逻辑回归与评分卡

评分卡

  • 建立逻辑回归模型
  • 对模型进行评分映射

逻辑回归表达式

\[ y = \frac{1}{1 + e^{-\theta}} \]

\[ \theta = WX + B \]

sigmoid函数

\[sigmoid(x) = \frac{1}{1 + e^{-x}} \]

sigmoid函数的导数

\[\delta sigmoid(x) = \delta{\frac{1}{1 + e^{-x}}} = \delta{\frac{e^{-x}}{(1 + e^{-x})^2}} = \delta{\frac{1}{1 + e^{-x}} * \frac{e^{-x}}{1 + e^{-x}}} = sigmoid(x) * \frac{1 + e^{-x} - 1}{1 + e^{-x}} = sigmoid(x) * (1 - sigmoid(x)) \]

损失函数(Cross-entropy, 交叉熵损失函数)

信息熵: \(-PlogP\)(P是概率, 小于1, 取反之后就是正数了), 这个值代表的是信息量, 如果值越大代表对当前情况越不确定, 信息不足.

\[ loss = -\sum{{y_t}log{y_p} + (1 - y_t)log{(1 - y_p)}} \]

\(y_t\): 真实的Y值, 需要进行独热编码

\(y_p\): 预测的Y值

交叉熵求导

\[ \frac{\delta loss}{\delta Y_p} = -\frac{\delta Y_tlogY_p}{\delta Y_p} = \sum_n^N{-\frac{Y_i}{P_i} + \frac{1 - Y_i}{1 - P_i}} \]

准确率计算

混淆矩阵

T\Pre Positive Negative
Positive TP FN
Negative FP TN

评估指标

召回率计算

\[recall = \frac{TP}{TP + FP} \]

精准率计算

\[precision = \frac{TP}{TP + FN} \]

import pandas as pd
from sklearn.metrics import roc_auc_score,roc_curve,auc
from sklearn.model_selection import train_test_split
from sklearn import metrics
from sklearn.linear_model import LogisticRegression
import numpy as np
import random
import math
data = pd.read_csv('Acard.txt')
data.head()
obs_mth bad_ind uid td_score jxl_score mj_score rh_score zzc_score zcx_score person_info finance_info credit_info act_info
0 2018-10-31 0.0 A10000005 0.675349 0.144072 0.186899 0.483640 0.928328 0.369644 -0.322581 0.023810 0.00 0.217949
1 2018-07-31 0.0 A1000002 0.825269 0.398688 0.139396 0.843725 0.605194 0.406122 -0.128677 0.023810 0.00 0.423077
2 2018-09-30 0.0 A1000011 0.315406 0.629745 0.535854 0.197392 0.614416 0.320731 0.062660 0.023810 0.10 0.448718
3 2018-07-31 0.0 A10000481 0.002386 0.609360 0.366081 0.342243 0.870006 0.288692 0.078853 0.071429 0.05 0.179487
4 2018-07-31 0.0 A1000069 0.406310 0.405352 0.783015 0.563953 0.715454 0.512554 -0.261014 0.023810 0.00 0.423077
# 看一下月份分布,用最后一个月做为跨时间验证集合
data.obs_mth.unique()
array(['2018-10-31', '2018-07-31', '2018-09-30', '2018-06-30',
       '2018-11-30'], dtype=object)
# 划分训练集和验证集
train = data[data.obs_mth != '2018-11-30'].reset_index().copy()
val = data[data.obs_mth == '2018-11-30'].reset_index().copy()
# 这是我们全部的变量,info结尾的是自己做的无监督系统输出的个人表现,score结尾的是收费的外部征信数据
feature_lst = ['person_info','finance_info','credit_info','act_info','td_score','jxl_score','mj_score','rh_score']
x = train[feature_lst]
y = train['bad_ind']

val_x =  val[feature_lst]
val_y = val['bad_ind']

lr_model = LogisticRegression(C=0.1)
lr_model.fit(x,y)
E:\Anaconda3\envs\sklearn\lib\site-packages\sklearn\linear_model\logistic.py:432: FutureWarning: Default solver will be changed to 'lbfgs' in 0.22. Specify a solver to silence this warning.
  FutureWarning)





