不使用sklearn库实现以上问题

以下是不使用sklearn库的解决方案：

1. 逻辑回归：

首先，定义sigmoid函数：

def sigmoid(z):
    return 1 / (1 + np.exp(-z))

接着，定义代价函数和梯度下降算法：

def cost_function(theta, X, y):
    m = len(y)
    h = sigmoid(X @ theta)
    cost = (1 / m) * np.sum(-y * np.log(h) - (1 - y) * np.log(1 - h))
    grad = (1 / m) * (X.T @ (h - y))
    return cost, grad

def gradient_descent(X, y, theta, alpha, num_iters):
    m = len(y)
    J_history = []
    for i in range(num_iters):
        cost, grad = cost_function(theta, X, y)
        theta = theta - (alpha * grad)
        J_history.append(cost)
    return theta, J_history

最后，将数据集拆分为训练集和测试集，然后执行以下代码：

# 构建特征矩阵和标签向量
X = np.column_stack((np.ones(m), data[:, :-1]))
y = data[:, -1]

# 初始化参数
initial_theta = np.zeros(X.shape[1])

# 训练模型
theta, J_history = gradient_descent(X_train, y_train, initial_theta, 0.01, 1000)

# 预测测试集
y_pred = sigmoid(X_test @ theta)

# 计算准确率
accuracy = np.mean((y_pred >= 0.5) == y_test)
print(f"Accuracy: {accuracy}")

2. 线性回归：

首先，定义代价函数和梯度下降算法：

def cost_function(theta, X, y):
    m = len(y)
    h = X @ theta
    cost = (1 / (2 * m)) * np.sum((h - y) ** 2)
    grad = (1 / m) * (X.T @ (h - y))
    return cost, grad

def gradient_descent(X, y, theta, alpha, num_iters):
    m = len(y)
    J_history = []
    for i in range(num_iters):
        cost, grad = cost_function(theta, X, y)
        theta = theta - (alpha * grad)
        J_history.append(cost)
    return theta, J_history

最后，将数据集拆分为训练集和测试集，然后执行以下代码：

# 构建特征矩阵和标签向量
X = np.column_stack((np.ones(m), data[:, :-1]))
y = data[:, -1]

# 初始化参数
initial_theta = np.zeros(X.shape[1])

# 训练模型
theta, J_history = gradient_descent(X_train, y_train, initial_theta, 0.01, 1000)

# 预测测试集
y_pred = X_test @ theta

# 计算均方误差
mse = np.mean((y_pred - y_test) ** 2)
print(f"MSE: {mse}")