分箱、WOE、IV的计算

%matplotlib inline

import pandas as pd
import numpy as np
import matplotlib.pyplot as plt 
import seaborn as sns
import warnings
from scipy import stats
warnings.filterwarnings('ignore')

plt.style.use("seaborn")
plt.rc('font', family='SimHei', size=13)  # 显示中文
plt.rcParams['axes.unicode_minus'] = False  # 用来正常显示负

# 载入数据
data = pd.read_csv(r"./cs_training.csv",encoding='gbk')
# 查看数据集
# data.head(10)

# 将特征名字改为中文
column={'SeriousDlqin2yrs':'好坏客户',
        'RevolvingUtilizationOfUnsecuredLines':'可用额度比值',
        'age':'年龄',
        'NumberOfTime30-59DaysPastDueNotWorse':'逾期30-59天笔数',
        'DebtRatio':'负债率',
        'MonthlyIncome':'月收入',
        'NumberOfOpenCreditLinesAndLoans':'信贷数量',
        'NumberOfTimes90DaysLate':'逾期90天笔数',
        'NumberRealEstateLoansOrLines':'固定资产贷款量',
        'NumberOfTime60-89DaysPastDueNotWorse':'逾期60-89天笔数',
        'NumberOfDependents':'家属数量'}
data.rename(columns=column,inplace=True)
data.head()

	好坏客户	可用额度比值	年龄	逾期30-59天笔数	负债率	月收入	信贷数量	逾期90天笔数	固定资产贷款量	逾期60-89天笔数	家属数量
0	1	0.766127	45	2	0.802982	9120.0	13	0	6	0	2.0
1	0	0.957151	40	0	0.121876	2600.0	4	0	0	0	1.0
2	0	0.658180	38	1	0.085113	3042.0	2	1	0	0	0.0
3	0	0.233810	30	0	0.036050	3300.0	5	0	0	0	0.0
4	0	0.907239	49	1	0.024926	63588.0	7	0	1	0	0.0

from sklearn.ensemble import RandomForestRegressor

# 用随机森林对缺失值预测填充函数
def set_missing(df):
    # 把已有的数值型特征取出来
    process_df = df.iloc[:,[5,0,1,2,3,4,6,7,8,9]]
    # 分成已知该特征和未知该特征两部分
    # dataframe.values获取的是dataframe中的数据为数组array
    known = process_df[process_df['月收入'].notnull()].values
    unknown = process_df[process_df['月收入'].isnull()].values
    # X为已知月收入的特征属性值
    X = known[:, 1:]
    # y为结果标签值月收入
    y = known[:, 0]
    # X与y用于训练随机森林模型，fit到RandomForestRegressor之中
    rfr = RandomForestRegressor(random_state=0, n_estimators=200,max_depth=3,n_jobs=-1)
    rfr.fit(X,y)
    # 用得到的模型进行未知特征值预测
    predicted = rfr.predict(unknown[:, 1:]).round(0)
    # 用得到的预测结果填补原缺失数据
    df.loc[df['月收入'].isnull(), '月收入'] = predicted
    return df

# 用随机森林填补比较多的缺失值
data=set_missing(data)   

# 删除比较少的缺失值
data=data.dropna()   

# 删除重复项
data = data.drop_duplicates()    

data.info()

Int64Index: 145563 entries, 0 to 149999
Data columns (total 11 columns):
好坏客户          145563 non-null int64
可用额度比值        145563 non-null float64
年龄            145563 non-null int64
逾期30-59天笔数    145563 non-null int64
负债率           145563 non-null float64
月收入           145563 non-null float64
信贷数量          145563 non-null int64
逾期90天笔数       145563 non-null int64
固定资产贷款量       145563 non-null int64
逾期60-89天笔数    145563 non-null int64
家属数量          145563 non-null float64
dtypes: float64(4), int64(7)
memory usage: 13.3 MB

# 删除逾期30-59天笔数、逾期90天笔数、逾期60-89天笔数大于80的数据
data = data[data['逾期30-59天笔数'] < 80]
data = data[data['逾期90天笔数'] < 80]
data = data[data['逾期60-89天笔数'] < 80]

data = data[data['年龄'] > 0]

col_list = data.columns.values
col_list

array(['好坏客户', '可用额度比值', '年龄', '逾期30-59天笔数', '负债率', '月收入', '信贷数量',
       '逾期90天笔数', '固定资产贷款量', '逾期60-89天笔数', '家属数量'], dtype=object)

new_col_list = []
for i in range(len(col_list)):
    if i != 0 and i != 3 and i != 7 and i != 9:
        new_col_list.append(col_list[i])

