信用评分卡（A卡） 基于LR模型的数据处理及建模过程
数据来自：魔镜杯风控算法大赛（拍拍贷）。有关数据的具体描述可以看比赛页面。
0. 数据集的关键字段及描述：
- Master：每一行代表一个样本（一笔成功成交借款），每个样本包含200多个各类字段。
- idx：每一笔贷款的unique key，可以与另外2个文件里的idx相匹配。
- UserInfo_*：借款人特征字段
- WeblogInfo_*：Info网络行为字段
- Education_Info*：学历学籍字段
- ThirdParty_Info_PeriodN_*：第三方数据时间段N字段
- SocialNetwork_*：社交网络字段
- LinstingInfo：借款成交时间
- Target：违约标签（1 = 贷款违约，0 = 正常还款）。测试集里不包含target字段。
- Log_Info：借款人的登陆信息。
- ListingInfo：借款成交时间
- LogInfo1：操作代码
- LogInfo2：操作类别
- LogInfo3：登陆时间
- idx：每一笔贷款的unique key
- Userupdate_Info：借款人修改信息
- ListingInfo1：借款成交时间
- UserupdateInfo1：修改内容
- UserupdateInfo2：修改时间
- idx：每一笔贷款的unique key
Logistic Regression的优点在于简单、稳定可解释，作为初次实践，用这个模型比较好上手。
1. 数据预处理
提炼特征的方法有求和、比例、频率、平均。
对Log_Info的处理
对于本数据中的登录时间，登录日期与放款日期的间隔天数，大部分在180天以内。
选取半年内的时间切片：30、60、90、120、150、180
可以计算不同时间切片下的：
- 登录次数
- 不同登录方式的个数
- 不同登录方式的平均个数
缺失值处理
- 缺失值占比超过80%做删除处理，否则按特殊值处理
- Master中的UserInfo_的缺失值根据相关性较高的字段进行填充
异常值处理
为了不丢失重要信息，先不做处理，在分箱过程中进行处理。
数据一致性
数据格式差异：Master中的LinstingInfo，统一转成时间戳形式；大小写不一致的数据；手机号格式统一等
2. 特征工程
变量分箱使用卡方分箱法，并通过
来判断分箱后的分布均匀性。
同时：
- 处理异常值：占比低于5%，将特殊值与正常值中的最大的一箱进行合并。
- 类别型变量分箱：
- 学历等有序的：按照排序赋值
- 省份城市等无序的：用该类型的坏样本率代替
分箱后编码：WOE=ln(GoodPercent/BadPercent)
挑选特征：
特征信息值IV = （GoodPercent-BadPercent）*WOE
IV衡量的是特征总体的重要性，也与分箱方式有关。
由上图可知，变量的IV普遍较低，稍微放宽IV选择的条件，以0.02为阈值进行粗筛。
线性相关性：通过相关矩阵来判断
多重共线性：VIF（方差膨胀因子）如果大于10，则存在
部分变量的p值不显著，WOE也存在正值，因此要检查显著性和正确性。
对所有p值超过0.1的变量单独做一元逻辑回归模型，p值全部低于0.1，说明不显著的p值是由于线性相关性引起的。
对上述所有正系数的变量单独做一元逻辑回归模型，系数全部为-1。
将变量根据IV进行降序排列，从IV最高的变量开始，逐个放入，如仍 满足p小于0.1，则继续加入，否则剔除新加入的变量。
变量选择后，符号都为负，且p值小于阈值0.1
3. 尺度化
将概率转化成分数，违约概率越低，资质越好，分数越高。
y = log(p/(1-p))
PDO:好坏比上升1倍时，分数上升PDO个单位。
评分分布较均匀。
附：
代码1-数据处理、建模代码
import pandas as pd
import datetime
import collections
import numpy as np
import numbers
import random
import sys
import pickle
from itertools import combinations
from sklearn.linear_model import LinearRegression
from sklearn.ensemble import RandomForestClassifier
from sklearn.model_selection import train_test_split
from sklearn.metrics import roc_curve
from sklearn.metrics import roc_auc_score
import statsmodels.api as sm
from importlib import reload
from matplotlib import pyplot as plt
reload(sys)
sys.setdefaultencoding( "utf-8")
from scorecard_functions import *
from sklearn.linear_model import LogisticRegressionCV
# -*- coding: utf-8 -*-
################################
######## UDF: 自定义函数 ########
################################
### 对时间窗口，计算累计产比 ###
def TimeWindowSelection(df, daysCol, time_windows):
'''
:param df: the dataset containg variabel of days
:param daysCol: the column of days
:param time_windows: the list of time window
:return:
'''
freq_tw = {}
for tw in time_windows:
freq = sum(df[daysCol].apply(lambda x: int(x<=tw)))
freq_tw[tw] = freq
return freq_tw
def DeivdedByZero(nominator, denominator):
'''
当分母为0时，返回0；否则返回正常值
'''
if denominator == 0:
return 0
else:
return nominator*1.0/denominator
#对某些统一的字段进行统一
def ChangeContent(x):
y = x.upper()
if y == '_MOBILEPHONE':
y = '_PHONE'
return y
def MissingCategorial(df,x):
missing_vals = df[x].map(lambda x: int(x!=x))
return sum(missing_vals)*1.0/df.shape[0]
def MissingContinuous(df,x):
missing_vals = df[x].map(lambda x: int(np.isnan(x)))
return sum(missing_vals) * 1.0 / df.shape[0]
def MakeupRandom(x, sampledList):
if x==x:
return x
else:
randIndex = random.randint(0, len(sampledList)-1)
return sampledList[randIndex]
############################################################
#Step 0: 数据分析的初始工作, 包括读取数据文件、检查用户Id的一致性等#
############################################################
folderOfData = '/Users/Code/Data Collections/bank default/'
data1 = pd.read_csv(folderOfData+'PPD_LogInfo_3_1_Training_Set.csv', header = 0)
data2 = pd.read_csv(folderOfData+'PPD_Training_Master_GBK_3_1_Training_Set.csv', header = 0,encoding = 'gbk')
data3 = pd.read_csv(folderOfData+'PPD_Userupdate_Info_3_1_Training_Set.csv', header = 0)
#############################################################################################
# Step 1: 从PPD_LogInfo_3_1_Training_Set & PPD_Userupdate_Info_3_1_Training_Set数据中衍生特征#
#############################################################################################
# compare whether the four city variables match
data2['city_match'] = data2.apply(lambda x: int(x.UserInfo_2 == x.UserInfo_4 == x.UserInfo_8 == x.UserInfo_20),axis = 1)
del data2['UserInfo_2']
del data2['UserInfo_4']
del data2['UserInfo_8']
del data2['UserInfo_20']
### 提取申请日期，计算日期差，查看日期差的分布
data1['logInfo'] = data1['LogInfo3'].map(lambda x: datetime.datetime.strptime(x,'%Y-%m-%d'))
data1['Listinginfo'] = data1['Listinginfo1'].map(lambda x: datetime.datetime.strptime(x,'%Y-%m-%d'))
data1['ListingGap'] = data1[['logInfo','Listinginfo']].apply(lambda x: (x[1]-x[0]).days,axis = 1)
plt.hist(data1['ListingGap'],bins=200)
plt.title('Days between login date and listing date')
ListingGap2 = data1['ListingGap'].map(lambda x: min(x,365))
plt.hist(ListingGap2,bins=200)
timeWindows = TimeWindowSelection(data1, 'ListingGap', range(30,361,30))
'''
使用180天作为最大的时间窗口计算新特征
所有可以使用的时间窗口可以有7 days, 30 days, 60 days, 90 days, 120 days, 150 days and 180 days.
