基于 Decision Tree ID3 算法实现贷款风险审批

    1. import numpy as np
    2. import pandas as pd
    1. data = pd.read_csv('dataset/loans.csv').sample(40000)
    2. display(data.sample(10))
    3. display(data.shape)
    4. # grade: 贷款级别
    5. # sub_grade: 贷款细分级别
    6. # short_emp: 一年以内短期雇佣
    7. # emp_length_num:　受雇年限
    8. # home_ownership:居住状态（自有，按揭，租住）
    9. # dti：贷款占收入比例
    10. # purpose:贷款用途
    11. # term:贷款周期
    12. # last_delinq_none:贷款申请人是否有不良记录　
    13. # last_major_derog_none:贷款申请人是否有还款逾期90天以上记录
    14. # reｖol_util：透支额度占信用比例
    15. # total_rec_late_fee:逾期罚款总额
    16. # safe_loans:贷款是否安全
    2.         grade    sub_grade    short_emp    emp_length_num    home_ownership    dti        purpose                term        last_delinq_none    last_major_derog_none    revol_util    total_rec_late_fee    safe_loans
    3. 19847    C        C3            0            11                MORTGAGE        20.18    debt_consolidation    36 months    1                    1                        80.2        0.0000                1
    4. 24186    D        D2            0            5                RENT            21.19    debt_consolidation    36 months    0                    1                        47.6        0.0000                1
    5. 40013    B        B5            0            11                MORTGAGE        30.90    debt_consolidation    36 months    0                    1                        90.1        0.0000                -1
    6. 5405    B        B1            1            1                MORTGAGE        22.17    debt_consolidation    36 months    1                    1                        40.1        0.0000                -1
    7. 20203    F        F4            0            3                RENT            21.94    debt_consolidation    60 months    0                    1                        34.5        0.0000                1
    8. 12408    B        B5            0            7                RENT            8.56    debt_consolidation    36 months    1                    1                        34.4        0.0000                -1
    9. 31935    B        B2            0            3                RENT            9.68    debt_consolidation    36 months    1                    1                        78.6        0.0000                1
    10. 4621    D        D1            0            3                MORTGAGE        3.78    other                36 months    1                    1                        70.4        14.9409                -1
    11. 30060    C        C3            0            6                MORTGAGE        14.60    debt_consolidation    60 months    0                    1                        39.0        0.0000                1
    12. 8089    D        D3            0            7                OWN                17.85    debt_consolidation    36 months    1                    1                        70.7        0.3500                -1

    (40000, 13)

    1. from sklearn.preprocessing import LabelEncoder
    2. from collections import defaultdict
    1. # 将以上非数值数据映射为数值型。
    2. d = defaultdict(LabelEncoder)
    3. data = data.apply(lambda x : d[x.name].fit_transform(x))
    4. X_train = data.iloc[:800, :-1]
    5. y_train = data.iloc[:800, -1]
    6. test_X = data.iloc[800:, :-1]
    7. test_y = data.iloc[800:, -1]
    8. display(X_train)
    10. # 特征二值化
    11. # 由于特征值太过复杂，不利于处理。我们根据均值来二值化
    12. for i in X_train.columns:
    13.     mean = np.mean(X_train[i])
    14.     for j in range(len(X_train[i])):
    15.         X_train[i].values[j] = 1 if X_train[i].values[j]>mean else 0
    16. for i in test_X.columns:
    17.     mean = np.mean(test_X[i])
    18.     for j in range(len(test_X[i])):
    19.         test_X[i].values[j] = 1 if test_X[i].values[j]>mean else 0
    20. display(X_train)
    1.         grade    sub_grade    short_emp    emp_length_num    home_ownership    dti        purpose    term    last_delinq_none    last_major_derog_none    revol_util    total_rec_late_fee
    2. 4104    2        12            1            1                0                1764    2        0        0                    1                        320            0
    3. 18915    0        3            0            2                3                1608    1        0        0                    1                        240            0
    4. 44721    1        8            0            7                3                3240    1        0        1                    1                        561            0
    5. 21933    0        2            0            3                0                2328    1        0        1                    1                        325            0
    6. 20111    2        14            0            8                0                1048    2        0        0                    1                        280            0
    7. ...        ...        ...            ...            ...                ...                ...        ...        ...        ...                    ...                        ...            ...
    8. 13839    3        16            0            11                0                2025    2        1        1                    1                        817            0
    9. 15773    3        17            0            11                2                1483    2        0        0                    1                        958            0
    10. 27211    2        14            0            6                3                1701    2        1        1                    1                        157            0
    11. 17372    2        11            0            11                0                489        2        1        0                    0                        279            0
    12. 5410    2        14            0            11                0                1785    1        1        0                    0                        755            0