# 去除单侧99%上部分异常值
for item in new_col_list:
    data = data[data[item] < data[item].quantile(0.99)]
    

import woe.feature_process as fp
import woe.eval as eval

data.columns

Index(['好坏客户', '可用额度比值', '年龄', '逾期30-59天笔数', '负债率', '月收入', '信贷数量', '逾期90天笔数',
       '固定资产贷款量', '逾期60-89天笔数', '家属数量'],
      dtype='object')

data.rename(columns={'好坏客户': 'target'}, inplace=True)

# woe分箱, iv and transform
data_woe = data # 用于存储所有数据的woe值
civ_list = []
n_positive = sum(data['target'])
n_negtive = len(data) - n_positive
for column in list(data.columns[1:]):
    if data[column].dtypes == 'object':
        civ = fp.proc_woe_discrete(data, column, n_positive, n_negtive, 0.05 * len(data), alpha=0.05)
    else:            
        civ = fp.proc_woe_continuous(data, column, n_positive, n_negtive, 0.05 * len(data), alpha=0.05)
    civ_list.append(civ)
    data_woe[column] = fp.woe_trans(data[column], civ)
    
civ_df = eval.eval_feature_detail(civ_list,'output_feature_detail_0315.csv')
# 删除iv值过小的变量
iv_thre = 0.001
iv = civ_df[['var_name','iv']].drop_duplicates()
x_columns = iv.var_name[iv.iv > iv_thre]

-------------process continuous variable:可用额度比值-------------
---------------process continuous variable:年龄---------------
-----------process continuous variable:逾期30-59天笔数-----------
--------------process continuous variable:负债率---------------
--------------process continuous variable:月收入---------------
--------------process continuous variable:信贷数量--------------
------------process continuous variable:逾期90天笔数-------------
------------process continuous variable:固定资产贷款量-------------
-----------process continuous variable:逾期60-89天笔数-----------
--------------process continuous variable:家属数量--------------
可用额度比值
年龄
逾期30-59天笔数
负债率
月收入
信贷数量
逾期90天笔数
固定资产贷款量
逾期60-89天笔数
家属数量

civ_df

	var_name	split_list	sub_total_sample_num	positive_sample_num	negative_sample_num	sub_total_num_percentage	positive_rate_in_sub_total	woe_list	iv_list	iv
0	可用额度比值	(-INF,0.0]	9352	281	9071	0.071213	0.030047	-0.757463	0.029626	1.097527
1	可用额度比值	(0.0,0.04215617400000002]	29188	365	28823	0.222259	0.012505	-1.652011	0.312454	1.097527
2	可用额度比值	(0.04215617400000002,0.0596119858]	6962	111	6851	0.053014	0.015944	-1.405599	0.059004	1.097527
3	可用额度比值	(0.0596119858,0.13857709429999995]	17901	375	17526	0.136312	0.020949	-1.127495	0.108464	1.097527
4	可用额度比值	(0.13857709429999995,0.21535932080000003]	10113	295	9818	0.077008	0.029170	-0.787977	0.034242	1.097527
5	可用额度比值	(0.21535932080000003,0.30067412204]	8274	296	7978	0.063004	0.035775	-0.577063	0.016386	1.097527
6	可用额度比值	(0.30067412204,0.3974544458]	7510	371	7139	0.057187	0.049401	-0.240106	0.002970	1.097527
7	可用额度比值	(0.3974544458,0.5331554074]	8506	586	7920	0.064771	0.068893	0.113193	0.000872	1.097527
8	可用额度比值	(0.5331554074,0.74050784496]	9985	996	8989	0.076033	0.099750	0.517010	0.025541	1.097527
9	可用额度比值	(0.74050784496,0.90349439404]	7295	1103	6192	0.055550	0.151199	0.991796	0.084555	1.097527
10	可用额度比值	(0.90349439404,+INF)	16238	3360	12878	0.123648	0.206922	1.373441	0.423411	1.097527
0	年龄	(-INF,32.0]	13531	1392	12139	0.103035	0.102875	0.551338	0.039964	0.046040
1	年龄	(32.0,+INF)	117793	6747	111046	0.896965	0.057278	-0.083827	0.006076	0.046040
0	逾期30-59天笔数	(-INF,0.0]	111119	4279	106840	0.846144	0.038508	-0.500593	0.170989	0.606073
1	逾期30-59天笔数	(0.0,+INF)	20205	3860	16345	0.153856	0.191042	1.273765	0.435084	0.606073
0	负债率	(-INF,0.018495376]	10665	466	10199	0.081211	0.043694	-0.368839	0.009420	0.088009
1	负债率	(0.018495376,0.087064379]	8843	577	8266	0.067337	0.065249	0.054956	0.000208	0.088009
2	负债率	(0.087064379,0.138218834]	7533	449	7084	0.057362	0.059604	-0.041551	0.000097	0.088009
3	负债率	(0.138218834,0.191269577]	9148	493	8655	0.069660	0.053892	-0.148363	0.001437	0.088009
4	负债率	(0.191269577,0.229044637]	6898	383	6515	0.052527	0.055523	-0.116807	0.000681	0.088009
5	负债率	(0.229044637,0.26480176767999997]	6780	313	6467	0.051628	0.046165	-0.311244	0.004370	0.088009
6	负债率	(0.26480176767999997,0.33095571454]	12054	618	11436	0.091788	0.051269	-0.201013	0.003398	0.088009
7	负债率	(0.33095571454,0.37664756308]	7551	440	7111	0.057499	0.058270	-0.065603	0.000240	0.088009
8	负债率	(0.37664756308,0.4237495164599999]	6696	406	6290	0.050988	0.060633	-0.023343	0.000028	0.088009
9	负债率	(0.4237495164599999,0.54743575044]	12664	918	11746	0.096433	0.072489	0.167949	0.002929	0.088009
10	负债率	(0.54743575044,0.7263413320000001]	9112	825	8287	0.069386	0.090540	0.409960	0.013976	0.088009
11	负债率	(0.7263413320000001,2.6823588614000204]	9111	1026	8085	0.069378	0.112611	0.652677	0.039439	0.088009
12	负债率	(2.6823588614000204,1009.0]	10925	639	10286	0.083191	0.058490	-0.061614	0.000307	0.088009
13	负债率	(1009.0,+INF)	13344	586	12758	0.101611	0.043915	-0.363574	0.011478	0.088009
0	月收入	(-INF,1159.0]	13281	877	12404	0.101132	0.066034	0.067753	0.000478	0.114078
...	...	...	...	...	...	...	...	...	...	...
7	月收入	(4831.0,5332.68]	7201	470	6731	0.054834	0.065269	0.055274	0.000172	0.114078
8	月收入	(5332.68,5917.0]	7381	432	6949	0.056205	0.058529	-0.060907	0.000203	0.114078
9	月收入	(5917.0,6667.0]	8647	533	8114	0.065845	0.061640	-0.005805	0.000002	0.114078
10	月收入	(6667.0,7916.0]	10306	519	9787	0.078478	0.050359	-0.219886	0.003448	0.114078
11	月收入	(7916.0,8333.0]	7967	327	7640	0.060667	0.041044	-0.434172	0.009484	0.114078
12	月收入	(8333.0,10300.0]	10340	489	9851	0.078737	0.047292	-0.285946	0.005687	0.114078
13	月收入	(10300.0,+INF)	13361	474	12887	0.101741	0.035476	-0.585747	0.027165	0.114078
0	信贷数量	(-INF,3.0]	18632	1864	16768	0.141878	0.100043	0.520272	0.048333	0.067247
1	信贷数量	(3.0,4.0]	10396	620	9776	0.079163	0.059638	-0.040946	0.000130	0.067247
2	信贷数量	(4.0,5.0]	11689	671	11018	0.089009	0.057404	-0.081496	0.000570	0.067247
3	信贷数量	(5.0,6.0]	12373	651	11722	0.094217	0.052615	-0.173693	0.002635	0.067247
4	信贷数量	(6.0,7.0]	12102	629	11473	0.092154	0.051975	-0.186600	0.002958	0.067247
5	信贷数量	(7.0,8.0]	11422	518	10904	0.086976	0.045351	-0.329890	0.008205	0.067247
6	信贷数量	(8.0,9.0]	10219	568	9651	0.077815	0.055583	-0.115675	0.000990	0.067247
7	信贷数量	(9.0,10.0]	8745	488	8257	0.066591	0.055803	-0.111481	0.000788	0.067247
8	信贷数量	(10.0,11.0]	7431	405	7026	0.056585	0.054501	-0.136466	0.000993	0.067247
9	信贷数量	(11.0,13.0]	11199	615	10584	0.085278	0.054916	-0.128456	0.001330	0.067247
10	信贷数量	(13.0,+INF)	17116	1110	16006	0.130334	0.064852	0.048416	0.000312	0.067247
0	逾期90天笔数	(-INF,0.0]	124488	5426	119062	0.947946	0.043587	-0.371422	0.111376	0.800610
1	逾期90天笔数	(0.0,+INF)	6836	2713	4123	0.052054	0.396870	2.298494	0.689234	0.800610
0	固定资产贷款量	(-INF,0.0]	49471	3805	45666	0.376710	0.076914	0.231982	0.022454	0.043142
1	固定资产贷款量	(0.0,1.0]	48153	2429	45724	0.366673	0.050443	-0.218124	0.015867	0.043142
2	固定资产贷款量	(1.0,2.0]	28413	1538	26875	0.216358	0.054130	-0.143694	0.004196	0.043142
3	固定资产贷款量	(2.0,+INF)	5287	367	4920	0.040259	0.069416	0.121318	0.000625	0.043142
0	逾期60-89天笔数	(-INF,0.0]	125162	6053	119109	0.953078	0.048361	-0.262465	0.058584	0.515526
1	逾期60-89天笔数	(0.0,+INF)	6162	2086	4076	0.046922	0.338526	2.047152	0.456942	0.515526
0	家属数量	(-INF,0.0]	79954	4351	75603	0.608830	0.054419	-0.138070	0.010928	0.028199
1	家属数量	(0.0,1.0]	24473	1683	22790	0.186356	0.068770	0.111276	0.002423	0.028199
2	家属数量	(1.0,2.0]	18117	1377	16740	0.137957	0.076006	0.219126	0.007295	0.028199
3	家属数量	(2.0,3.0]	8780	728	8052	0.066858	0.082916	0.313645	0.007553	0.028199

66 rows × 10 columns

x_columns

0        可用额度比值
0            年龄
0    逾期30-59天笔数
0           负债率
0           月收入
0          信贷数量
0       逾期90天笔数
0       固定资产贷款量
0    逾期60-89天笔数
0          家属数量
Name: var_name, dtype: object

iv

	var_name	iv
0	可用额度比值	1.097527
0	年龄	0.046040
0	逾期30-59天笔数	0.606073
0	负债率	0.088009
0	月收入	0.114078
0	信贷数量	0.067247
0	逾期90天笔数	0.800610
0	固定资产贷款量	0.043142
0	逾期60-89天笔数	0.515526
0	家属数量	0.028199

data_woe.head()

	target	可用额度比值	年龄	逾期30-59天笔数	负债率	月收入	信贷数量	逾期90天笔数	固定资产贷款量	逾期60-89天笔数	家属数量
1	0	1.373441	-0.083827	-0.500593	-0.041551	0.461028	-0.040946	-0.371422	0.231982	-0.262465	0.111276
2	0	0.517010	-0.083827	1.273765	0.054956	0.461028	0.520272	2.298494	0.231982	-0.262465	-0.138070
3	0	-0.577063	0.551338	-0.500593	0.054956	0.461028	-0.081496	-0.371422	0.231982	-0.262465	-0.138070
5	0	-0.787977	-0.083827	-0.500593	-0.065603	0.243014	0.520272	-0.371422	-0.218124	-0.262465	0.111276
7	0	0.991796	-0.083827	-0.500593	-0.116807	0.243014	-0.329890	-0.371422	0.231982	-0.262465	-0.138070

模型建立

信用评分卡模型在国外是一种成熟的预测方法，尤其在信用风险评估以及金融风险控制领域更是得到了比较广泛的使用，其原理是将模型变量WOE编码方式离散化之后运用logistic回归模型进行的一种二分类变量的广义线性模型，下面将模型目标标量为1记为违约用户，对于目标变量为0记为正常用户，采用sklearn中LogisticRegression进行建模

from sklearn.model_selection import train_test_split
from sklearn.linear_model import LogisticRegression
# 模型评估
from sklearn.metrics import accuracy_score
from sklearn import metrics
from sklearn.metrics import auc