在每个时间窗口内，计算总的登录次数，不同的登录方式，以及每种登录方式的平均次数
'''
time_window = [7, 30, 60, 90, 120, 150, 180]
var_list = ['LogInfo1','LogInfo2']
data1GroupbyIdx = pd.DataFrame({'Idx':data1['Idx'].drop_duplicates()})
for tw in time_window:
data1['TruncatedLogInfo'] = data1['Listinginfo'].map(lambda x: x + datetime.timedelta(-tw))
temp = data1.loc[data1['logInfo'] >= data1['TruncatedLogInfo']]
for var in var_list:
#count the frequences of LogInfo1 and LogInfo2
count_stats = temp.groupby(['Idx'])[var].count().to_dict()
data1GroupbyIdx[str(var)+'_'+str(tw)+'_count'] = data1GroupbyIdx['Idx'].map(lambda x: count_stats.get(x,0))
# count the distinct value of LogInfo1 and LogInfo2
Idx_UserupdateInfo1 = temp[['Idx', var]].drop_duplicates()
uniq_stats = Idx_UserupdateInfo1.groupby(['Idx'])[var].count().to_dict()
data1GroupbyIdx[str(var) + '_' + str(tw) + '_unique'] = data1GroupbyIdx['Idx'].map(lambda x: uniq_stats.get(x,0))
# calculate the average count of each value in LogInfo1 and LogInfo2
data1GroupbyIdx[str(var) + '_' + str(tw) + '_avg_count'] = data1GroupbyIdx[[str(var)+'_'+str(tw)+'_count',str(var) + '_' + str(tw) + '_unique']].\
apply(lambda x: DeivdedByZero(x[0],x[1]), axis=1)
data3['ListingInfo'] = data3['ListingInfo1'].map(lambda x: datetime.datetime.strptime(x,'%Y/%m/%d'))
data3['UserupdateInfo'] = data3['UserupdateInfo2'].map(lambda x: datetime.datetime.strptime(x,'%Y/%m/%d'))
data3['ListingGap'] = data3[['UserupdateInfo','ListingInfo']].apply(lambda x: (x[1]-x[0]).days,axis = 1)
collections.Counter(data3['ListingGap'])
hist_ListingGap = np.histogram(data3['ListingGap'])
hist_ListingGap = pd.DataFrame({'Freq':hist_ListingGap[0],'gap':hist_ListingGap[1][1:]})
hist_ListingGap['CumFreq'] = hist_ListingGap['Freq'].cumsum()
hist_ListingGap['CumPercent'] = hist_ListingGap['CumFreq'].map(lambda x: x*1.0/hist_ListingGap.iloc[-1]['CumFreq'])
'''
对 QQ和qQ, Idnumber和idNumber,MOBILEPHONE和PHONE 进行统一
在时间切片内，计算
(1) 更新的频率
(2) 每种更新对象的种类个数
(3) 对重要信息如IDNUMBER,HASBUYCAR, MARRIAGESTATUSID, PHONE的更新
'''
data3['UserupdateInfo1'] = data3['UserupdateInfo1'].map(ChangeContent)
data3GroupbyIdx = pd.DataFrame({'Idx':data3['Idx'].drop_duplicates()})
time_window = [7, 30, 60, 90, 120, 150, 180]
for tw in time_window:
data3['TruncatedLogInfo'] = data3['ListingInfo'].map(lambda x: x + datetime.timedelta(-tw))
temp = data3.loc[data3['UserupdateInfo'] >= data3['TruncatedLogInfo']]
#frequency of updating
freq_stats = temp.groupby(['Idx'])['UserupdateInfo1'].count().to_dict()
data3GroupbyIdx['UserupdateInfo_'+str(tw)+'_freq'] = data3GroupbyIdx['Idx'].map(lambda x: freq_stats.get(x,0))
# number of updated types
Idx_UserupdateInfo1 = temp[['Idx','UserupdateInfo1']].drop_duplicates()
uniq_stats = Idx_UserupdateInfo1.groupby(['Idx'])['UserupdateInfo1'].count().to_dict()
data3GroupbyIdx['UserupdateInfo_' + str(tw) + '_unique'] = data3GroupbyIdx['Idx'].map(lambda x: uniq_stats.get(x, x))
#average count of each type
data3GroupbyIdx['UserupdateInfo_' + str(tw) + '_avg_count'] = data3GroupbyIdx[['UserupdateInfo_'+str(tw)+'_freq', 'UserupdateInfo_' + str(tw) + '_unique']]. \
apply(lambda x: x[0] * 1.0 / x[1], axis=1)
#whether the applicant changed items like IDNUMBER,HASBUYCAR, MARRIAGESTATUSID, PHONE
Idx_UserupdateInfo1['UserupdateInfo1'] = Idx_UserupdateInfo1['UserupdateInfo1'].map(lambda x: [x])
Idx_UserupdateInfo1_V2 = Idx_UserupdateInfo1.groupby(['Idx'])['UserupdateInfo1'].sum()
for item in ['_IDNUMBER','_HASBUYCAR','_MARRIAGESTATUSID','_PHONE']:
item_dict = Idx_UserupdateInfo1_V2.map(lambda x: int(item in x)).to_dict()
data3GroupbyIdx['UserupdateInfo_' + str(tw) + str(item)] = data3GroupbyIdx['Idx'].map(lambda x: item_dict.get(x, x))
# Combine the above features with raw features in PPD_Training_Master_GBK_3_1_Training_Set
allData = pd.concat([data2.set_index('Idx'), data3GroupbyIdx.set_index('Idx'), data1GroupbyIdx.set_index('Idx')],axis= 1)
allData.to_csv(folderOfData+'allData_0.csv',encoding = 'gbk')
#######################################
# Step 2: 对类别型变量和数值型变量进行补缺#
######################################
allData = pd.read_csv(folderOfData+'allData_0.csv',header = 0,encoding = 'gbk')
allFeatures = list(allData.columns)
allFeatures.remove('target')
if 'Idx' in allFeatures:
allFeatures.remove('Idx')
allFeatures.remove('ListingInfo')
#检查是否有常数型变量，并且检查是类别型还是数值型变量
numerical_var = []
for col in allFeatures:
if len(set(allData[col])) == 1:
print('delete {} from the dataset because it is a constant'.format(col))
del allData[col]
allFeatures.remove(col)
else:
uniq_valid_vals = [i for i in allData[col] if i == i]
uniq_valid_vals = list(set(uniq_valid_vals))
if len(uniq_valid_vals) >= 10 and isinstance(uniq_valid_vals[0], numbers.Real):
numerical_var.append(col)
categorical_var = [i for i in allFeatures if i not in numerical_var]
#检查变量的最多值的占比情况,以及每个变量中占比最大的值
records_count = allData.shape[0]
col_most_values,col_large_value = {},{}
for col in allFeatures:
value_count = allData[col].groupby(allData[col]).count()
col_most_values[col] = max(value_count)/records_count
large_value = value_count[value_count== max(value_count)].index[0]
col_large_value[col] = large_value
col_most_values_df = pd.DataFrame.from_dict(col_most_values, orient = 'index')
col_most_values_df.columns = ['max percent']
col_most_values_df = col_most_values_df.sort_values(by = 'max percent', ascending = False)
pcnt = list(col_most_values_df[:500]['max percent'])
vars = list(col_most_values_df[:500].index)
plt.bar(range(len(pcnt)), height = pcnt)
plt.title('Largest Percentage of Single Value in Each Variable')
#计算多数值占比超过90%的字段中，少数值的坏样本率是否会显著高于多数值
large_percent_cols = list(col_most_values_df[col_most_values_df['max percent']>=0.9].index)
bad_rate_diff = {}
for col in large_percent_cols:
large_value = col_large_value[col]