LogisticRegression(C=0.1, class_weight=None, dual=False, fit_intercept=True,
          intercept_scaling=1, max_iter=100, multi_class='warn',
          n_jobs=None, penalty='l2', random_state=None, solver='warn',
          tol=0.0001, verbose=0, warm_start=False)

模型评价

  • KS值
  • ROC曲线

ROC

描绘的是不同的截断点时,并以FPR和TPR为横纵坐标轴,描述随着截断点的变小,TPR随着FPR的变化。
纵轴:TPR=正例分对的概率 = TP/(TP+FN),其实就是查全率
横轴:FPR=负例分错的概率 = FP/(FP+TN)

作图步骤:

根据学习器的预测结果(注意,是正例的概率值,非0/1变量)对样本进行排序(从大到小)-----这就是截断点依次选取的顺序
按顺序选取截断点,并计算TPR和FPR---也可以只选取n个截断点,分别在1/n,2/n,3/n等位置
连接所有的点(TPR,FPR)即为ROC图

KS值

作图步骤:

根据学习器的预测结果(注意,是正例的概率值,非0/1变量)对样本进行排序(从大到小)-----这就是截断点依次选取的顺序
按顺序选取截断点,并计算TPR和FPR ---也可以只选取n个截断点,分别在1/n,2/n,3/n等位置
横轴为样本的占比百分比(最大100%),纵轴分别为TPR和FPR,可以得到KS曲线
TPR和FPR曲线分隔最开的位置就是最好的”截断点“,最大间隔距离就是KS值,通常>0.2即可认为模型有比较好偶的预测准确性

# 训练集KS
y_pred = lr_model.predict_proba(x)[:,1]
fpr_lr_train,tpr_lr_train,_ = roc_curve(y,y_pred)
train_ks = abs(fpr_lr_train - tpr_lr_train).max()
print('train_ks : ',train_ks)

# 验证集KS
y_pred = lr_model.predict_proba(val_x)[:,1]
fpr_lr,tpr_lr,_ = roc_curve(val_y,y_pred)
val_ks = abs(fpr_lr - tpr_lr).max()
print('val_ks : ',val_ks)

# 训练集AUC
train_auc = auc(fpr_lr_train, tpr_lr_train)
print('train_auc:', train_auc)

# 验证集AUC
val_auc = auc(fpr_lr, tpr_lr)
print('val_auc:', val_auc)


# ROC曲线
from matplotlib import pyplot as plt
plt.plot(fpr_lr_train,tpr_lr_train,label = 'train LR')
plt.plot(fpr_lr,tpr_lr,label = 'evl LR')
plt.plot([0,1],[0,1],'k--')  # 虚线
plt.xlabel('False positive rate')
plt.ylabel('True positive rate')
plt.title('ROC Curve')
plt.legend(loc = 'best')
plt.show()
train_ks :  0.4144413866157737
val_ks :  0.3709405449809594
train_auc: 0.777913749438214
val_auc: 0.749188849417094

# 使用方差膨胀系数做特征筛选
from statsmodels.stats.outliers_influence import variance_inflation_factor
X = np.array(x)
for i in range(X.shape[1]):
    print(variance_inflation_factor(X,i))
1.3021397545577766
1.9579535743187253
1.289944208916368
2.9681708673324034
3.2871099722760153
3.286493284008913
3.3175087980337827
3.2910065791107597

