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300字范文 > 【风控模型】神经网络DNN算法构建信用评分卡模型

【风控模型】神经网络DNN算法构建信用评分卡模型

时间:2024-07-06 03:46:07

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【风控模型】神经网络DNN算法构建信用评分卡模型

【博客地址】:/sunyaowu315

【博客大纲地址】:/sunyaowu315/article/details/82905347

数据集介绍:

本次案例分析所用的数据,是拍拍贷发起的一次与信贷申请审核工作相关的竞赛数据集。其中共有3份文件:

PPD_Training_Master_GBK_3_1_Training_Set.csv :信贷用户在拍拍贷上的申报信息和部分三方数据信息,以及需要预测的目标变量。PPD_LogInfo_3_1_Training_Set : 信贷客户的登录信息PPD_Userupdate_Info_3_1_Training_Set :部分客户的信息修改行为

建模工作就是从上述三个文件中对数据进行加工,提取特征并且建立合适的模型,对贷后表现做预测。

关键字段

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主程序

import pandas as pdimport datetimeimport collectionsimport numpy as npimport numbersimport randomimport sys_path = r'C:\Users\A3\Desktop\DNN_scorecard'sys.path.append(_path)import picklefrom sklearn.model_selection import train_test_splitfrom sklearn.metrics import roc_auc_scorefrom importlib import reloadfrom matplotlib import pyplot as pltimport operatorreload(sys)#sys.setdefaultencoding( "utf-8")# -*- coding: utf-8 -*-### 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] = freqreturn freq_twdef DeivdedByZero(nominator, denominator):'''当分母为0时,返回0;否则返回正常值'''if denominator == 0:return 0else: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 ydef 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 xelse:randIndex = random.randint(0, len(sampledList)-1)return sampledList[randIndex]def Outlier_Dectection(df,x):''':param df::param x::return:'''p25, p75 = np.percentile(df[x], 25),np.percentile(df[x], 75)d = p75 - p25upper, lower = p75 + 1.5*d, p25-1.5*dtruncation = df[x].map(lambda x: max(min(upper, x), lower))return truncation#############################################################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 = 'C:/Users/A3/Desktop/DNN_scorecard/'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)#将数据集分为训练集与测试集all_ids = data2['Idx']train_ids, test_ids = train_test_split(all_ids, test_size=0.3)train_ids = pd.DataFrame(train_ids)test_ids = pd.DataFrame(test_ids)data1_train = pd.merge(left=train_ids,right = data1, on='Idx', how='inner')data2_train = pd.merge(left=train_ids,right = data2, on='Idx', how='inner')data3_train = pd.merge(left=train_ids,right = data3, on='Idx', how='inner')data1_test = pd.merge(left=test_ids,right = data1, on='Idx', how='inner')data2_test = pd.merge(left=test_ids,right = data2, on='Idx', how='inner')data3_test = pd.merge(left=test_ids,right = data3, on='Idx', how='inner')############################################################################################## 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 matchdata2_train['city_match'] = data2_train.apply(lambda x: int(x.UserInfo_2 == x.UserInfo_4 == x.UserInfo_8 == x.UserInfo_20),axis = 1)del data2_train['UserInfo_2']del data2_train['UserInfo_4']del data2_train['UserInfo_8']del data2_train['UserInfo_20']### 提取申请日期,计算日期差,查看日期差的分布data1_train['logInfo'] = data1_train['LogInfo3'].map(lambda x: datetime.datetime.strptime(x,'%Y-%m-%d'))data1_train['Listinginfo'] = data1_train['Listinginfo1'].map(lambda x: datetime.datetime.strptime(x,'%Y-%m-%d'))data1_train['ListingGap'] = data1_train[['logInfo','Listinginfo']].apply(lambda x: (x[1]-x[0]).days,axis = 1)### 提取申请日期,计算日期差,查看日期差的分布'''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_train['Idx'].drop_duplicates()})for tw in time_window:data1_train['TruncatedLogInfo'] = data1_train['Listinginfo'].map(lambda x: x + datetime.timedelta(-tw))temp = data1_train.loc[data1_train['logInfo'] >= data1_train['TruncatedLogInfo']]for var in var_list:#count the frequences of LogInfo1 and LogInfo2count_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 LogInfo2Idx_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 LogInfo2data1GroupbyIdx[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_train['ListingInfo'] = data3_train['ListingInfo1'].map(lambda x: datetime.datetime.strptime(x,'%Y/%m/%d'))data3_train['UserupdateInfo'] = data3_train['UserupdateInfo2'].map(lambda x: datetime.datetime.strptime(x,'%Y/%m/%d'))data3_train['ListingGap'] = data3_train[['UserupdateInfo','ListingInfo']].apply(lambda x: (x[1]-x[0]).days,axis = 1)collections.Counter(data3_train['ListingGap'])hist_ListingGap = np.histogram(data3_train['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_train['UserupdateInfo1'] = data3_train['UserupdateInfo1'].map(ChangeContent)data3GroupbyIdx = pd.DataFrame({'Idx':data3_train['Idx'].drop_duplicates()})time_window = [7, 30, 60, 90, 120, 150, 180]for tw in time_window:data3_train['TruncatedLogInfo'] = data3_train['ListingInfo'].map(lambda x: x + datetime.timedelta(-tw))temp = data3_train.loc[data3_train['UserupdateInfo'] >= data3_train['TruncatedLogInfo']]#frequency of updatingfreq_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 typesIdx_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 typedata3GroupbyIdx['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, PHONEIdx_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_SetallData = pd.concat([data2_train.