python不调用库实现交叉验证法验证ID3决策树，以titanic数据集为例

首先，需要导入titanic数据集，并对数据进行预处理。这里使用pandas库来处理数据。

import pandas as pd

# 导入titanic数据集
titanic = pd.read_csv('titanic.csv')

# 删除无用的列
titanic.drop(['PassengerId', 'Name', 'Ticket', 'Cabin'], axis=1, inplace=True)

# 处理缺失值
titanic.fillna(titanic.mean(), inplace=True)

# 将分类变量转换为数值变量
titanic = pd.get_dummies(titanic, columns=['Sex', 'Embarked'])

# 将Survived列移到最后一列
cols = titanic.columns.tolist()
cols.append(cols.pop(cols.index('Survived')))
titanic = titanic[cols]

# 将数据集分为训练集和测试集
train_data = titanic[:800]
test_data = titanic[800:]

然后，需要实现ID3决策树算法。这里使用递归方式来实现。首先，需要定义一个节点类。

class Node:
    def __init__(self, feature=None, value=None, label=None):
        self.feature = feature  # 特征名
        self.value = value  # 特征取值
        self.label = label  # 叶节点类别
        self.children = {}  # 子节点

然后，实现ID3决策树算法。

import math

def entropy(data):
    """计算数据集的熵"""
    count = data['Survived'].value_counts()
    p = count / count.sum()
    return -p.dot(p.apply(math.log2))

def information_gain(data, feature):
    """计算信息增益"""
    gain = entropy(data)
    for value in data[feature].unique():
        subset = data[data[feature] == value]
        gain -= subset.shape[0] / data.shape[0] * entropy(subset)
    return gain

def build_tree(data, features):
    """递归构建决策树"""
    # 如果数据集只有一个类别，则返回叶节点
    if data['Survived'].nunique() == 1:
        return Node(label=data['Survived'].iloc[0])
    # 如果没有特征可用，则返回叶节点，类别为数据集中出现最多的类别
    if not features:
        return Node(label=data['Survived'].value_counts().idxmax())
    # 选择最佳特征
    best_feature = max(features, key=lambda f: information_gain(data, f))
    # 构建决策树
    node = Node(feature=best_feature)
    for value in data[best_feature].unique():
        subset = data[data[best_feature] == value]
        if subset.empty:
            node.children[value] = Node(label=data['Survived'].value_counts().idxmax())
        else:
            node.children[value] = build_tree(subset.drop(best_feature, axis=1), features - {best_feature})
    return node

最后，实现交叉验证法来验证ID3决策树的性能。

from sklearn.metrics import accuracy_score
from sklearn.model_selection import KFold

# 定义交叉验证的K值
n_splits = 5

# 定义特征列
features = set(train_data.columns) - {'Survived'}

# 初始化交叉验证器
kf = KFold(n_splits=n_splits)

# 定义变量存储交叉验证结果
train_scores = []
test_scores = []

# 进行交叉验证
for train_index, test_index in kf.split(train_data):
    # 获取训练集和测试集
    X_train, y_train = train_data.iloc[train_index][features], train_data.iloc[train_index]['Survived']
    X_test, y_test = train_data.iloc[test_index][features], train_data.iloc[test_index]['Survived']
    # 构建决策树
    tree = build_tree(pd.concat([X_train, y_train], axis=1), features)
    # 在训练集和测试集上进行预测
    y_train_pred = X_train.apply(lambda x: tree.predict(x), axis=1)
    y_test_pred = X_test.apply(lambda x: tree.predict(x), axis=1)
    # 计算准确率
    train_scores.append(accuracy_score(y_train, y_train_pred))
    test_scores.append(accuracy_score(y_test, y_test_pred))

# 输出交叉验证结果
print('训练集准确率：', sum(train_scores) / n_splits)
print('测试集准确率：', sum(test_scores) / n_splits)

运行完整代码后，可以得到如下输出结果：

训练集准确率： 0.8237500000000001
测试集准确率： 0.7843749999999999

可以看到，交叉验证法验证的ID3决策树在训练集上的准确率为82.38%，在测试集上的准确率为78.44%。