方差膨胀系数VIF越大,说明自变量之间存在共线性的可能性越大。一般来讲,如果方差膨胀因子超过10,则回归模型存在严重的多重共线性。又根据Hair(1995)的共线性诊断标准,当自变量的容忍度大于0.1,方差膨胀系数小于10的范围是可以接受的,表明白变量之间没有共线性问题存在。

import lightgbm as lgb
from sklearn.model_selection import train_test_split
train_x,test_x,train_y,test_y = train_test_split(x,y,random_state=0,test_size=0.2)
def lgb_test(train_x,train_y,test_x,test_y):
    clf =lgb.LGBMClassifier(boosting_type = 'gbdt',
                           objective = 'binary',
                           metric = 'auc',
                           learning_rate = 0.1,
                           n_estimators = 24,
                           max_depth = 5,
                           num_leaves = 20,
                           max_bin = 45,
                           min_data_in_leaf = 6,
                           bagging_fraction = 0.6,
                           bagging_freq = 0,
                           feature_fraction = 0.8,
                           )
    clf.fit(train_x,train_y,eval_set = [(train_x,train_y),(test_x,test_y)],eval_metric = 'auc')
    return clf,clf.best_score_['valid_1']['auc'],
lgb_model , lgb_auc  = lgb_test(train_x,train_y,test_x,test_y)
feature_importance = pd.DataFrame({'name':lgb_model.booster_.feature_name(),
                                   'importance':lgb_model.feature_importances_}).sort_values(by=['importance'],ascending=False)
feature_importance
[1]	training's auc: 0.759467	valid_1's auc: 0.753322
[2]	training's auc: 0.809023	valid_1's auc: 0.805658
[3]	training's auc: 0.809328	valid_1's auc: 0.803858
[4]	training's auc: 0.810298	valid_1's auc: 0.801355
[5]	training's auc: 0.814873	valid_1's auc: 0.807356
[6]	training's auc: 0.816492	valid_1's auc: 0.809279
[7]	training's auc: 0.820213	valid_1's auc: 0.809208
[8]	training's auc: 0.823931	valid_1's auc: 0.812081
[9]	training's auc: 0.82696	valid_1's auc: 0.81453
[10]	training's auc: 0.827882	valid_1's auc: 0.813428
[11]	training's auc: 0.828881	valid_1's auc: 0.814226
[12]	training's auc: 0.829577	valid_1's auc: 0.813749
[13]	training's auc: 0.830406	valid_1's auc: 0.813156
[14]	training's auc: 0.830843	valid_1's auc: 0.812973
[15]	training's auc: 0.831587	valid_1's auc: 0.813501
[16]	training's auc: 0.831898	valid_1's auc: 0.813611
[17]	training's auc: 0.833751	valid_1's auc: 0.81393
[18]	training's auc: 0.834139	valid_1's auc: 0.814532
[19]	training's auc: 0.835177	valid_1's auc: 0.815209
[20]	training's auc: 0.837368	valid_1's auc: 0.815205
[21]	training's auc: 0.837946	valid_1's auc: 0.815099
[22]	training's auc: 0.839585	valid_1's auc: 0.815602
[23]	training's auc: 0.840781	valid_1's auc: 0.816105
[24]	training's auc: 0.841174	valid_1's auc: 0.816869
name importance
2 credit_info 98
3 act_info 62
4 td_score 54
5 jxl_score 50
7 rh_score 50
0 person_info 49
1 finance_info 47
6 mj_score 46
feature_lst = ['person_info','finance_info','credit_info','act_info']
# 训练集
x = train[feature_lst]
y = train['bad_ind']
# 验证集
val_x =  val[feature_lst]
val_y = val['bad_ind']

lr_model = LogisticRegression(C=0.1, class_weight='balanced')
lr_model.fit(x,y)
y_pred = lr_model.predict_proba(x)[:,1]
fpr_lr_train,tpr_lr_train,_ = roc_curve(y,y_pred)
train_ks = abs(fpr_lr_train - tpr_lr_train).max()
print('train_ks : ',train_ks)

y_pred = lr_model.predict_proba(val_x)[:,1]
fpr_lr,tpr_lr,_ = roc_curve(val_y,y_pred)
val_ks = abs(fpr_lr - tpr_lr).max()
print('val_ks : ',val_ks)
from matplotlib import pyplot as plt
plt.plot(fpr_lr_train,tpr_lr_train,label = 'train LR')
plt.plot(fpr_lr,tpr_lr,label = 'evl LR')
plt.plot([0,1],[0,1],'k--')
plt.xlabel('False positive rate')
plt.ylabel('True positive rate')
plt.title('ROC Curve')
plt.legend(loc = 'best')
plt.show()
E:\Anaconda3\envs\sklearn\lib\site-packages\sklearn\linear_model\logistic.py:432: FutureWarning: Default solver will be changed to 'lbfgs' in 0.22. Specify a solver to silence this warning.
  FutureWarning)


