汽车金融评分卡

from sklearn.preprocessing import Imputer#只支持 均值，中位数，众数填补
inputer = Imputer(missing_values = ‘NaN‘, strategy = ‘mean‘, axis = 0)
inputer = inputer.fit(X)
X = inputer.transform(X)
主要参数说明：
missing_values：缺失值，可以为整数或NaN(缺失值numpy.nan用字符串‘NaN’表示)，默认为NaN

strategy：替换策略，字符串，默认用均值‘mean’替换

①若为mean时，用特征列的均值替换

②若为median时，用特征列的中位数替换

③若为most_frequent时，用特征列的众数替换

axis：指定轴数，默认axis=0代表列，axis=1代表行

copy：设置为True代表不在原数据集上修改，设置为False时，就地修改，存在如下情况时，即使设置为False时，也不会就地修改

①X不是浮点值数组

②X是稀疏且missing_values=0

③axis=0且X为CRS矩阵

④axis=1且X为CSC矩阵

statistics_属性：axis设置为0时，每个特征的填充值数组，axis=1时，报没有该属性错误

for i in rej_Floatcolumns:
    rej[i]=acc[i].fillna(rej[i].mean())
#mean()均值
#mode（）众数
#median（）中位数

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0<x<3,x映射成3
    ‘‘‘
    N = len(bin)
    if x<=min(bin):
        return min(bin)
    elif x>max(bin):
        return 10e10
    else:
        for i in range(N-1):
            if bin[i] < x <= bin[i+1]:
                return bin[i+1]

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 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

#计算woe 和IV 值
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}

 ‘‘‘
    :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 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

#该包中对变量进行分箱的原理类似于二叉决策树，只是决定如何划分的目标函数是iv值

import woe.feature_process as fp
import woe.eval as eva
continuous=[]
for i in col:
    con=fp.proc_woe_continuous(df=df3,
                           var=i,#变量名
                           global_bt=df3.target.sum(),#正样本总量
                           global_gt=df3.shape[0]-df3.target.sum(),#负样本总量
                           min_sample=0.05*df3.shape[0],#单个区间最小样本量阈值
                           alpha=0.05)
    continuous.append(con)
#该分箱方法缺点分箱太多，不利于业务开展，且手动调整过于繁琐，且容易照成失误

df_continuous=eva.eval_feature_detail(Info_Value_list=continuous)
df_continuous.to_csv(‘fx2.csv‘,index=False)