temp = allData[[col,'target']]
temp[col] = temp.apply(lambda x: int(x[col]==large_value),axis=1)
bad_rate = temp.groupby(col).mean()
if bad_rate.iloc[0]['target'] == 0:
bad_rate_diff[col] = 0
continue
bad_rate_diff[col] = np.log(bad_rate.iloc[0]['target']/bad_rate.iloc[1]['target'])
bad_rate_diff_sorted = sorted(bad_rate_diff.items(),key=lambda x: x[1], reverse=True)
bad_rate_diff_sorted_values = [x[1] for x in bad_rate_diff_sorted]
plt.bar(x = range(len(bad_rate_diff_sorted_values)), height = bad_rate_diff_sorted_values)
#由于所有的少数值的坏样本率并没有显著高于多数值，意味着这些变量可以直接剔除
for col in large_percent_cols:
if col in numerical_var:
numerical_var.remove(col)
else:
categorical_var.remove(col)
del allData[col]
'''
对类别型变量，如果缺失超过80%, 就删除，否则当成特殊的状态
'''
missing_pcnt_threshould_1 = 0.8
for col in categorical_var:
missingRate = MissingCategorial(allData,col)
print('{0} has missing rate as {1}'.format(col,missingRate))
if missingRate > missing_pcnt_threshould_1:
categorical_var.remove(col)
del allData[col]
if 0 < missingRate < missing_pcnt_threshould_1:
uniq_valid_vals = [i for i in allData[col] if i == i]
uniq_valid_vals = list(set(uniq_valid_vals))
if isinstance(uniq_valid_vals[0], numbers.Real):
missing_position = allData.loc[allData[col] != allData[col]][col].index
not_missing_sample = [-1]*len(missing_position)
allData.loc[missing_position, col] = not_missing_sample
else:
# In this way we convert NaN to NAN, which is a string instead of np.nan
allData[col] = allData[col].map(lambda x: str(x).upper())
allData_bk = allData.copy()
'''
检查数值型变量
'''
missing_pcnt_threshould_2 = 0.8
deleted_var = []
for col in numerical_var:
missingRate = MissingContinuous(allData, col)
print('{0} has missing rate as {1}'.format(col, missingRate))
if missingRate > missing_pcnt_threshould_2:
deleted_var.append(col)
print('we delete variable {} because of its high missing rate'.format(col))
else:
if missingRate > 0:
not_missing = allData.loc[allData[col] == allData[col]][col]
#makeuped = allData[col].map(lambda x: MakeupRandom(x, list(not_missing)))
missing_position = allData.loc[allData[col] != allData[col]][col].index
not_missing_sample = random.sample(list(not_missing), len(missing_position))
allData.loc[missing_position,col] = not_missing_sample
#del allData[col]
#allData[col] = makeuped
missingRate2 = MissingContinuous(allData, col)
print('missing rate after making up is:{}'.format(str(missingRate2)))
if deleted_var != []:
for col in deleted_var:
numerical_var.remove(col)
del allData[col]
allData.to_csv(folderOfData+'allData_1.csv', header=True,encoding='gbk', columns = allData.columns, index=False)
allData = pd.read_csv(folderOfData+'allData_1.csv', header=0,encoding='gbk')
###################################
# Step 3: 基于卡方分箱法对变量进行分箱#
###################################
'''
对不同类型的变量，分箱的处理是不同的：
（1）数值型变量可直接分箱
（2）取值个数较多的类别型变量，需要用bad rate做编码转换成数值型变量，再分箱
（3）取值个数较少的类别型变量不需要分箱，但是要检查是否每个类别都有好坏样本。如果有类别只有好或坏，需要合并
'''
#for each categorical variable, if it has distinct values more than 5, we use the ChiMerge to merge it
trainData = pd.read_csv(folderOfData+'allData_1.csv',header = 0, encoding='gbk',dtype=object)
allFeatures = list(trainData.columns)
allFeatures.remove('ListingInfo')
allFeatures.remove('target')
#allFeatures.remove('Idx')
#将特征区分为数值型和类别型
numerical_var = []
for var in allFeatures:
uniq_vals = list(set(trainData[var]))
if np.nan in uniq_vals:
uniq_vals.remove( np.nan)
if len(uniq_vals) >= 10 and isinstance(uniq_vals[0],numbers.Real):
numerical_var.append(var)
categorical_var = [i for i in allFeatures if i not in numerical_var]
for col in categorical_var:
#for Chinese character, upper() is not valid
if col not in ['UserInfo_7','UserInfo_9','UserInfo_19']:
trainData[col] = trainData[col].map(lambda x: str(x).upper())
'''
对于类别型变量，按照以下方式处理
1，如果变量的取值个数超过5，计算bad rate进行编码
2，除此之外，其他任何类别型变量如果有某个取值中，对应的样本全部是坏样本或者是好样本，进行合并。
'''
deleted_features = [] #将处理过的变量删除，防止对后面建模的干扰
encoded_features = {} #将bad rate编码方式保存下来，在以后的测试和生产环境中需要使用
merged_features = {} #将类别型变量合并方案保留下来
var_IV = {} #save the IV values for binned features #将IV值保留和WOE值
var_WOE = {}
for col in categorical_var:
print('we are processing {}'.format(col))
if len(set(trainData[col]))>5:
print('{} is encoded with bad rate'.format(col))
col0 = str(col)+'_encoding'
#(1), 计算坏样本率并进行编码
encoding_result = BadRateEncoding(trainData, col, 'target')
trainData[col0], br_encoding = encoding_result['encoding'],encoding_result['bad_rate']
#(2), 将（1）中的编码后的变量也加入数值型变量列表中，为后面的卡方分箱做准备
numerical_var.append(col0)
#(3), 保存编码结果
encoded_features[col] = [col0, br_encoding]
#(4), 删除原始值
deleted_features.append(col)
else:
bad_bin = trainData.groupby([col])['target'].sum()
#对于类别数少于5个，但是出现0坏样本的特征需要做处理
if min(bad_bin) == 0:
print('{} has 0 bad sample!'.format(col))
col1 = str(col) + '_mergeByBadRate'
#(1), 找出最优合并方式，使得每一箱同时包含好坏样本
mergeBin = MergeBad0(trainData, col, 'target')
#(2), 依照（1）的结果对值进行合并
trainData[col1] = trainData[col].map(mergeBin)
maxPcnt = MaximumBinPcnt(trainData, col1)
#如果合并后导致有箱占比超过90%，就删除。
if maxPcnt > 0.9:
print('{} is deleted because of large percentage of single bin'.format(col))
deleted_features.append(col)
categorical_var.remove(col)
del trainData[col]
continue
#(3) 如果合并后的新的变量满足要求，就保留下来
merged_features[col] = [col1, mergeBin]
WOE_IV = CalcWOE(trainData, col1, 'target')
var_WOE[col1] = WOE_IV['WOE']
var_IV[col1] = WOE_IV['IV']
#del trainData[col]
deleted_features.append(col)
else:
WOE_IV = CalcWOE(trainData, col, 'target')
var_WOE[col] = WOE_IV['WOE']
var_IV[col] = WOE_IV['IV']
'''
对于连续型变量，处理方式如下：
1，利用卡方分箱法将变量分成5个箱
2，检查坏样本率的单带性，如果发现单调性不满足，就进行合并，直到满足单调性
'''
var_cutoff = {}
for col in numerical_var:
print("{} is in processing".format(col))
col1 = str(col) + '_Bin'
#(1),用卡方分箱法进行分箱，并且保存每一个分割的端点。例如端点=[10,20,30]表示将变量分为x<10,1030.