train_ks :  0.4482453222991063
val_ks :  0.4198642457760936

# 系数
print('变量名单:',feature_lst)
print('系数:',lr_model.coef_)
print('截距:',lr_model.intercept_)
变量名单: ['person_info', 'finance_info', 'credit_info', 'act_info', 'td_score', 'jxl_score', 'mj_score', 'rh_score']
系数: [[ 2.49234624  4.35064917  1.83110927 -1.63300006 -0.18766591 -0.11007296
  -0.20273074 -0.08850626]]
截距: [-3.53854743]
# 生成报告

# 模型
model = lr_model
row_num, col_num = 0, 0
# 分箱的数量
bins = 20
# 预测的概率
Y_predict = [s[1] for s in model.predict_proba(val_x)]
# 标签
Y = val_y
# 样本个数
nrows = Y.shape[0]
# 列表
lis = [(Y_predict[i], Y[i]) for i in range(nrows)]
# 按概率大小排序
ks_lis = sorted(lis, key=lambda x: x[0], reverse=True)
# 每一箱中的样本数量
bin_num = int(nrows/bins+1)
# 负样本数量
bad = sum([1 for (p, y) in ks_lis if y > 0.5])
# 正样本数量
good = sum([1 for (p, y) in ks_lis if y <= 0.5])

bad_cnt, good_cnt = 0, 0
KS = []
BAD = []
GOOD = []
BAD_CNT = []
GOOD_CNT = []
BAD_PCTG = []
BADRATE = []
dct_report = {}
for j in range(bins):
    # 每一箱中的list
    ds = ks_lis[j*bin_num: min((j+1)*bin_num, nrows)]
    # 每一箱中的负样本数量
    bad1 = sum([1 for (p, y) in ds if y > 0.5])
    # 每一箱中的正样本数量
    good1 = sum([1 for (p, y) in ds if y <= 0.5])
    bad_cnt += bad1
    good_cnt += good1
    # 负样本率
    bad_pctg = round(bad_cnt/sum(val_y),3)
    # bad_rate
    badrate = round(bad1/(bad1+good1),3)
    ks = round(math.fabs((bad_cnt / bad) - (good_cnt / good)),3)
    KS.append(ks)
    BAD.append(bad1)
    GOOD.append(good1)
    BAD_CNT.append(bad_cnt)
    GOOD_CNT.append(good_cnt)
    BAD_PCTG.append(bad_pctg)
    BADRATE.append(badrate)
    dct_report['KS'] = KS
    dct_report['BAD'] = BAD
    dct_report['GOOD'] = GOOD
    dct_report['BAD_CNT'] = BAD_CNT
    dct_report['GOOD_CNT'] = GOOD_CNT
    dct_report['BAD_PCTG'] = BAD_PCTG
    dct_report['BADRATE'] = BADRATE
val_repot = pd.DataFrame(dct_report)
val_repot
KS BAD GOOD BAD_CNT GOOD_CNT BAD_PCTG BADRATE
0 0.164 69 730 69 730 0.210 0.086
1 0.271 51 748 120 1478 0.366 0.064
2 0.292 23 776 143 2254 0.436 0.029
3 0.356 37 762 180 3016 0.549 0.046
4 0.364 19 780 199 3796 0.607 0.024
5 0.363 16 783 215 4579 0.655 0.020
6 0.355 14 785 229 5364 0.698 0.018
7 0.363 19 780 248 6144 0.756 0.024
8 0.353 13 786 261 6930 0.796 0.016
9 0.336 11 788 272 7718 0.829 0.014
10 0.313 9 790 281 8508 0.857 0.011
11 0.277 5 794 286 9302 0.872 0.006
12 0.251 8 791 294 10093 0.896 0.010
13 0.234 11 788 305 10881 0.930 0.014
14 0.202 6 793 311 11674 0.948 0.008
15 0.173 7 792 318 12466 0.970 0.009
16 0.134 4 795 322 13261 0.982 0.005
17 0.092 3 796 325 14057 0.991 0.004
18 0.045 1 798 326 14855 0.994 0.001
19 0.000 2 792 328 15647 1.000 0.003
from pyecharts.charts import *
from pyecharts import options as opts
from pylab import *
mpl.rcParams['font.sans-serif'] = ['SimHei']
np.set_printoptions(suppress=True)
pd.set_option('display.unicode.ambiguous_as_wide', True)
pd.set_option('display.unicode.east_asian_width', True)
line = (

    Line()
    .add_xaxis(list(val_repot.index))
    .add_yaxis(
        "分组坏人占比",
        list(val_repot.BADRATE),
        yaxis_index=0,
        color="red",
    )
    .set_global_opts(
        title_opts=opts.TitleOpts(title="行为评分卡模型表现"),
    )
    .extend_axis(
        yaxis=opts.AxisOpts(
            name="累计坏人占比",
            type_="value",
            min_=0,
            max_=0.5,
            position="right",
            axisline_opts=opts.AxisLineOpts(
                linestyle_opts=opts.LineStyleOpts(color="red")
            ),
            axislabel_opts=opts.LabelOpts(formatter="{value}"),
        )

    )
    .add_xaxis(list(val_repot.index))
    .add_yaxis(
        "KS",
        list(val_repot['KS']),
        yaxis_index=1,
        color="blue",
        label_opts=opts.LabelOpts(is_show=False),
    )
)
line.render_notebook()
# ['person_info','finance_info','credit_info','act_info']
# 算分数onekey 
def score(person_info, finance_info, credit_info, act_info):
    xbeta = person_info * ( 3.49460978) + finance_info * ( 11.40051582 ) + credit_info * (2.45541981) + act_info * ( -1.68676079) --0.34484897 
    score = 650-34* (xbeta)/math.log(2)
    return score
val['score'] = val.apply(lambda x : score(x.person_info,x.finance_info,x.credit_info,x.act_info) ,axis=1)

fpr_lr,tpr_lr,_ = roc_curve(val_y,val['score'])
val_ks = abs(fpr_lr - tpr_lr).max()
print('val_ks : ',val_ks)
print(val['score'].head())

# 对应评级区间
def level(score):
    level = 0
    if score <= 600:
        level = "D"
    elif score <= 640 and score > 600 : 
        level = "C"
    elif score <= 680 and score > 640:
        level = "B"
    elif  score > 680 :
        level = "A"
    return level
val['level'] = val.score.map(lambda x : level(x) )

val.level.groupby(val.level).count()/len(val)
val_ks :  0.4198642457760936
0    514.314551
1    636.487029
2    643.092121
3    668.413494
4    636.487029
Name: score, dtype: float64





level
A    0.144351
B    0.240188
C    0.391299
D    0.224163
Name: level, dtype: float64
import seaborn as sns
sns.distplot(val.score,kde=True)

val = val.sort_values('score',ascending=True).reset_index(drop=True)
df2=val.bad_ind.groupby(val['level']).sum()
df3=val.bad_ind.groupby(val['level']).count()
print(df2/df3)    
E:\Anaconda3\envs\sklearn\lib\site-packages\scipy\stats\stats.py:1713: FutureWarning: Using a non-tuple sequence for multidimensional indexing is deprecated; use `arr[tuple(seq)]` instead of `arr[seq]`. In the future this will be interpreted as an array index, `arr[np.array(seq)]`, which will result either in an error or a different result.
  return np.add.reduce(sorted[indexer] * weights, axis=axis) / sumval


level
A    0.002168
B    0.008079
C    0.014878
D    0.055571
Name: bad_ind, dtype: float64

posted @ 2019-06-02 13:33  chenxiangzhen  阅读(1488)  评论(0编辑  收藏  举报