1个类别标签，我们使用ID3算法构建决策树，并对一个测试样本进行分类。