如何删除数据透视表数据框的 VALUES 列

Question

需要删除从数据透视表创建的多索引数据框的子列

只需要在特定列（月）动态删除子列

我有一个从数据透视表创建的数据框，需要动态删除特定列的子列...
如果今天的日期小于 15，我需要删除除 Sep-19 以外所有月份的子栏 Bill1（当月）
如果今天的日期大于 15，则应删除除 Oct-19（下个月)

data_frame1 = pd.pivot_table(data_frame, index=['PC', 'Geo', 'Comp'], values=['Bill1', 'Bill2'], columns=['Month'], fill_value=0)
data_frame1 = data_frame1.swaplevel(0,1, axis=1).sort_index(axis=1)
tuples = [(a.strftime('%b-%y'), b) if a!= 'All' else (a,b) for a,b in data_frame1.columns]
data_frame1.columns = pd.MultiIndex.from_tuples(tuples)

输出：

              Sep-19             OCT-19        Nov-19
             Bill1 Bill2      Bill1 Bill2     Bill1 Bill2     
PC Geo Comp
A  Ind   OS   1     1.28        1    1.28      1    1.28

所需的输出：
如果今天的日期小于 15

               Sep-19          OCT-19        Nov-19
              Bill1  Bill2       Bill2         Bill2     
PC Geo Comp
A  Ind   OS    1      1.28        1.28          1.28

如果今天的日期大于 15

               Sep-19       OCT-19            Nov-19
                Bill2     Bill1  Bill2        Bill2     
PC Geo Comp
A  Ind   OS     1.28        1     1.28         1.28

Answer 1

用途：

#convert first level for datetimes and to month periods
level0 = pd.to_datetime(df.columns.get_level_values(0), format='%b-%y').to_period('m')
#get second level
level1 = df.columns.get_level_values(1)
print (level0)
PeriodIndex(['2019-09', '2019-09', '2019-10', '2019-10', '2019-11', '2019-11'],
             dtype='period[M]', freq='M')

print (level1)
Index(['Bill1', 'Bill2', 'Bill1', 'Bill2', 'Bill1', 'Bill2'], dtype='object')

#test for next 15 days
#dat = pd.to_datetime('2019-09-20')
#get today timestamp
dat = pd.to_datetime('now')
print (dat)

#convert timestamp to period
today_per = dat.to_period('m')

#compare day and filter
if dat.day < 15:
    df = df.loc[:, (level0 == today_per) | (level1 != 'Bill1')]
else:
    #test with add 1 month to today period
    df = df.loc[:, (level0 == today_per + 1) | (level1 != 'Bill1')]
print (df)
         Sep-19       Oct-19 Nov-19
          Bill1 Bill2  Bill2  Bill2
A Ind OS      1  1.28   1.28   1.28

下个月测试：

#convert first level for datetimes and to month periods
level0 = pd.to_datetime(df.columns.get_level_values(0), format='%b-%y').to_period('m')
#get second level
level1 = df.columns.get_level_values(1)
print (level0)
PeriodIndex(['2019-09', '2019-09', '2019-10', '2019-10', '2019-11', '2019-11'],
             dtype='period[M]', freq='M')

print (level1)
Index(['Bill1', 'Bill2', 'Bill1', 'Bill2', 'Bill1', 'Bill2'], dtype='object')

#test for next 15 days
dat = pd.to_datetime('2019-09-20')
#get today timestamp
#dat = pd.to_datetime('now')
print (dat)

#convert timestamp to period
today_per = dat.to_period('m')

#compare day and filter
if dat.day < 15:
    df = df.loc[:, (level0 == today_per) | (level1 != 'Bill1')]
else:
    #test with add 1 month to today period
    df = df.loc[:, (level0 == today_per + 1) | (level1 != 'Bill1')]
print (df)
         Sep-19 Oct-19       Nov-19
          Bill2  Bill1 Bill2  Bill2
A Ind OS   1.28      1  1.28   1.28