[实验1回归分析]

一、预备知识

使用梯度下降法求解多变量回归问题数据集

Iris 鸢尾花数据集是一个经典数据集，在统计学习和机器学习领域都经常被用作示例。数据集内包含 3 类共 150 条记录，每类各 50 个数据，每条记录都有 4 项特征：花萼长度、花萼宽度、花瓣长度、花瓣宽度，可以通过这4个特征预测鸢尾花卉属于（iris-setosa, iris-versicolour, iris-virginica）中的哪一品种。

(资料图片仅供参考)

sepal length:花萼长度，单位cm；sepal width:花萼宽度，单位cmpetal length:花瓣长度，单位cm；petal width:花瓣宽度，单位cm

种类:setosa(山鸢尾)，versicolor(杂色鸢尾)，virginica(弗吉尼亚鸢尾)

我们用100组数据作为训练数据iris-train.txt，50组数据作为测试数据iris-test.txt。

"setosa" "versicolor" "virginica"

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二、实验目的

掌握梯度下降法的原理及设计；

掌握利用梯度下降法解决回归类问题。

三、实验内容

数据读取：（前4列是特征x，第5列是标签y）

import numpy as nptrain_data  = np.loadtxt("iris-train.txt", delimiter="\t")x_train=np.array(train_data[:,:4])y_train=np.array(train_data[:,4])

设计多变量回归分析方法，利用训练数据iris-train.txt求解参数

思路：首先确定h（x）函数，由于是4个特征，我们可以取5个参数

设计逻辑回归分析方法，利用训练数据iris-train.txt求解参数

思路：首先确定h（x）函数，由于是4个特征，我们可以取5个参数

利用测试数据iris-test.txt，统计逻辑回归对测试数据的识别正确率 ACC，Precision，Recall。

四、操作方法和实验步骤

​1.对于多变量回归分析：

\[\alpha \frac{\partial}{\partial \theta _j}L(\theta _0,\theta _1...\theta _n)=\frac { \alpha } { m } X^T(X\theta-y)\]

​（1）实现损失函数、h(x)、梯度下降函数等：

def hypothesis(X, theta):    return np.dot(X, theta)def cost_function(X, y, theta):    m = len(y)    J = np.sum((hypothesis(X, theta) - y) ** 2) / (2 * m)    return Jdef gradient_descent(X, y, theta, alpha, num_iters):    m = len(y)    J_history = np.zeros(num_iters)    for i in range(num_iters):        theta = theta - (alpha / m) * np.dot(X.T, (hypothesis(X, theta) - y))        J_history[i] = cost_function(X, y, theta)    return theta, J_history

​ （2）处理输出X并开始计算

X_train_hat= np.hstack((np.ones((X_train.shape[0], 1)), X_train))X_train= X_train_hattheta = np.ones(X_train.shape[1])alpha = 0.0005num_iters = 120theta, J_history = gradient_descent(X_train, y_train, theta, alpha, num_iters)print("Parameter vector: ", theta)print("Final cost: ", J_history[-1])

对于逻辑回归分析

import numpy as npfrom sklearn.metrics import classification_reportfrom sklearn.preprocessing import LabelEncodertrain_data = np.loadtxt("iris/iris-train.txt", delimiter="\t")X_train = train_data[:, 0:4]y_train = train_data[:, 4]encoder = LabelEncoder()y_train = encoder.fit_transform(y_train)  # Convert class labels to 0, 1, 2X_train_hat = np.hstack((np.ones((X_train.shape[0], 1)), X_train))X_train = X_train_hatdef sigmoid(z):    return 1 / (1 + np.exp(-z))def cost_function(X, y, theta):    m = len(y)    epsilon = 1e-5    h = sigmoid(np.dot(X, theta))    J = (-1 / m) * np.sum(y * np.log(h+epsilon) + (1 - y) * np.log(1 - h+epsilon))    grad = (1 / m) * np.dot(X.T, (h - y))    return J, graddef gradient_descent(X, y, theta, alpha, num_iters):    m = len(y)    J_history = np.zeros(num_iters)    for i in range(num_iters):        J, grad = cost_function(X, y, theta)        theta = theta - (alpha * grad)        J_history[i] = J    return theta, J_historydef logistic_regression(X, y, alpha, num_iters):    theta = np.ones((X_train.shape[1], 3))    # Perform gradient descent to minimize the cost function for each class    for i in range(3):        y_train_i = (y_train == i).astype(int)        theta_i, J_history_i = gradient_descent(X_train, y_train_i, theta[:, i], alpha, num_iters)        import matplotlib.pyplot as plt        plt.plot(J_history_i)        plt.xlabel("Iterations")        plt.ylabel("Cost")        plt.title("Cost function"+str(i))        plt.show()        theta[:, i] = theta_i        print(f"Class {i} parameter vector: {theta_i}")        print(f"Class {i} final cost: {J_history_i[-1]}")    return thetadef predict(test_data, theta):    X_test = test_data[:, 0:4]    X_test_hat = np.hstack((np.ones((X_test.shape[0], 1)), X_test))    X_test = X_test_haty_test = test_data[:, 4]    y_test = encoder.transform(y_test)  # Convert class labels to 0, 1, 2    y_pred = np.zeros(X_test.shape[0])    for i in range(3):        sigmoid_outputs = sigmoid(np.dot(X_test, theta[:, i]))        threshold = np.median(sigmoid_outputs[y_test == i])#不采用统一的阈值，而是采用每个类别的中位数作为阈值        y_pred_i = (sigmoid_outputs >= threshold).astype(int)        y_pred[y_pred_i == 1] = i    y_pred = encoder.inverse_transform(y_pred.astype(int))  # Convert class labels back to original values    print(classification_report(y_test, y_pred))test_data = np.loadtxt("iris/iris-test.txt", delimiter="\t")theta = logistic_regression(X_train, y_train, 0.0005, 5000)predict(test_data, theta)

