机器学习第五次作业

第一题

高斯贝叶斯分类代码如下：

import numpy as np
def gaussian_bayes(x_0, x_1):
    n = len(data)
    c_0 = 0
    c_1 = 0
    mu_00 = 0
    mu_01 = 0
    mu_10 = 0
    mu_11 = 0
    var_00 = 0
    var_01 = 0
    var_10 = 0
    var_11 = 0
    var_0 = 0
    var_1 = 0

    for d in data:
        if d[2] == 0:
            mu_00 += d[0]
            mu_01 += d[1]
            c_0 += 1
        else:
            mu_10 += d[0]
            mu_11 += d[1]
            c_1 += 1
    mu_00 /= c_0
    mu_01 /= c_0
    mu_10 /= c_1
    mu_11 /= c_1

    for d in data:
        if d[2] == 0:
            var_00 += (d[0] - mu_00) * (d[0] - mu_00)
            var_01 += (d[1] - mu_01) * (d[1] - mu_01)
            var_0 += (d[0] - mu_00) * (d[1] - mu_01)
        else:
            var_10 += (d[0] - mu_10) * (d[0] - mu_10)
            var_11 += (d[1] - mu_11) * (d[1] - mu_11)
            var_1 += (d[0] - mu_10) * (d[1] - mu_11)
    # 协方差
    var_0 = np.array([[var_00, var_0], [var_0, var_01]]) / c_0
    var_1 = np.array([[var_10, var_1], [var_1, var_11]]) / c_1

    x = [x_0, x_1]
    mu_0 = np.array([[mu_00, mu_01]])
    mu_1 = np.array([[mu_10, mu_11]])

    p_0 = 1 / (2 * np.pi ** (2 / 2) * np.sqrt(var_00 * var_01)) * \
          np.exp(-1 / 2 * np.dot(np.dot((x - mu_0), np.linalg.inv(var_0)), (x - mu_0).T))
    p_1 = 1 / (2 * np.pi ** (2 / 2) * np.sqrt(var_10 * var_11)) * \
          np.exp(-1 / 2 * np.dot(np.dot((x - mu_1), np.linalg.inv(var_1)), (x - mu_1).T))
    print("好瓜：", p_1)
    print("坏瓜：", p_0)

高斯贝叶斯分类结果如下：

好瓜： [[0.75333906]]
坏瓜： [[0.30333779]]

高斯朴素贝叶斯分类代码如下：

import numpy as np

data = [[0.697, 0.460, 1], [0.774, 0.376, 1], [0.634, 0.264, 1], [0.608, 0.318, 1], [0.556, 0.215, 1],
        [0.403, 0.237, 1], [0.481, 0.149, 1], [0.437, 0.211, 1],
        [0.666, 0.091, 0], [0.243, 0.267, 0], [0.245, 0.057, 0], [0.343, 0.099, 0], [0.639, 0.161, 0],
        [0.657, 0.198, 0], [0.360, 0.370, 0], [0.593, 0.042, 0], [0.719, 0.103, 0]]

label_1_x1 = [i[0] for i in data if i[2]==1]
label_1_x2 = [i[1] for i in data if i[2]==1]
label_0_x1 = [i[0] for i in data if i[2]==0]
label_0_x2 = [i[1] for i in data if i[2]==0]

# 上面那组
label_0_x1_mean = np.mean(label_0_x1)
label_0_x1_std = np.std(label_0_x1)
label_0_x2_mean = np.mean(label_0_x2)
label_0_x2_std = np.std(label_0_x2)

# 下面那组
label_1_x1_mean = np.mean(label_1_x1)
label_1_x1_std = np.std(label_1_x1)
label_1_x2_mean = np.mean(label_1_x2)
label_1_x2_std = np.std(label_1_x2)