#特别地，缺失值-1不参与分箱
if -1 in set(trainData[col]):
special_attribute = [-1]
else:
special_attribute = []
cutOffPoints = ChiMerge(trainData, col, 'target',special_attribute=special_attribute)
var_cutoff[col] = cutOffPoints
trainData[col1] = trainData[col].map(lambda x: AssignBin(x, cutOffPoints,special_attribute=special_attribute))
#(2), check whether the bad rate is monotone
BRM = BadRateMonotone(trainData, col1, 'target',special_attribute=special_attribute)
if not BRM:
if special_attribute == []:
bin_merged = Monotone_Merge(trainData, 'target', col1)
removed_index = []
for bin in bin_merged:
if len(bin)>1:
indices = [int(b.replace('Bin ','')) for b in bin]
removed_index = removed_index+indices[0:-1]
removed_point = [cutOffPoints[k] for k in removed_index]
for p in removed_point:
cutOffPoints.remove(p)
var_cutoff[col] = cutOffPoints
trainData[col1] = trainData[col].map(lambda x: AssignBin(x, cutOffPoints, special_attribute=special_attribute))
else:
cutOffPoints2 = [i for i in cutOffPoints if i not in special_attribute]
temp = trainData.loc[~trainData[col].isin(special_attribute)]
bin_merged = Monotone_Merge(temp, 'target', col1)
removed_index = []
for bin in bin_merged:
if len(bin) > 1:
indices = [int(b.replace('Bin ', '')) for b in bin]
removed_index = removed_index + indices[0:-1]
removed_point = [cutOffPoints2[k] for k in removed_index]
for p in removed_point:
cutOffPoints2.remove(p)
cutOffPoints2 = cutOffPoints2 + special_attribute
var_cutoff[col] = cutOffPoints2
trainData[col1] = trainData[col].map(lambda x: AssignBin(x, cutOffPoints2, special_attribute=special_attribute))
#(3), 分箱后再次检查是否有单一的值占比超过90%。如果有，删除该变量
maxPcnt = MaximumBinPcnt(trainData, col1)
if maxPcnt > 0.9:
# del trainData[col1]
deleted_features.append(col)
numerical_var.remove(col)
print('we delete {} because the maximum bin occupies more than 90%'.format(col))
continue
WOE_IV = CalcWOE(trainData, col1, 'target')
var_IV[col] = WOE_IV['IV']
var_WOE[col] = WOE_IV['WOE']
#del trainData[col]
trainData.to_csv(folderOfData+'allData_2.csv', header=True,encoding='gbk', columns = trainData.columns, index=False)
with open(folderOfData+'var_WOE.pkl',"wb") as f:
f.write(pickle.dumps(var_WOE))
with open(folderOfData+'var_IV.pkl',"wb") as f:
f.write(pickle.dumps(var_IV))
with open(folderOfData+'var_cutoff.pkl',"wb") as f:
f.write(pickle.dumps(var_cutoff))
with open(folderOfData+'merged_features.pkl',"wb") as f:
f.write(pickle.dumps(merged_features))
########################################
# Step 4: WOE编码后的单变量分析与多变量分析#
########################################
trainData = pd.read_csv(folderOfData+'allData_2.csv', header=0, encoding='gbk')
with open(folderOfData+'var_WOE.pkl',"rb") as f:
var_WOE = pickle.load(f)
with open(folderOfData+'var_IV.pkl',"rb") as f:
var_IV = pickle.load(f)
with open(folderOfData+'var_cutoff.pkl',"rb") as f:
var_cutoff = pickle.load(f)
with open(folderOfData+'merged_features.pkl',"rb") as f:
merged_features = pickle.load(f)
#将一些看起来像数值变量实际上是类别变量的字段转换成字符
num2str = ['SocialNetwork_13','SocialNetwork_12','UserInfo_6','UserInfo_5','UserInfo_10','UserInfo_17']
for col in num2str:
trainData[col] = trainData[col].map(lambda x: str(x))
for col in var_WOE.keys():
print(col)
col2 = str(col)+"_WOE"
if col in var_cutoff.keys():
cutOffPoints = var_cutoff[col]
special_attribute = []
if - 1 in cutOffPoints:
special_attribute = [-1]
binValue = trainData[col].map(lambda x: AssignBin(x, cutOffPoints,special_attribute=special_attribute))
trainData[col2] = binValue.map(lambda x: var_WOE[col][x])
else:
trainData[col2] = trainData[col].map(lambda x: var_WOE[col][x])
trainData.to_csv(folderOfData+'allData_3.csv', header=True,encoding='gbk', columns = trainData.columns, index=False)
### (i) 选择IV高于阈值的变量
trainData = pd.read_csv(folderOfData+'allData_3.csv', header=0,encoding='gbk')
all_IV = list(var_IV.values())
all_IV = sorted(all_IV, reverse=True)
plt.bar(x=range(len(all_IV)), height = all_IV)
iv_threshould = 0.02
varByIV = [k for k, v in var_IV.items() if v > iv_threshould]
### (ii) 检查WOE编码后的变量的两两线性相关性
var_IV_selected = {k:var_IV[k] for k in varByIV}
var_IV_sorted = sorted(var_IV_selected.items(), key=lambda d:d[1], reverse = True)
var_IV_sorted = [i[0] for i in var_IV_sorted]
removed_var = []
roh_thresould = 0.6
for i in range(len(var_IV_sorted)-1):
if var_IV_sorted[i] not in removed_var:
x1 = var_IV_sorted[i]+"_WOE"
for j in range(i+1,len(var_IV_sorted)):
if var_IV_sorted[j] not in removed_var:
x2 = var_IV_sorted[j] + "_WOE"
roh = np.corrcoef([trainData[x1], trainData[x2]])[0, 1]
if abs(roh) >= roh_thresould:
print('the correlation coeffient between {0} and {1} is {2}'.format(x1, x2, str(roh)))
if var_IV[var_IV_sorted[i]] > var_IV[var_IV_sorted[j]]:
removed_var.append(var_IV_sorted[j])
else:
removed_var.append(var_IV_sorted[i])
var_IV_sortet_2 = [i for i in var_IV_sorted if i not in removed_var]
### (iii）检查是否有变量与其他所有变量的VIF > 10
for i in range(len(var_IV_sortet_2)):