# 数据提取与数据分割
col_names = data_woe.columns.values
X = data_woe[col_names[1:]]  # 特征列
y = data_woe[col_names[0]]  # 标签列
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.3,random_state=0)

lr = LogisticRegression(C=1000.0, random_state=0)
result = lr.fit(X_train, y_train)

result

LogisticRegression(C=1000.0, class_weight=None, dual=False, fit_intercept=True,
                   intercept_scaling=1, l1_ratio=None, max_iter=100,
                   multi_class='warn', n_jobs=None, penalty='l2',
                   random_state=0, solver='warn', tol=0.0001, verbose=0,
                   warm_start=False)

# 模型预测
y_pred = lr.predict(X_test)
y_pred

array([0, 0, 0, ..., 0, 0, 0], dtype=int64)

# 预测为坏的客户的概率
prob_pred = [round(u[1], 5) for u in lr.predict_proba(X_test)]

# 预测的准确率
accuracy_score(y_test, y_pred)

0.9387532362048835

# 样本类别不平衡，用PR不好评价，采用ROC曲线
FPR, TPR, thresholds = metrics.roc_curve(y_test, prob_pred, pos_label=1)
metrics.auc(FPR, TPR)

0.8499778184241903

# 画图对预测值和实际值进行比较
plt.plot(FPR, TPR, 'b', label='AUC = %0.2f' % metrics.auc(FPR, TPR)) # 生成ROC曲线
plt.legend(loc='lower right')
plt.plot([0, 1], [0, 1], 'r--')
plt.xlim([0, 1])
plt.ylim([0, 1])
plt.ylabel('真正率')
plt.xlabel('假正率')
plt.show()

[外链图片转存失败,源站可能有防盗链机制,建议将图片保存下来直接上传(img-ZuGQDu7G-1577332256208)(output_33_0.png)]

从上图可知，AUC值为0.85，说明该模型的预测效果还是不错的，正确率较高

评分卡计算方法

odds为good用户概率（p）与bad用户概率（1-p）的比值

$\mathrm{odds}=\frac{p}{1-p}$

评分卡设定的分值刻度可以通过将分值表示为比率对数的现行表达式来定义。公式如下：

$scor{e}_{总}=A+B\ast \ln (odds)$

常数 A 和 B 通常被称为补偿和刻度，它们的值可以通过将两个已知或者假设的分值带入$scor{e}_{总}=A+B\ast \ln (odds)$ 中得到。通常，需要两个假设：

• 在某个特定的比率设定特定的预期分值$P_0$
• 指定比率翻番的分数（PDO，Point-to-Double Odds）

首先，设定比率为odds的特定点的分值为$P_0$。然后，比率为 2odds的点分值为$P_0-PDO$，带入可以得到
$B=\frac{PDO}{\log (2)}$

$A=P_0+B\log \left(odds\right)$

import math

# PDO为比率翻番的分数，P0为特定比例的预期分值，B为刻度
PDO = 20
P0 = 600
B = PDO / math.log(2)
B

28.85390081777927

# A为补偿
A = P0 + B * math.log(1 / 60)
A

481.8621880878296

基于Logistic的评分卡构建

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最终，评分卡的分值可以写成下列形式：

Score$=A-B\left({\beta }_0+{\beta }_1{x}_1+\cdots +{\beta }_p{x}_p\right)$

变量$x_1$,⋯, $x_p$为自变量对应WOE, ${\beta }_0$,⋯,${\beta }_p$为逻辑斯蒂回归方程的系数

# 逻辑斯蒂回归的系数列表
coef_list = list(result.coef_[0])
coef_list.insert(0, result.intercept_[0])

# 计算信用评分
def credit_socre(data, coef): 
    score_list = []
    for i in range(data.shape[0]):
        tmp_score = coef[0]
        for j in range(data.shape[1]):
            tmp_score += data.iat[i, j] * coef[j + 1]
        score = A - B * tmp_score
        score_list.append(score)
    return score_list

score_list = credit_socre(data_woe.iloc[:, 1:], coef_list)

data_woe.insert(11, 'credit_score', score_list)

data_woe.head().append(data_woe.tail())

# 在原始数据中插入信用评分
data.insert(11, 'credit_socre', score_list)

data.head().append(data.tail())