以机器学习西瓜的好坏为例，写出朴素贝叶斯算法判断西瓜好坏的程序

假设用以下5个特征来描述西瓜：
1. 色泽：青绿（0）、乌黑（1）、浅白（2）
2. 根蒂：蜷缩（0）、稍蜷（1）、硬挺（2）
3. 敲声：清脆（0）、浊响（1）、沉闷（2）
4. 纹理：清晰（0）、稍糊（1）、模糊（2）
5. 脐部：凹陷（0）、稍凹（1）、平坦（2）

好瓜标记为1，坏瓜标记为0。

训练集如下：

|编号 | 色泽 | 根蒂 | 敲声 | 纹理 | 脐部 | 好瓜 |
|:-:|:-:|:-:|:-:|:-:|:-:|:-:|
|1 | 2 | 2 | 0 | 2 | 0 | 0 |
|2 | 2 | 2 | 0 | 2 | 0 | 0 |
|3 | 1 | 2 | 0 | 2 | 0 | 1 |
|4 | 0 | 2 | 0 | 2 | 0 | 1 |
|5 | 0 | 1 | 0 | 2 | 0 | 1 |
|6 | 0 | 0 | 0 | 1 | 0 | 0 |
|7 | 1 | 0 | 0 | 1 | 0 | 1 |
|8 | 2 | 1 | 1 | 1 | 1 | 1 |
|9 | 0 | 0 | 1 | 1 | 1 | 1 |
|10 | 1 | 1 | 0 | 0 | 2 | 1 |
|11 | 0 | 1 | 1 | 0 | 2 | 1 |
|12 | 1 | 1 | 0 | 1 | 2 | 1 |
|13 | 1 | 2 | 2 | 2 | 1 | 0 |
|14 | 2 | 1 | 1 | 0 | 2 | 1 |
|15 | 0 | 1 | 1 | 2 | 1 | 0 |

代码实现如下：

# 好瓜的先验概率
p_good = 0.5
# 坏瓜的先验概率
p_bad = 0.5

# 色泽特征值与标记的映射关系
color_dict = {0: 'qing_lv', 1: 'wu_hei', 2: 'qian_bai'}
# 根蒂特征值与标记的映射关系
root_dict = {0: 'quan_su', 1: 'shao_quan', 2: 'ying_ting'}
# 敲声特征值与标记的映射关系
knock_dict = {0: 'qing_cui', 1: 'zhuo_xiang', 2: 'chen_men'}
# 纹理特征值与标记的映射关系
texture_dict = {0: 'qing_xi', 1: 'shao_hu', 2: 'mo_hu'}
# 脐部特征值与标记的映射关系
navel_dict = {0: 'ao_xian', 1: 'shao_ao', 2: 'ping_tan'}

# 训练集
X_train = [[2, 2, 0, 2, 0],
           [2, 2, 0, 2, 0],
           [1, 2, 0, 2, 0],
           [0, 2, 0, 2, 0],
           [0, 1, 0, 2, 0],
           [0, 0, 0, 1, 0],
           [1, 0, 0, 1, 0],
           [2, 1, 1, 1, 1],
           [0, 0, 1, 1, 1],
           [1, 1, 0, 0, 2],
           [0, 1, 1, 0, 2],
           [1, 1, 0, 1, 2],
           [1, 2, 2, 2, 1],
           [2, 1, 1, 0, 2],
           [0, 1, 1, 2, 1]]

# 训练集样本对应的标记
y_train = [0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0]

# 待判断的西瓜
x_pred = [0, 2, 0, 2, 1]

# 坏瓜的条件概率
p0 = 1
# 好瓜的条件概率
p1 = 1

# 计算坏瓜和好瓜的条件概率
for i in range(len(x_pred)):
    if i == 0:
        p0 *= X_train.count([x_pred[0], X_train[i][1], X_train[i][2], X_train[i][3], X_train[i][4]]) / y_train.count(0)
        p1 *= X_train.count([x_pred[0], X_train[i][1], X_train[i][2], X_train[i][3], X_train[i][4]]) / y_train.count(1)
    elif i == 1:
        p0 *= X_train.count([X_train[i][0], x_pred[1], X_train[i][2], X_train[i][3], X_train[i][4]]) / y_train.count(0)
        p1 *= X_train.count([X_train[i][0], x_pred[1], X_train[i][2], X_train[i][3], X_train[i][4]]) / y_train.count(1)
    elif i == 2:
        p0 *= X_train.count([X_train[i][0], X_train[i][1], x_pred[2], X_train[i][3], X_train[i][4]]) / y_train.count(0)
        p1 *= X_train.count([X_train[i][0], X_train[i][1], x_pred[2], X_train[i][3], X_train[i][4]]) / y_train.count(1)
    elif i == 3:
        p0 *= X_train.count([X_train[i][0], X_train[i][1], X_train[i][2], x_pred[3], X_train[i][4]]) / y_train.count(0)
        p1 *= X_train.count([X_train[i][0], X_train[i][1], X_train[i][2], x_pred[3], X_train[i][4]]) / y_train.count(1)
    else:
        p0 *= X_train.count([X_train[i][0], X_train[i][1], X_train[i][2], X_train[i][3], x_pred[4]]) / y_train.count(0)
        p1 *= X_train.count([X_train[i][0], X_train[i][1], X_train[i][2], X_train[i][3], x_pred[4]]) / y_train.count(1)

# 计算后验概率
posterior_bad = p0 * p_bad
posterior_good = p1 * p_good

# 判断西瓜好坏
if posterior_bad > posterior_good:
    print('坏瓜')
else:
    print('好瓜')