x0 = trainData[var_IV_sortet_2[i]+'_WOE']
x0 = np.array(x0)
X_Col = [k+'_WOE' for k in var_IV_sortet_2 if k != var_IV_sortet_2[i]]
X = trainData[X_Col]
X = np.matrix(X)
regr = LinearRegression()
clr= regr.fit(X, x0)
x_pred = clr.predict(X)
R2 = 1 - ((x_pred - x0) ** 2).sum() / ((x0 - x0.mean()) ** 2).sum()
vif = 1/(1-R2)
if vif > 10:
print("Warning: the vif for {0} is {1}".format(var_IV_sortet_2[i], vif))
#########################
# Step 5: 应用逻辑回归模型#
#########################
multi_analysis = [i+'_WOE' for i in var_IV_sortet_2]
y = trainData['target']
X = trainData[multi_analysis].copy()
X['intercept'] = [1]*X.shape[0]
LR = sm.Logit(y, X).fit()
summary = LR.summary2()
pvals = LR.pvalues.to_dict()
params = LR.params.to_dict()
#发现有变量不显著，因此需要单独检验显著性
varLargeP = {k: v for k,v in pvals.items() if v >= 0.1}
varLargeP = sorted(varLargeP.items(), key=lambda d:d[1], reverse = True)
varLargeP = [i[0] for i in varLargeP]
p_value_list = {}
for var in varLargeP:
X_temp = trainData[var].copy().to_frame()
X_temp['intercept'] = [1] * X_temp.shape[0]
LR = sm.Logit(y, X_temp).fit()
p_value_list[var] = LR.pvalues[var]
for k,v in p_value_list.items():
print("{0} has p-value of {1} in univariate regression".format(k,v))
#发现有变量的系数为正，因此需要单独检验正确性
varPositive = [k for k,v in params.items() if v >= 0]
coef_list = {}
for var in varPositive:
X_temp = trainData[var].copy().to_frame()
X_temp['intercept'] = [1] * X_temp.shape[0]
LR = sm.Logit(y, X_temp).fit()
coef_list[var] = LR.params[var]
for k,v in coef_list.items():
print("{0} has coefficient of {1} in univariate regression".format(k,v))
selected_var = [multi_analysis[0]]
for var in multi_analysis[1:]:
try_vars = selected_var+[var]
X_temp = trainData[try_vars].copy()
X_temp['intercept'] = [1] * X_temp.shape[0]
LR = sm.Logit(y, X_temp).fit()
#summary = LR.summary2()
pvals, params = LR.pvalues, LR.params
del params['intercept']
if max(pvals)<0.1 and max(params)<0:
selected_var.append(var)
LR.summary2()
y_pred = LR.predict(X_temp)
y_result = pd.DataFrame({'y_pred':y_pred, 'y_real':list(trainData['target'])})
KS(y_result,'y_pred','y_real')
roc_auc_score(trainData['target'], y_pred)
################
# Step 6: 尺度化#
################
scores = Prob2Score(y_pred, 200, 100)
plt.hist(score,bins=100) 代码-计算函数自定义
import numpy as np
import pandas as pd
def SplitData(df, col, numOfSplit, special_attribute=[]):
'''
:param df: 按照col排序后的数据集
:param col: 待分箱的变量
:param numOfSplit: 切分的组别数
:param special_attribute: 在切分数据集的时候，某些特殊值需要排除在外
:return: 在原数据集上增加一列，把原始细粒度的col重新划分成粗粒度的值，便于分箱中的合并处理
'''
df2 = df.copy()
if special_attribute != []:
df2 = df.loc[~df[col].isin(special_attribute)]
N = df2.shape[0]
n = int(N/numOfSplit)
splitPointIndex = [i*n for i in range(1,numOfSplit)]
rawValues = sorted(list(df2[col]))
splitPoint = [rawValues[i] for i in splitPointIndex]
splitPoint = sorted(list(set(splitPoint)))
return splitPoint
def MaximumBinPcnt(df,col):
'''
:return: 数据集df中，变量col的分布占比
'''
N = df.shape[0]
total = df.groupby([col])[col].count()
pcnt = total*1.0/N
return max(pcnt)
def Chi2(df, total_col, bad_col):
'''
:param df: 包含全部样本总计与坏样本总计的数据框
:param total_col: 全部样本的个数
:param bad_col: 坏样本的个数
:return: 卡方值
'''
df2 = df.copy()
# 求出df中，总体的坏样本率和好样本率
badRate = sum(df2[bad_col])*1.0/sum(df2[total_col])
# 当全部样本只有好或者坏样本时，卡方值为0
if badRate in [0,1]:
return 0
df2['good'] = df2.apply(lambda x: x[total_col] - x[bad_col], axis = 1)
goodRate = sum(df2['good']) * 1.0 / sum(df2[total_col])
# 期望坏（好）样本个数＝全部样本个数*平均坏（好）样本占比
df2['badExpected'] = df[total_col].apply(lambda x: x*badRate)
df2['goodExpected'] = df[total_col].apply(lambda x: x * goodRate)
badCombined = zip(df2['badExpected'], df2[bad_col])
goodCombined = zip(df2['goodExpected'], df2['good'])
badChi = [(i[0]-i[1])**2/i[0] for i in badCombined]
goodChi = [(i[0] - i[1]) ** 2 / i[0] for i in goodCombined]
chi2 = sum(badChi) + sum(goodChi)
return chi2
def BinBadRate(df, col, target, grantRateIndicator=0):
'''
:param df: 需要计算好坏比率的数据集
:param col: 需要计算好坏比率的特征
:param target: 好坏标签
:param grantRateIndicator: 1返回总体的坏样本率，0不返回
:return: 每箱的坏样本率，以及总体的坏样本率（当grantRateIndicator＝＝1时）
'''
total = df.groupby([col])[target].count()
total = pd.DataFrame({'total': total})
bad = df.groupby([col])[target].sum()
bad = pd.DataFrame({'bad': bad})
regroup = total.merge(bad, left_index=True, right_index=True, how='left')
regroup.reset_index(level=0, inplace=True)
regroup['bad_rate'] = regroup.apply(lambda x: x.bad / x.total, axis=1)
dicts = dict(zip(regroup[col],regroup['bad_rate']))
if grantRateIndicator==0:
return (dicts, regroup)
N = sum(regroup['total'])
B = sum(regroup['bad'])
overallRate = B * 1.0 / N
return (dicts, regroup, overallRate)
def AssignGroup(x, bin):
'''
:return: 数值x在区间映射下的结果。例如，x=2，bin=[0,3,5], 由于0max(bin):
return 10e10
else:
for i in range(N-1):
if bin[i] < x <= bin[i+1]:
return bin[i+1]
def ChiMerge(df, col, target, max_interval=5,special_attribute=[],minBinPcnt=0):
