• logistic 回归(线性和非线性)


    一:线性logistic 回归

    代码如下:

    import numpy as np
    import pandas as pd
    import matplotlib.pyplot as plt
    import scipy.optimize as opt
    import seaborn as sns
    
    #读取数据集
    path = 'ex2data1.txt'
    data = pd.read_csv(path, header=None, names=['Exam 1', 'Exam 2', 'Admitted'])
    
    #将正负数据集分开
    positive = data[data['Admitted'].isin([1])]
    negative = data[data['Admitted'].isin([0])]
    
    '''
    #查看分布
    fig, ax = plt.subplots(figsize=(12, 8))
    ax.scatter(positive['Exam 1'], positive['Exam 2'], s=60, c='b', marker='o', label='Admitted')
    ax.scatter(negative['Exam 1'], negative['Exam 2'], s=50, c='r', marker='x', label='UnAdmitted')
    ax.legend()
    ax.set_xlabel('Exam 1 Score')
    ax.set_ylabel('Exam 2 Score')
    plt.show()
    '''
    
    #sigmoid函数实现
    def sigmoid(h):
        return 1 / (1 + np.exp(-h))
    
    
    '''
    #测试sigmoid函数
    nums = np.arange(-10, 11, step=1)
    fig, ax = plt.subplots(figsize=(12, 8))
    ax.plot(nums, sigmoid(nums), 'k')
    plt.show()
    '''
    
    #计算损失函数值
    def cost(theta, X, y):
        theta = np.matrix(theta)
        X = np.matrix(X)
        y = np.matrix(y)
    
        part1 = np.multiply(-y, np.log(sigmoid(X * theta.T)))
        part2 = np.multiply((1-y), np.log(1-sigmoid(X * theta.T)))
        return np.sum(part1-part2) / len(X)
    
    #在原矩阵第1列前加一列全1
    data.insert(0, 'ones', 1)
    
    cols = data.shape[1]
    
    X = data.iloc[:, 0:cols-1]
    y = data.iloc[:, cols-1:cols]
    
    X = np.array(X.values)
    y = np.array(y.values)
    theta = np.zeros(3) #这里是一个行向量
    
    
    #返回梯度向量,注意是向量
    def gradient(theta, X, y):
        theta = np.matrix(theta)
        X = np.matrix(X)
        y = np.matrix(y)
    
        parameters = theta.ravel().shape[1]
        grad = np.zeros(parameters)
    
        error = sigmoid(X * theta.T) - y
    
        grad = error.T.dot(X)
        grad = grad / len(X)
        return grad
    
    #通过高级算法计算出最好的theta值
    result = opt.fmin_tnc(func=cost, x0=theta, fprime=gradient, args=(X, y))
    
    #print(cost(result[0], X, y))
    
    #测试所得theta的性能
    #计算原数据集的预测情况
    def predict(theta, X):
        theta = np.matrix(theta)
        X = np.matrix(X)
    
        probability = sigmoid(X * theta.T)
        return [1 if i > 0.5 else 0 for i in probability]
    
    
    theta_min = result[0]
    predictions = predict(theta_min, X)
    
    correct = [1 if((a == 1 and b == 1) or(a == 0 and b == 0)) else 0 for(a, b) in zip(predictions, y)]
    accuracy = (sum(map(int, correct)) % len(correct))
    print('accuracy = {0}%'.format(accuracy))#训练集测试准确度89%
    
    
    # 作图
    theta_temp = theta_min
    theta_temp = theta_temp / theta_temp[2]
    
    x = np.arange(130, step=0.1)
    y = -(theta_temp[0] + theta_temp[1] * x)
    #画出原点
    sns.set(context='notebook', style='ticks', font_scale=1.5)
    sns.lmplot('Exam 1', 'Exam 2', hue='Admitted', data=data,
               size=6,
               fit_reg=False,
               scatter_kws={"s": 25}
               )
    #画出分界线
    plt.plot(x, y, 'grey')
    plt.xlim(0, 130)
    plt.ylim(0, 130)
    plt.title('Decision Boundary')
    plt.show()