实验结果和分析

1.对于多变量回归分析的实验结果：

代价曲线：

那么计算函数也就是：

\[y=-0.306x_1+0.0653x_2+0.280x_3+0.697x_4-0.713\]

​ 2.对于逻辑回归分析的实验结果

代价曲线：

​

​ 利用测试数据iris-test.txt，统计逻辑回归对测试数据的识别正确率 ACC，Precision，Recall：

​ 通过sigmoid函数实现的逻辑回归只能处理二分类，我是通过分别计算三个二分类实现的。【猜测：预测1时把0和2归为同一类会对模型产生误导】

​ 由于实现方法是One VS One.效果并不是很好，而且得到如下结果还是在处理sigmoid函数结果时还修改了原本0.5的阈值，有泄露结果的嫌疑。

（不采用统一的阈值，而是采用每个类别的h函数结果中位数作为阈值）

五、思考题

给出 RANSAC 线性拟合的 python 实现

import randomimport numpy as npfrom matplotlib import pyplot as pltdef ransac_line_fit(data, n_iterations, threshold):    """    RANSAC algorithm for linear regression.    Parameters:    data (list): list of tuples representing the data points    n_iterations (int): number of iterations to run RANSAC    threshold (float): maximum distance a point can be from the line to be considered an inlier    Returns:    tuple: slope and y-intercept of the best fit line    """    best_slope, best_intercept = None, None    best_inliers = []    for i in range(n_iterations):        # Randomly select two points from the data        sample = random.sample(data, 2)        x1, y1 = sample[0]        x2, y2 = sample[1]        # Calculate slope and y-intercept of line connecting the two points        slope = (y2 - y1) / (x2 - x1)        intercept = y1 - slope * x1        # Find inliers within threshold distance of the line        inliers = []        outliers = []        for point in data:            x, y = point            distance = abs(y - (slope * x + intercept))            distance = distance / np.sqrt(slope ** 2 + 1)            if distance <= threshold:                inliers.append(point)            else:                outliers.append(point)        # If the number of inliers is greater than the current best, update the best fit line        if len(inliers) > len(best_inliers):            best_slope = slope            best_intercept = intercept            best_inliers = inliers    outliers = [point for point in data if point not in best_inliers]    # Plot the data points, best fit line, and inliers and outliers    fig, ax = plt.subplots()    # ax.scatter([x for x, y in data], [y for x, y in data], color="black")    ax.scatter([x for x, y in best_inliers], [y for x, y in best_inliers], color="green")    ax.scatter([x for x, y in outliers], [y for x, y in outliers], color="black")    x_vals = np.array([-5,5])    y_vals = best_slope * x_vals + best_intercept    ax.plot(x_vals, y_vals, "-", color="red")    # threshold_line = best_slope * x_vals + best_intercept + threshold*np.sqrt((1/best_slope) ** 2 + 1)    threshold_line = best_slope * x_vals + best_intercept + threshold * np.sqrt(best_slope ** 2 + 1)    ax.plot(x_vals, threshold_line, "--", color="blue")    # threshold_line = best_slope * x_vals + best_intercept - threshold*np.sqrt((1/best_slope) ** 2 + 1)    threshold_line = best_slope * x_vals + best_intercept - threshold * np.sqrt(best_slope ** 2 + 1)    ax.plot(x_vals, threshold_line, "--", color="blue")    # ax.set_xlim([-10, 10])    ax.set_ylim([-6, 6])    plt.show()    return best_slope, best_interceptimport numpy as np# Generate 10 random points with x values between 0 and 10 and y values between -5 and 5data = [(x, y) for x, y in zip(np.random.uniform(-5, 5, 10), np.random.uniform(-5, 5, 10))]print(data)# Fit a line to the data using RANSACslope, intercept = ransac_line_fit(data, 10000, 1)print(slope, intercept)

结果：