'''
:param df: 包含目标变量与分箱属性的数据框
:param col: 需要分箱的属性
:param target: 目标变量，取值0或1
:param max_interval: 最大分箱数。如果原始属性的取值个数低于该参数，不执行这段函数
:param special_attribute: 不参与分箱的属性取值
:param minBinPcnt：最小箱的占比，默认为0
:return: 分箱结果
'''
colLevels = sorted(list(set(df[col])))
N_distinct = len(colLevels)
if N_distinct <= max_interval: #如果原始属性的取值个数低于max_interval，不执行这段函数
print("The number of original levels for {} is less than or equal to max intervals".format(col))
return colLevels[:-1]
else:
if len(special_attribute)>=1:
df1 = df.loc[df[col].isin(special_attribute)]
df2 = df.loc[~df[col].isin(special_attribute)]
else:
df2 = df.copy()
N_distinct = len(list(set(df2[col])))
# 步骤一: 通过col对数据集进行分组，求出每组的总样本数与坏样本数
if N_distinct > 100:
split_x = SplitData(df2, col, 100)
df2['temp'] = df2[col].map(lambda x: AssignGroup(x, split_x))
else:
df2['temp'] = df2[col]
# 总体bad rate将被用来计算expected bad count
(binBadRate, regroup, overallRate) = BinBadRate(df2, 'temp', target, grantRateIndicator=1)
# 首先，每个单独的属性值将被分为单独的一组
# 对属性值进行排序，然后两两组别进行合并
colLevels = sorted(list(set(df2['temp'])))
groupIntervals = [[i] for i in colLevels]
# 步骤二：建立循环，不断合并最优的相邻两个组别，直到：
# 1，最终分裂出来的分箱数<＝预设的最大分箱数
# 2，每箱的占比不低于预设值（可选）
# 3，每箱同时包含好坏样本
# 如果有特殊属性，那么最终分裂出来的分箱数＝预设的最大分箱数－特殊属性的个数
split_intervals = max_interval - len(special_attribute)
while (len(groupIntervals) > split_intervals): # 终止条件: 当前分箱数＝预设的分箱数
# 每次循环时, 计算合并相邻组别后的卡方值。具有最小卡方值的合并方案，是最优方案
chisqList = []
for k in range(len(groupIntervals)-1):
temp_group = groupIntervals[k] + groupIntervals[k+1]
df2b = regroup.loc[regroup['temp'].isin(temp_group)]
chisq = Chi2(df2b, 'total', 'bad')
chisqList.append(chisq)
best_comnbined = chisqList.index(min(chisqList))
groupIntervals[best_comnbined] = groupIntervals[best_comnbined] + groupIntervals[best_comnbined+1]
# 当将最优的相邻的两个变量合并在一起后，需要从原来的列表中将其移除。例如，将[3,4,5] 与[6,7]合并成[3,4,5,6,7]后，需要将[3,4,5] 与[6,7]移除，保留[3,4,5,6,7]
groupIntervals.remove(groupIntervals[best_comnbined+1])
groupIntervals = [sorted(i) for i in groupIntervals]
cutOffPoints = [max(i) for i in groupIntervals[:-1]]
# 检查是否有箱没有好或者坏样本。如果有，需要跟相邻的箱进行合并，直到每箱同时包含好坏样本
groupedvalues = df2['temp'].apply(lambda x: AssignBin(x, cutOffPoints))
df2['temp_Bin'] = groupedvalues
(binBadRate,regroup) = BinBadRate(df2, 'temp_Bin', target)
[minBadRate, maxBadRate] = [min(binBadRate.values()),max(binBadRate.values())]
while minBadRate ==0 or maxBadRate == 1:
# 找出全部为好／坏样本的箱
indexForBad01 = regroup[regroup['bad_rate'].isin([0,1])].temp_Bin.tolist()
bin=indexForBad01[0]
# 如果是最后一箱，则需要和上一个箱进行合并，也就意味着分裂点cutOffPoints中的最后一个需要移除
if bin == max(regroup.temp_Bin):
cutOffPoints = cutOffPoints[:-1]
# 如果是第一箱，则需要和下一个箱进行合并，也就意味着分裂点cutOffPoints中的第一个需要移除
elif bin == min(regroup.temp_Bin):
cutOffPoints = cutOffPoints[1:]
# 如果是中间的某一箱，则需要和前后中的一个箱进行合并，依据是较小的卡方值
else:
# 和前一箱进行合并，并且计算卡方值
currentIndex = list(regroup.temp_Bin).index(bin)
prevIndex = list(regroup.temp_Bin)[currentIndex - 1]
df3 = df2.loc[df2['temp_Bin'].isin([prevIndex, bin])]
(binBadRate, df2b) = BinBadRate(df3, 'temp_Bin', target)
chisq1 = Chi2(df2b, 'total', 'bad')
# 和后一箱进行合并，并且计算卡方值
laterIndex = list(regroup.temp_Bin)[currentIndex + 1]
df3b = df2.loc[df2['temp_Bin'].isin([laterIndex, bin])]
(binBadRate, df2b) = BinBadRate(df3b, 'temp_Bin', target)
chisq2 = Chi2(df2b, 'total', 'bad')
if chisq1 < chisq2:
cutOffPoints.remove(cutOffPoints[currentIndex - 1])
else:
cutOffPoints.remove(cutOffPoints[currentIndex])
# 完成合并之后，需要再次计算新的分箱准则下，每箱是否同时包含好坏样本
groupedvalues = df2['temp'].apply(lambda x: AssignBin(x, cutOffPoints))
df2['temp_Bin'] = groupedvalues
(binBadRate, regroup) = BinBadRate(df2, 'temp_Bin', target)
[minBadRate, maxBadRate] = [min(binBadRate.values()), max(binBadRate.values())]
# 需要检查分箱后的最小占比
if minBinPcnt > 0:
groupedvalues = df2['temp'].apply(lambda x: AssignBin(x, cutOffPoints))
df2['temp_Bin'] = groupedvalues
valueCounts = groupedvalues.value_counts().to_frame()
N = sum(valueCounts['temp'])
valueCounts['pcnt'] = valueCounts['temp'].apply(lambda x: x * 1.0 / N)
valueCounts = valueCounts.sort_index()
minPcnt = min(valueCounts['pcnt'])
while minPcnt < minBinPcnt and len(cutOffPoints) > 2:
# 找出占比最小的箱
indexForMinPcnt = valueCounts[valueCounts['pcnt'] == minPcnt].index.tolist()[0]
# 如果占比最小的箱是最后一箱，则需要和上一个箱进行合并，也就意味着分裂点cutOffPoints中的最后一个需要移除
if indexForMinPcnt == max(valueCounts.index):
cutOffPoints = cutOffPoints[:-1]
# 如果占比最小的箱是第一箱，则需要和下一个箱进行合并，也就意味着分裂点cutOffPoints中的第一个需要移除
elif indexForMinPcnt == min(valueCounts.index):
cutOffPoints = cutOffPoints[1:]
# 如果占比最小的箱是中间的某一箱，则需要和前后中的一个箱进行合并，依据是较小的卡方值
else:
# 和前一箱进行合并，并且计算卡方值
currentIndex = list(valueCounts.index).index(indexForMinPcnt)
prevIndex = list(valueCounts.index)[currentIndex - 1]
df3 = df2.loc[df2['temp_Bin'].isin([prevIndex, indexForMinPcnt])]
(binBadRate, df2b) = BinBadRate(df3, 'temp_Bin', target)
chisq1 = Chi2(df2b, 'total', 'bad')
# 和后一箱进行合并，并且计算卡方值
laterIndex = list(valueCounts.index)[currentIndex + 1]
df3b = df2.loc[df2['temp_Bin'].isin([laterIndex, indexForMinPcnt])]