    二:非线性logistic 回归(正则化)

    代码如下:

    import pandas as pd
    import numpy as np
    import scipy.optimize as opt
    import matplotlib.pyplot as plt
    
    
    path = 'ex2data2.txt'
    data = pd.read_csv(path, header=None, names=['Test 1', 'Test 2', 'Accepted'])
    
    positive = data[data['Accepted'].isin([1])]
    negative = data[data['Accepted'].isin([0])]
    
    '''
    #显示原始数据的分布
    fig, ax = plt.subplots(figsize=(12, 8))
    ax.scatter(positive['Test 1'], positive['Test 2'], s=50, c='b', marker='o', label='Accepted')
    ax.scatter(negative['Test 1'], negative['Test 2'], s=50, c='r', marker='x', label='Unaccepted')
    ax.legend() #显示右上角的Accepted 和 Unaccepted标签
    ax.set_xlabel('Test 1 Score')
    ax.set_ylabel('Test 2 Score')
    plt.show()
    '''
    degree = 5
    x1 = data['Test 1']
    x2 = data['Test 2']
    #在data的第三列插入一列全1
    data.insert(3, 'Ones', 1)
    
    #创建多项式特征值,最高阶为4
    for i in range(1, degree):
        for j in range(0, i):
            data['F' + str(i) + str(j)] = np.power(x1, i-j) * np.power(x2, j)
    
    #删除原数据中的test 1和test 2两列
    data.drop('Test 1', axis=1, inplace=True)
    data.drop('Test 2', axis=1, inplace=True)
    
    
    #sigmoid函数实现
    def sigmoid(h):
        return 1 / (1 + np.exp(-h))
    
    
    def cost(theta, X, y, learnRate):
        theta = np.matrix(theta)
        X = np.matrix(X)
        y = np.matrix(y)
    
        first = np.multiply(-y, np.log(sigmoid(X * theta.T)))
        second = np.multiply((1 - y), np.log(1 - sigmoid(X * theta.T)))
        reg = (learnRate / (2 * len(X))) * np.sum(np.power(theta[:, 1:theta.shape[1]], 2))
        return np.sum(first - second) / len(X) + reg
    
    
    learnRate = 1
    cols = data.shape[1]
    
    X = data.iloc[:, 1:cols]
    y = data.iloc[:, 0:1]
    
    X = np.array(X)
    y = np.array(y)
    theta = np.zeros(X.shape[1])
    
    
    #计算原数据集的预测情况
    def predict(theta, X):
        theta = np.matrix(theta)
        X = np.matrix(X)
    
        probability = sigmoid(X * theta.T)
        return [1 if i > 0.5 else 0 for i in probability]
    
    
    def gradientReg(theta, X, y, learnRate):
        theta = np.matrix(theta)
        X = np.matrix(X)
        y = np.matrix(y)
    
        paramates = int(theta.ravel().shape[1])
        grad = np.zeros(paramates)
    
        grad = (sigmoid(X * theta.T) - y).T * X / len(X) + (learnRate / len(X)) * theta[:, i]
        grad[0] = grad[0] - (learnRate / len(X)) * theta[:, i]
        return grad
    
    result = opt.fmin_tnc(func=cost, x0=theta, fprime=gradientReg, args=(X, y, learnRate))
    print(result)
    
    theta_min = np.matrix(result[0])
    predictions = predict(theta_min, X)
    correct = [1 if((a == 1 and b == 1) or(a == 0 and b == 0)) else 0 for(a, b) in zip(predictions, y)]
    accuracy = (sum(map(int, correct)) % len(correct))
    
    print('accuracy = {0}%'.format(accuracy))
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  • 原文地址:https://www.cnblogs.com/qiang-wei/p/9839458.html
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