# 先算是bad瓜的概率
p1 = 1/(np.sqrt(2*np.pi)*label_0_x1_std)*np.exp((-(0.5-label_0_x1_mean)**2)/(2*(label_0_x1_std**2)))
p2 = 1/(np.sqrt(2*np.pi)*label_0_x2_std)*np.exp((-(0.3-label_0_x2_mean)**2)/(2*(label_0_x2_std**2)))
print("坏瓜：", p1*p2*(9/17))

# 再算是good瓜的概率
p1x = 1/(np.sqrt(2*np.pi)*label_1_x1_std)*np.exp((-(0.5-label_1_x1_mean)**2)/(2*label_1_x1_std**2))
p2x = 1/(np.sqrt(2*np.pi)*label_1_x2_std)*np.exp((-(0.3-label_1_x2_mean)**2)/(2*label_1_x2_std**2))
print("好瓜：", p1x*p2x*(8/17))

高斯朴素贝叶斯分类结果如下：

坏瓜： 1.6139228137336026
好瓜： 5.312634923114169

第二题

代码如下：

import numpy as np

data = [1.0, 1.3, 2.2, 2.6, 2.8, 5.0, 7.3, 7.4, 7.5, 7.7, 7.9]

# 均值 方差 pc
theta1 = [6, 1, 0.5]
theta2 = [7.5, 1, 0.5]

while True:
    # E-step
    # pxc: P(x|c)
    # pcx: P(c|x) pcx1:
    pc1x_list = []
    pc2x_list = []
    for d in data:
        pxc1 = 1 / np.sqrt(2 * np.pi * theta1[1]) * np.exp(-(d - theta1[0]) ** 2 / (2 * theta1[1]))
        pxc2 = 1 / np.sqrt(2 * np.pi * theta2[1]) * np.exp(-(d - theta2[0]) ** 2 / (2 * theta2[1]))
        pc1x = theta1[2] * pxc1 / (theta1[2] * pxc1 + theta2[2] * pxc2)
        pc2x = theta2[2] * pxc2 / (theta1[2] * pxc1 + theta2[2] * pxc2)
        pc1x_list.append(pc1x)
        pc2x_list.append(pc2x)

    # M-step
    mu1_temp1 = 0
    mu1_temp2 = 0
    mu2_temp1 = 0
    mu2_temp2 = 0
    sigma1_temp1 = 0
    sigma1_temp2 = 0
    sigma2_temp1 = 0
    sigma2_temp2 = 0
    pc1 = 0
    pc2 = 0
    for index, d in enumerate(data):
        # print(index, d)
        # mu
        mu1_temp1 += pc1x_list[index] * d
        mu1_temp2 += pc1x_list[index]
        mu2_temp1 += pc2x_list[index] * d
        mu2_temp2 += pc2x_list[index]

        # sigma
        sigma1_temp1 += pc1x_list[index] * (d - theta1[0]) ** 2
        sigma1_temp2 += pc1x_list[index]
        sigma2_temp1 += pc2x_list[index] * (d - theta2[0]) ** 2
        sigma2_temp2 += pc2x_list[index]

        # pc
        pc1 += pc1x_list[index]
        pc2 += pc2x_list[index]

    theta = theta1[:]

    # 更新均值
    theta1[0] = mu1_temp1 / mu1_temp2
    theta2[0] = mu2_temp1 / mu2_temp2
    # 更新方差
    theta1[1] = sigma1_temp1 / sigma1_temp2
    theta2[1] = sigma2_temp1 / sigma2_temp2
    # 更新P(C)
    theta1[2] = pc1 / len(data)
    theta2[2] = pc2 / len(data)

    if np.abs(theta[2] - theta1[2]) < 1e-7:
        break
print(theta1)
print(theta2)

结果如下：

[2.484129369953009, 1.6917479527139188, 0.5455419134418865]
[7.560020390897808, 0.046398844567893, 0.4544580865581134]

第三题

09, 1.6917479527139188, 0.5455419134418865]
[7.560020390897808, 0.046398844567893, 0.4544580865581134]
```

第三题

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