(binBadRate, df2b) = BinBadRate(df3b, 'temp_Bin', target)
chisq2 = Chi2(df2b, 'total', 'bad')
if chisq1 < chisq2:
cutOffPoints.remove(cutOffPoints[currentIndex - 1])
else:
cutOffPoints.remove(cutOffPoints[currentIndex])
groupedvalues = df2['temp'].apply(lambda x: AssignBin(x, cutOffPoints))
df2['temp_Bin'] = groupedvalues
valueCounts = groupedvalues.value_counts().to_frame()
valueCounts['pcnt'] = valueCounts['temp'].apply(lambda x: x * 1.0 / N)
valueCounts = valueCounts.sort_index()
minPcnt = min(valueCounts['pcnt'])
cutOffPoints = special_attribute + cutOffPoints
return cutOffPoints
def BadRateEncoding(df, col, target):
'''
:return: 在数据集df中，用坏样本率给col进行编码。target表示坏样本标签
'''
regroup = BinBadRate(df, col, target, grantRateIndicator=0)[1]
br_dict = regroup[[col,'bad_rate']].set_index([col]).to_dict(orient='index')
for k, v in br_dict.items():
br_dict[k] = v['bad_rate']
badRateEnconding = df[col].map(lambda x: br_dict[x])
return {'encoding':badRateEnconding, 'bad_rate':br_dict}
def AssignBin(x, cutOffPoints,special_attribute=[]):
'''
:param x: 某个变量的某个取值
:param cutOffPoints: 上述变量的分箱结果，用切分点表示
:param special_attribute: 不参与分箱的特殊取值
:return: 分箱后的对应的第几个箱，从0开始
例如, cutOffPoints = [10,20,30], 对于 x = 7, 返回 Bin 0；对于x=23，返回Bin 2； 对于x = 35, return Bin 3。
对于特殊值，返回的序列数前加"-"
'''
cutOffPoints2 = [i for i in cutOffPoints if i not in special_attribute]
numBin = len(cutOffPoints2)
if x in special_attribute:
i = special_attribute.index(x)+1
return 'Bin {}'.format(0-i)
if x<=cutOffPoints2[0]:
return 'Bin 0'
elif x > cutOffPoints2[-1]:
return 'Bin {}'.format(numBin)
else:
for i in range(0,numBin):
if cutOffPoints2[i] < x <= cutOffPoints2[i+1]:
return 'Bin {}'.format(i+1)
def CalcWOE(df, col, target):
'''
:param df: 包含需要计算WOE的变量和目标变量
:param col: 需要计算WOE、IV的变量，必须是分箱后的变量，或者不需要分箱的类别型变量
:param target: 目标变量，0、1表示好、坏
:return: 返回WOE和IV
'''
total = df.groupby([col])[target].count()
total = pd.DataFrame({'total': total})
bad = df.groupby([col])[target].sum()
bad = pd.DataFrame({'bad': bad})
regroup = total.merge(bad, left_index=True, right_index=True, how='left')
regroup.reset_index(level=0, inplace=True)
N = sum(regroup['total'])
B = sum(regroup['bad'])
regroup['good'] = regroup['total'] - regroup['bad']
G = N - B
regroup['bad_pcnt'] = regroup['bad'].map(lambda x: x*1.0/B)
regroup['good_pcnt'] = regroup['good'].map(lambda x: x * 1.0 / G)
regroup['WOE'] = regroup.apply(lambda x: np.log(x.good_pcnt*1.0/x.bad_pcnt),axis = 1)
WOE_dict = regroup[[col,'WOE']].set_index(col).to_dict(orient='index')
for k, v in WOE_dict.items():
WOE_dict[k] = v['WOE']
IV = regroup.apply(lambda x: (x.good_pcnt-x.bad_pcnt)*np.log(x.good_pcnt*1.0/x.bad_pcnt),axis = 1)
IV = sum(IV)
return {"WOE": WOE_dict, 'IV':IV}
def FeatureMonotone(x):
'''
:return: 返回序列x中有几个元素不满足单调性，以及这些元素的位置。
例如，x=[1,3,2,5], 元素3比前后两个元素都大，不满足单调性；元素2比前后两个元素都小，也不满足单调性。
故返回的不满足单调性的元素个数为2，位置为1和2.
'''
monotone = [x[i]x[i+1] and x[i] > x[i-1] for i in range(1,len(x)-1)]
index_of_nonmonotone = [i+1 for i in range(len(monotone)) if monotone[i]]
return {'count_of_nonmonotone':monotone.count(True), 'index_of_nonmonotone':index_of_nonmonotone}
## 判断某变量的坏样本率是否单调
def BadRateMonotone(df, sortByVar, target,special_attribute = []):
'''
:param df: 包含检验坏样本率的变量，和目标变量
:param sortByVar: 需要检验坏样本率的变量
:param target: 目标变量，0、1表示好、坏
:param special_attribute: 不参与检验的特殊值
:return: 坏样本率单调与否
'''
df2 = df.loc[~df[sortByVar].isin(special_attribute)]
if len(set(df2[sortByVar])) <= 2:
return True
regroup = BinBadRate(df2, sortByVar, target)[1]
combined = zip(regroup['total'],regroup['bad'])
badRate = [x[1]*1.0/x[0] for x in combined]
badRateNotMonotone = FeatureMonotone(badRate)['count_of_nonmonotone']
if badRateNotMonotone > 0:
return False
else:
return True
def MergeBad0(df,col,target, direction='bad'):
'''
:param df: 包含检验0％或者100%坏样本率
:param col: 分箱后的变量或者类别型变量。检验其中是否有一组或者多组没有坏样本或者没有好样本。如果是，则需要进行合并
:param target: 目标变量，0、1表示好、坏
:return: 合并方案，使得每个组里同时包含好坏样本
'''
regroup = BinBadRate(df, col, target)[1]
if direction == 'bad':
# 如果是合并0坏样本率的组，则跟最小的非0坏样本率的组进行合并
regroup = regroup.sort_values(by = 'bad_rate')
else:
# 如果是合并0好样本率的组，则跟最小的非0好样本率的组进行合并
regroup = regroup.sort_values(by='bad_rate',ascending=False)
regroup.index = range(regroup.shape[0])
col_regroup = [[i] for i in regroup[col]]
del_index = []
for i in range(regroup.shape[0]-1):
col_regroup[i+1] = col_regroup[i] + col_regroup[i+1]
del_index.append(i)
if direction == 'bad':
if regroup['bad_rate'][i+1] > 0:
break
else:
if regroup['bad_rate'][i+1] < 1:
break
col_regroup2 = [col_regroup[i] for i in range(len(col_regroup)) if i not in del_index]
newGroup = {}
for i in range(len(col_regroup2)):
for g2 in col_regroup2[i]:
newGroup[g2] = 'Bin '+str(i)
return newGroup
def Monotone_Merge(df, target, col):
'''
:return:将数据集df中，不满足坏样本率单调性的变量col进行合并，使得合并后的新的变量中，坏样本率单调，输出合并方案。
例如，col=[Bin 0, Bin 1, Bin 2, Bin 3, Bin 4]是不满足坏样本率单调性的。合并后的col是：
[Bin 0&Bin 1, Bin 2, Bin 3, Bin 4].
