Question

我想进行rfm分析并返回特定段的所有ID。

class RfmTable:
quintiles = {}

def _init(self,rfm):
    self.rfm = rfm

def conn_data(self,table_name,start_time,end_time):        
    conn = psycopg2.connect(database=database_name, user=user_name,                           
                        password=password_info, host=host_info, port=port_info)
    cursor = conn.cursor()
    cursor.execute('select random_user_id,type,revenue,timestamp from '+table_name+ " where timestamp between EXTRACT(EPOCH FROM date '"+start_time+"') AND EXTRACT(EPOCH FROM date '"+end_time+"')")
    a = cursor.fetchall()

    df = pd.DataFrame(a, columns=[user_id, type,revenue,timestamp])
    df[date] = pd.DatetimeIndex(pd.to_datetime(df[timestamp], unit='s')).tz_localize('UTC').tz_convert('Asia/Shanghai')
    df[date] = df[date].apply(lambda x: x.strftime('%Y-%m-%d'))
    df = df.drop([timestamp],axis = 1)

    df_revenue = df[df[type].isin([revenue])]
    df_install = df[df[type].isin([install])]
    df_revenue = pd.merge(df_revenue,df_install,on=user_id,suffixes=('','_i'))
    df_revenue = df_revenue.loc[:,[user_id,type,revenue,date]]
    df_revenue[revenue]=pd.DataFrame(df_revenue[revenue],dtype=np.float)
    df_revenue[revenue] = df_revenue[revenue]/100
    df_revenue[date]=pd.to_datetime(df_revenue[date])

    NOW = df_revenue[date].max() + timedelta(days=1)

    rfm = df_revenue.groupby(user_id).agg({date: lambda x: (NOW - x.max()).days, # Recency (最近一次充值时间的间隔)
                                            type:  lambda x: len(x),      # Frequency                                                
                                            revenue: lambda x: x.sum()}) # Monetary

    rfm[date] = rfm[date].astype(int)
    rfm.rename(columns={date: 'Recency', 
                                type: 'Frequency', 
                                revenue: 'Monetary'}, inplace=True)
    quintiles = rfm.quantile(q=[0.2,0.4,0.6,0.8]).to_dict()

    def r_score(x,c):
        if x <= quintiles[c][.2]:
            return 5
        elif x <= quintiles[c][.4]:
            return 4
        elif x <= quintiles[c][.6]:
            return 3
        elif x <= quintiles[c][.8]:
            return 2
        else:
            return 1

    def fm_score(x, c):
        if x <= quintiles[c][.2]:
            return 1
        elif x <= quintiles[c][.4]:
            return 2
        elif x <= quintiles[c][.6]:
            return 3
        elif x <= quintiles[c][.8]:
            return 4
        else:
            return 5    

    rfm['R'] = rfm['Recency'].apply(lambda x: r_score(x,'Recency'))
    rfm['F'] = rfm['Frequency'].apply(lambda x: fm_score(x, 'Frequency'))
    rfm['M'] = rfm['Monetary'].apply(lambda x: fm_score(x, 'Monetary'))

    rfm['RFM Score'] = rfm['R'].map(str) + rfm['F'].map(str) + rfm['M'].map(str)

    segt_map = {
        r'[4-5][4-5][5]': '重要价值用户',
        r'[4-5][1-3][5]': '重要保持用户',
        r'[1-3][4-5][5]': '重要发展用户',
        r'[1-3][1-3][5]': '重要挽留用户',
        r'[4-5][4-5][1-4]':'一般价值用户',
        r'[4-5][1-3][1-4]': '一般发展用户',
        r'[1-3][4-5][1-4]': '一般保持用户',
        r'[1-3][1-3][1-4]': '一般挽留用户'
    }

    rfm['Segment'] = rfm['R'].map(str) + rfm['F'].map(str) + rfm['M'].map(str)
    rfm['Segment'] = rfm['Segment'].replace(segt_map, regex=True)
    rfm.reset_index(inplace=True)
    return rfm

def get(self,segment):
    return self.rfm[self.rfm['Segment'].isin([segment])][user_id].values.tolist()

conn_data方法创建一个名为rfm的数据帧，其中包含id，新近度，频率，货币，R，F，M，RFM Scole和Segment。

我想使用get方法返回某些输入段的所有ID。我写完之后：

data1 = RfmTable()
data1.conn_data('adjust_logs_lanyue','20190610','20190613')
data1.get('重要挽留用户')

有错误：'AttributeError：'RfmTable'对象没有属性'rfm'

那我该怎么做才能纠正呢？

Answer 1

与其他对象语言不同，Python在Java或C ++中没有像this这样的隐式对象的概念。这意味着当您在rfm中使用conn_data时，您只是在使用一个局部变量，该局部变量将在函数返回时立即消失。您必须显式地影响对象的属性：

def conn_data(self,table_name,start_time,end_time):        
    ...
    self.rfm = rfm             # store the object as an attribute of self
    return rfm

“ RfmTable”对象没有属性“ rfm”

1 个答案: