用于计算大型数据帧的更快的函数或脚本

Question

我有以下信息的在线用户数据

df.head()

USER Timestamp  day_of_week Busi_days   Busi_hours
AAD 2017-07-11 09:31:44 TRUE    TRUE
AAD 2017-07-11 23:24:43 TRUE    FALSE
AAD 2017-07-12 13:24:43 TRUE    TRUE
SAP 2017-07-23 14:24:34 FALSE   FALSE
SAP 2017-07-24 16:58:49 TRUE    TRUE
YAS 2017-07-31 21:10:35 TRUE    FALSE

我想计算USER列的活动并创建三个新列，即：1。Activity：信息基于用户的活跃程度，这意味着如果同一个用户点击了两次以上，那么就称之为TRUE或false。2.Multiple_days:如果用户点击网站超过一天，如果同一个用户点击超过2天，请调用TRUE else FALSE列。三。Busniess_days:用户是否在工作日点击，如果用户在工作时间内的工作日点击网站，则称之为True或FALSE

我有下面的脚本来执行上述任务，但是对于我庞大的数据帧来说它非常慢my data frame is 117Mb in size.任何更好的解决方案都是很好的

我的尝试：

df.Timestamp = pd.to_datetime(df.Timestamp)
df['date'] = [x.date() for x in df.Timestamp]
target_df = pd.DataFrame()
target_df['USER'] = df.USER.unique()
a = df.groupby(['USER', 'date']).size()
a = a[a>1]
UID=pd.DataFrame(a).reset_index().USER.values

target_df['Active'] = [True if x in UID else False for x in target_df.USER.values]
a = df.groupby('USER')['Timestamp'].nunique()
a = a[a>1]
UUID2=pd.DataFrame(a).reset_index().USER.values 
target_df['Multiple_days'] = [True if x in UUID2 else False for x in target_df.USER.values]

a = df[(df.Busi_days==True)&(df.Busi_hours==True)].USER.unique()

target_df['Busi_weekday'] = [True if x in a else False for x in target_df.USER.values]

target_df.head()


USER Active  Multiple_days   Busi_weekday
AAD TRUE    TRUE    TRUE
SAP FALSE   TRUE    FALSE
YAS FALSE   FALSE   FALSE

Answer 1

您可以使用：

df.Timestamp = pd.to_datetime(df.Timestamp)

df['date'] = df.Timestamp.dt.floor('d')

u = df.USER.unique()
a = df.groupby(['USER', 'date']).size().reset_index(level=1, drop=True)
a = a[a>1]
target_df = a[~a.index.duplicated()]
                .astype(bool).reindex(u, fill_value=False).to_frame(name='Active')

a = df.groupby('USER')['Timestamp'].nunique()
target_df['Multiple_days'] = a[a>1].astype(bool).reindex(u, fill_value=False)

a = df[(df.Busi_days==True)&(df.Busi_hours==True)].USER.unique()
target_df['Busi_weekday'] = target_df.index.isin(a)
print(target_df)

      Active  Multiple_days  Busi_weekday
USER                                     
AAD     True           True          True
SAP    False           True          True
YAS    False          False         False

编辑：

具有自定义功能的解决方案：

print (df1)
  USER   Timestamp day_of_week  Busi_days  Busi_hours
0  AAD  2017-07-11    09:31:44       True        True
1  AAD  2017-07-11    23:24:43       True       False
2  AAD  2017-07-12    13:24:43       True        True
3  SAP  2017-07-23    14:24:34      False       False
4  SAP  2017-07-24    16:58:49       True        True
5  YAS  2017-07-31    21:10:35       True       False

def func(df, time_col, user_col):
    df[time_col] = pd.to_datetime(df[time_col])

    df['date'] = df[time_col].dt.floor('d')

    u = df.USER.unique()
    a = df.groupby([user_col, 'date']).size().reset_index(level=1, drop=True)
    a = a[a>1]
    target_df = (a[~a.index.duplicated()]
                    .astype(bool).reindex(u, fill_value=False).to_frame(name='Active'))

    a = df.groupby(user_col)[time_col].nunique()
    target_df['Multiple_days'] = a[a>1].astype(bool).reindex(u, fill_value=False)

    a = df.loc[(df.Busi_days==True)&(df.Busi_hours==True), user_col].unique()
    target_df['Busi_weekday'] = target_df.index.isin(a)
    return target_df

#inputs are name of DataFrame, column for timestamp and column for user    
print (func(df1, 'Timestamp', 'USER'))
      Active  Multiple_days  Busi_weekday
USER                                     
AAD     True           True          True
SAP    False           True          True
YAS    False          False         False