    800 rows × 12 columns

    1.         grade    sub_grade    short_emp    emp_length_num    home_ownership    dti    purpose    term    last_delinq_none    last_major_derog_none    revol_util    total_rec_late_fee
    2. 4104    1        1            1            0                0                1    0        0        0                    1                        0            0
    3. 18915    0        0            0            0                1                1    0        0        0                    1                        0            0
    4. 44721    0        0            0            1                1                1    0        0        1                    1                        0            0
    5. 21933    0        0            0            0                0                1    0        0        1                    1                        0            0
    6. 20111    1        1            0            1                0                0    0        0        0                    1                        0            0
    7. ...        ...        ...            ...            ...                ...                ...    ...        ...        ...                    ...                        ...            ...
    8. 13839    1        1            0            1                0                1    0        1        1                    1                        1            0
    9. 15773    1        1            0            1                1                0    0        0        0                    1                        1            0
    10. 27211    1        1            0            0                1                1    0        1        1                    1                        0            0
    11. 17372    1        0            0            1                0                0    0        1        0                    0                        0            0
    12. 5410    1        1            0            1                0                1    0        1        0                    0                        1            0

    800 rows × 12 columns

    1. from collections import  Counter
    2. from tqdm import tqdm
    1. class DecisionTreeID3:
    2.     def __init__(self):
    3.         pass
    4.     def calc_entropy(self):
    5.         '''计算信息熵、条件熵和信息增益'''
    6.         # 把标签放在数据最后一列，方便create_tree处理
    7.         self.X = pd.concat([self.X, self.y], axis=1)
    8.         # ************ 计算信息熵 **************
    9.         yes = np.asarray(self.X[self.X[self.X.columns[-1]]==1])
    10.         no = np.asarray(self.X[self.X[self.X.columns[-1]]==0])
    11.         P_yes = len(yes)/len(self.X)
    12.         P_no = len(no)/len(self.X)
    13.         self.HX = - P_yes*np.log2(P_yes)- P_no*np.log2(P_no)
    14. #         display("信息熵 = " +str(self.HX))
    15.         # ************ 计算条件熵 **************
    16.         # H存放条件熵
    17.         self.Gda = []
    18.         # 遍历每一特征列。除了标签列
    19.         for i in self.X.columns[:-1]:
    20.             # 存放条件熵
    21.             Hi = 0
    22.             # 获取当前特征每种情况出现的次数，以便计算每个情况各自的概率
    23.             condProbCollections = Counter(self.X[i]).items()
    24.             # 每种特征可能有N中情况，累加
    25.             for k in condProbCollections:
    26.                 # 获取当前条件X发生的总样本(包含所有列)
    27.                 samples_of_current = self.X[self.X[i]==k[0]]
    28.                 # 获取当前条件X发生的总样本(仅包含当前特征列)
    29.                 samples_of_current_features = samples_of_current[i]
    30.                 # 获取当前条件X发生下，被判定为安全和不安全的总次数
    31.                 total = len(samples_of_current_features)
    32.                 # 安全总次数