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预处理#########################################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_countlarge_value = value_count[value_count== max(value_count)].index[0]col_large_value[col] = large_valuecol_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] = 0continuebad_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,否则保留。'''missing_pcnt_threshould_1 = 0.8for 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]allData_bk = allData.copy()'''用one-hot对类别型变量进行编码'''dummy_map = {}dummy_columns = []for raw_col in categorical_var:dummies = pd.get_dummies(allData.loc[:, raw_col], prefix=raw_col)col_onehot = pd.concat([allData[raw_col], dummies], axis=1)col_onehot = col_onehot.drop_duplicates()allData = pd.concat([allData, dummies], axis=1)del allData[raw_col]dummy_map[raw_col] = col_onehotdummy_columns = dummy_columns + list(dummies)with open(folderOfData+'dummy_map.pkl',"wb") as f:f.write(pickle.dumps(dummy_map))with open(folderOfData+'dummy_columns.pkl',"wb") as f:f.write(pickle.dumps(dummy_columns))'''检查数值型变量'''missing_pcnt_threshould_2 = 0.8deleted_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].indexnot_missing_sample = random.sample(list(not_missing), len(missing_position))allData.loc[missing_position,col] = not_missing_sample#del allData[col]#allData[col] = makeupedmissingRate2 = 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]'''对极端值变量做处理。'''max_min_standardized = {}for col in numerical_var:truncation = Outlier_Dectection(allData, col)upper, lower = max(truncation), min(truncation)d = upper - lowerif d == 0:print("{} is almost a constant".format(col))numerical_var.remove(col)continueallData[col] = truncation.map(lambda x: (upper - x)/d)max_min_standardized[col] = [lower, upper]with open(folderOfData+'max_min_standardized.pkl',"wb") as f:f.write(pickle.dumps(max_min_standardized))allData.to_csv(folderOfData+'allData_1_DNN.csv', header=True,encoding='gbk', columns = allData.columns, index=False)allData = pd.read_csv(folderOfData+'allData_1_DNN.csv', header=0,encoding='gbk')######################################### Step 3: 构建基于TensorFlow的神经网络模型 #########################################allFeatures = list(allData.columns)allFeatures.remove('target')with open(folderOfData+'allFeatures.pkl',"wb") as f:f.write(pickle.dumps(allFeatures))x_train = np.matrix(allData[allFeatures])y_train = np.matrix(allData['target']).T#进一步将训练集拆分成训练集和验证集。在新训练集上进行参数估计,在验证集上决定最优的参数x_train, x_validation, y_train, y_validation = train_test_split(x_train, y_train,test_size=0.4,random_state=9)#Example: select the best number of units in the 1-layer hidden layerimport tensorflow as tffrom tensorflow.contrib.learn.python.learn.estimators import SKCompatno_hidden_units_selection = {}feature_columns = [tf.contrib.layers.real_valued_column("", dimension = x_train.shape[1])]for no_hidden_units in range(50,101,10):print("the current choise of hidden units number is {}".format(no_hidden_units))clf0 = tf.contrib.learn.DNNClassifier(feature_columns = feature_columns,hidden_units=[no_hidden_units, no_hidden_units-10,no_hidden_units-20],n_classes=2,dropout = 0.5)clf = SKCompat(clf0)clf.fit(x_train, y_train, batch_size=256,steps = 100000)#monitor the performance of the model using AUC scoreclf_pred_proba = clf._estimator.predict_proba(x_validation)pred_proba = [i[1] for i in clf_pred_proba]auc_score = roc_auc_score(y_validation.getA(),pred_proba)no_hidden_units_selection[no_hidden_units] = auc_scorebest_hidden_units = max(no_hidden_units_selection.items(), key=operator.itemgetter(1))[0] #80#Example: check the dropout effectdropout_selection = {}feature_columns = [tf.contrib.layers.real_valued_column("", dimension = x_train.shape[1])]for dropout_prob in np.linspace(0,0.99,20):print("the current choise of drop out rate is {}".format(dropout_prob))clf0 = tf.contrib.learn.DNNClassifier(feature_columns = feature_columns,hidden_units = [best_hidden_units, best_hidden_units-10,best_hidden_units-20],n_classes=2,dropout = dropout_prob)clf = SKCompat(clf0)clf.fit(x_train, y_train, batch_size=256,steps = 100000)#monitor the performance of the model using AUC scoreclf_pred_proba = clf._estimator.predict_proba(x_validation)pred_proba = [i[1] for i in clf_pred_proba]auc_score = roc_auc_score(y_validation.getA(),pred_proba)dropout_selection[dropout_prob] = auc_scorebest_dropout_prob = max(dropout_selection.items(), key=operator.itemgetter(1))[0] #0.0#the best model isclf1 = tf.contrib.learn.DNNClassifier(feature_columns = feature_columns,hidden_units = [best_hidden_units, best_hidden_units-10,best_hidden_units-20],n_classes=2,dropout = best_dropout_prob)clf1.fit(x_train, y_train, batch_size=256,steps = 100000)clf_pred_proba = clf1.predict_proba(x_train)pred_proba = [i[1] for i in clf_pred_proba]auc_score = roc_auc_score(y_train.getA(),pred_proba) #0.773

功能模块

import numpy as npimport pandas as pddef 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 splitPointdef 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/Nreturn 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0if badRate in [0,1]:return 0df2['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 chi2def 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 * 1.0 / 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 / Nreturn (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0<x<3,x映射成3'''N = len(bin)if x<=min(bin):return min(bin)elif x>max(bin):return 10e10else: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'] = groupedvaluesvalueCounts = 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'] = groupedvaluesvalueCounts = 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 + cutOffPointsreturn cutOffPointsdef 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)+1return '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 - Bregroup['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] or 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 Trueregroup = 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 Falseelse:return Truedef 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:breakelse:if regroup['bad_rate'][i+1] < 1:breakcol_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 newGroupdef 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 mdef 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 - 1i_next = i + 1bins_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bif 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_bin2anot_monotone_count = not_monotone_count2abalance = 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aelif 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_bin2bnot_monotone_count = not_monotone_count2bbalance = 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_bin2bnot_monotone_count = not_monotone_count2bbalance = balance_belse:bins_list[i] = bins_list[i] + bins_list[i_next]bins_list.remove(bins_list[i_next])bad_by_bin = bad_by_bin2bnot_monotone_count = not_monotone_count2bbalance = balance_breturn {'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_listdef Prob2Score(prob, basePoint, PDO):#将概率转化成分数且为正整数y = np.log(prob/(1-prob))return (basePoint+PDO/np.log(2)*(-y)).map(lambda x: int(x))### 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'''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.indexall = 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)

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