合并只能在相邻的箱中进行。
迭代地寻找最优合并方案。每一步迭代时，都尝试将所有非单调的箱进行合并，每一次尝试的合并都是跟前后箱进行合并再做比较
'''
def MergeMatrix(m, i,j,k):
'''
:param m: 需要合并行的矩阵
:param i,j: 合并第i和j行
:param k: 删除第k行
:return: 合并后的矩阵
'''
m[i, :] = m[i, :] + m[j, :]
m = np.delete(m, k, axis=0)
return m
def Merge_adjacent_Rows(i, bad_by_bin_current, bins_list_current, not_monotone_count_current):
'''
:param i: 需要将第i行与前、后的行分别进行合并，比较哪种合并方案最佳。判断准则是，合并后非单调性程度减轻，且更加均匀
:param bad_by_bin_current:合并前的分箱矩阵，包括每一箱的样本个数、坏样本个数和坏样本率
:param bins_list_current: 合并前的分箱方案
:param not_monotone_count_current:合并前的非单调性元素个数
:return:分箱后的分箱矩阵、分箱方案、非单调性元素个数和衡量均匀性的指标balance
'''
i_prev = i - 1
i_next = i + 1
bins_list = bins_list_current.copy()
bad_by_bin = bad_by_bin_current.copy()
not_monotone_count = not_monotone_count_current
#合并方案a：将第i箱与前一箱进行合并
bad_by_bin2a = MergeMatrix(bad_by_bin.copy(), i_prev, i, i)
bad_by_bin2a[i_prev, -1] = bad_by_bin2a[i_prev, -2] / bad_by_bin2a[i_prev, -3]
not_monotone_count2a = FeatureMonotone(bad_by_bin2a[:, -1])['count_of_nonmonotone']
# 合并方案b：将第i行与后一行进行合并
bad_by_bin2b = MergeMatrix(bad_by_bin.copy(), i, i_next, i_next)
bad_by_bin2b[i, -1] = bad_by_bin2b[i, -2] / bad_by_bin2b[i, -3]
not_monotone_count2b = FeatureMonotone(bad_by_bin2b[:, -1])['count_of_nonmonotone']
balance = ((bad_by_bin[:, 1] / N).T * (bad_by_bin[:, 1] / N))[0, 0]
balance_a = ((bad_by_bin2a[:, 1] / N).T * (bad_by_bin2a[:, 1] / N))[0, 0]
balance_b = ((bad_by_bin2b[:, 1] / N).T * (bad_by_bin2b[:, 1] / N))[0, 0]
#满足下述2种情况时返回方案a：（1）方案a能减轻非单调性而方案b不能；（2）方案a和b都能减轻非单调性，但是方案a的样本均匀性优于方案b
if not_monotone_count2a < not_monotone_count_current and not_monotone_count2b >= not_monotone_count_current or \
not_monotone_count2a < not_monotone_count_current and not_monotone_count2b < not_monotone_count_current and balance_a < balance_b:
bins_list[i_prev] = bins_list[i_prev] + bins_list[i]
bins_list.remove(bins_list[i])
bad_by_bin = bad_by_bin2a
not_monotone_count = not_monotone_count2a
balance = balance_a
# 同样地，满足下述2种情况时返回方案b：（1）方案b能减轻非单调性而方案a不能；（2）方案a和b都能减轻非单调性，但是方案b的样本均匀性优于方案a
elif not_monotone_count2a >= not_monotone_count_current and not_monotone_count2b < not_monotone_count_current or \
not_monotone_count2a < not_monotone_count_current and not_monotone_count2b < not_monotone_count_current and balance_a > balance_b:
bins_list[i] = bins_list[i] + bins_list[i_next]
bins_list.remove(bins_list[i_next])
bad_by_bin = bad_by_bin2b
not_monotone_count = not_monotone_count2b
balance = balance_b
#如果方案a和b都不能减轻非单调性，返回均匀性更优的合并方案
else:
if balance_a< balance_b:
bins_list[i] = bins_list[i] + bins_list[i_next]
bins_list.remove(bins_list[i_next])
bad_by_bin = bad_by_bin2b
not_monotone_count = not_monotone_count2b
balance = balance_b
else:
bins_list[i] = bins_list[i] + bins_list[i_next]
bins_list.remove(bins_list[i_next])
bad_by_bin = bad_by_bin2b
not_monotone_count = not_monotone_count2b
balance = balance_b
return {'bins_list': bins_list, 'bad_by_bin': bad_by_bin, 'not_monotone_count': not_monotone_count,
'balance': balance}
N = df.shape[0]
[badrate_bin, bad_by_bin] = BinBadRate(df, col, target)
bins = list(bad_by_bin[col])
bins_list = [[i] for i in bins]
badRate = sorted(badrate_bin.items(), key=lambda x: x[0])
badRate = [i[1] for i in badRate]
not_monotone_count, not_monotone_position = FeatureMonotone(badRate)['count_of_nonmonotone'], FeatureMonotone(badRate)['index_of_nonmonotone']
#迭代地寻找最优合并方案，终止条件是:当前的坏样本率已经单调，或者当前只有2箱
while (not_monotone_count > 0 and len(bins_list)>2):
#当非单调的箱的个数超过1个时，每一次迭代中都尝试每一个箱的最优合并方案
all_possible_merging = []
for i in not_monotone_position:
merge_adjacent_rows = Merge_adjacent_Rows(i, np.mat(bad_by_bin), bins_list, not_monotone_count)
all_possible_merging.append(merge_adjacent_rows)
balance_list = [i['balance'] for i in all_possible_merging]
not_monotone_count_new = [i['not_monotone_count'] for i in all_possible_merging]
#如果所有的合并方案都不能减轻当前的非单调性，就选择更加均匀的合并方案
if min(not_monotone_count_new) >= not_monotone_count:
best_merging_position = balance_list.index(min(balance_list))
#如果有多个合并方案都能减轻当前的非单调性，也选择更加均匀的合并方案
else:
better_merging_index = [i for i in range(len(not_monotone_count_new)) if not_monotone_count_new[i] < not_monotone_count]
better_balance = [balance_list[i] for i in better_merging_index]
best_balance_index = better_balance.index(min(better_balance))
best_merging_position = better_merging_index[best_balance_index]
bins_list = all_possible_merging[best_merging_position]['bins_list']
bad_by_bin = all_possible_merging[best_merging_position]['bad_by_bin']
not_monotone_count = all_possible_merging[best_merging_position]['not_monotone_count']
not_monotone_position = FeatureMonotone(bad_by_bin[:, 3])['index_of_nonmonotone']
return bins_list
def Prob2Score(prob, basePoint, PDO):
#将概率转化成分数且为正整数
y = np.log(prob/(1-prob))
y2 = basePoint+PDO/np.log(2)*(-y)
score = y2.astype("int")
return score
### 计算KS值
def KS(df, score, target):
'''
:param df: 包含目标变量与预测值的数据集
:param score: 得分或者概率
:param target: 目标变量
:return: KS值
:return: KS值
'''
total = df.groupby([score])[target].count()
bad = df.groupby([score])[target].sum()
all = pd.DataFrame({'total':total, 'bad':bad})
all['good'] = all['total'] - all['bad']
all[score] = all.index
all = all.sort_values(by=score,ascending=False)
all.index = range(len(all))
all['badCumRate'] = all['bad'].cumsum() / all['bad'].sum()
all['goodCumRate'] = all['good'].cumsum() / all['good'].sum()
KS = all.apply(lambda x: x.badCumRate - x.goodCumRate, axis=1)
return max(KS)