    33.                 k_safe = len(samples_of_current[samples_of_current[samples_of_current.columns[-1]]==1])
    34.                 # 不安全总次数
    35.                 k_unsafe = total - k_safe
    36.                 # 计算安全和不安全的概率
    37.                 P_k_safe = k_safe/total
    38.                 P_k_unsafe = k_unsafe/total
    39.                 # 累加条件熵
    40.                 log_P_k_safe = 0 if P_k_safe==0 else np.log2(P_k_safe) # 防止出现0值报错
    41.                 log_P_k_unsafe = 0 if P_k_unsafe==0 else np.log2(P_k_unsafe) # 防止出现0值报错
    42.                 Hi +=  - (total/len(self.X))*(P_k_safe * log_P_k_safe + P_k_unsafe * log_P_k_unsafe)
    43.             # 保存信息增益
    44.             self.Gda.append({"value":self.HX - Hi, "feature":i})
    45. #         print("信息增益为")
    46. #         print(self.Gda)
    47.     def create_tree(self, node=False):
    48.         '''构建决策树结构。决策树信息存储在JSON字典当中
    50.         Parameters
    51.         -----
    52.         node: Series类型，用于传递当前节点包含的信息
    54.         Return
    55.         -----
    56.         tree: 构建好的树json结构
    57.         '''
    58.         # 递归出口
    59.         if len(self.Gda)==0:
    60.             return node.iloc[:1,-1].values[0]
    61.         # 获取第一个特征列
    62.         feature = self.Gda[0]['feature']
    63.         # 删除该列，以便递归时不会重复到达这里
    64.         del self.Gda[0]
    65.         # 获取当前特征每种情况出现的次数，以便计算每个情况各自的概率
    66.         condProbCollections = Counter(self.X[feature]).items()
    67. #         print(feature)
    68. #         print(condProbCollections)
    69.         # 定义树字典。必须使用特征feature作为键名
    70.         tree = {feature:{}}
    71.         for [value,counts] in condProbCollections:
    72. #             print(condProbCollections)
    73. #             print(value)
    74.             tree[feature][value] = self.create_tree(self.X[self.X[feature]==value])
    75.         # 保存树结构
    76.         self.tree = tree
    77.         return tree
    78.     def fit(self, X, y):
    79.         '''训练
    81.         Parameters
    82.         -----
    83.         X: 训练数据，形如 [样本数量，特征数量]
    84.         y: 类数组类型，形状为：[样本数量]
    85.         '''
    86.         self.X = X
    87.         self.y = y
    88.         self.calc_entropy()
    89.         self.create_tree()
    90.     def predict_item(self, x, node=False):
    91.         '''构建决策树结构。决策树信息存储在JSON字典当中
    93.         Parameters
    94.         -----
    95.         node: Series类型，用于传递当前节点包含的信息
    97.         Return
    98.         -----
    99.         tree: 构建好的树json结构
    100.         '''
    101.         if node==0 or node==1:
    102.             return node
    103.         label = -1
    104.         # 获取当前节点名
    105.         key = next(iter(node))
    106.         # 如果当前节点的值等于0，递归0下面的分支，否则1
    107.         if x[key].values[0] == 0:
    108.             label = self.predict_item(x, node=node[key][0])
    109.         else:
    110.             label = self.predict_item(x, node=node[key][1])
    111.         return label
    112.     def predict(self, X):
    113.         '''对样本进行预测
    114.         Parameters:
    115.         X: 类数组类型，可以是List也可以是Ndarray，形状为： [样本数量,特征数量]
    116.         Returns:
    117.         数组类型，预测结果
    118.         '''
    119.         result = []
    120.         for i in range(len(X)):
    121.             result.append(self.predict_item(X.iloc[i:i+1,], node=self.tree))
    122.         return result
    1. dt = DecisionTreeID3()
    2. dt.fit(X_train, y_train)
    3. result = dt.predict(test_X)
    4. display(np.sum(result==test_y)/len(result))
    1. a=[{"a":1},{"b":2},{"c":3}]
    2. c = iter(a)
    3. display(next(c))
    4. display(next(c))
    5. display(next(c))
    1. {'a': 1}
    2. {'b': 2}
    3. {'c': 3}