如何用多个级别按列过滤多索引数据帧？

Question

有了下面的dataframe，我喜欢将dataframe简化为只包含列（'count'，'LARGE'）和（'50%，'LARGE'），以及所有LabelX列。你知道吗

pd.DataFrame({('Label1', ''): {2363: 'D2',
2375: 'D2',
2387: 'D2',
2783: 'D2'},
('Label2', ''): {2363: 'D3',
2375: 'D3',
2387: 'D3',
2783: 'D3'},
('Label3', ''): {2363: 'D4',
2375: 'D4',
2387: 'D4',
2783: 'D4'},
('Label4', ''): {2363: 'na',
2375: 'na',
2387: 'na',
2783: 'na'},
('Label5', ''): {2363: 'False',
2375: 'False',
2387: 'False',
2783: 'False'},
('Label6', ''): {2363: 'D5',
2375: 'D5',
2387: 'D5',
2783: 'D5'},
('Label7', ''): {2363: 'A S',
2375: 'B S',
2387: 'C C',
2783: 'W I'},
('count', 'LARGE'): {2363: 777.0,
2375: 777.0,
2387: 777.0,
2783: 777.0},
          ('50%', 'LARGE'): {2363: pd.Timedelta('0 days 00:00:20'),
2375: pd.Timedelta('0 days 00:15:53'),
2387: pd.Timedelta('0 days 00:16:00'),
2783: pd.Timedelta('0 days 00:01:04')},
          ('50%', 'MEDIUM'): {2363: pd.Timedelta('0 days 00:00:20'),
2375: pd.Timedelta('0 days 00:12:49'),
2387: pd.Timedelta('0 days 00:13:54'),
2783: pd.Timedelta('0 days 00:01:01')},
         }
        )

已尝试使用以下方法删除列：

.drop(columns=[('count','LARGE'),('count','SMALL')])

我想知道的是，我是否可以指定要保留的内容，而不是删除不需要的列。我的用例有很多列，然后删除需要更多的代码。。。你知道吗

预期输出示例：

pd.DataFrame({('Label1', ''): {2363: 'D2',
2375: 'D2',
2387: 'D2',
2783: 'D2'},
('Label2', ''): {2363: 'D3',
2375: 'D3',
2387: 'D3',
2783: 'D3'},
('Label3', ''): {2363: 'D4',
2375: 'D4',
2387: 'D4',
2783: 'D4'},
('Label4', ''): {2363: 'na',
2375: 'na',
2387: 'na',
2783: 'na'},
('Label5', ''): {2363: 'False',
2375: 'False',
2387: 'False',
2783: 'False'},
('Label6', ''): {2363: 'D5',
2375: 'D5',
2387: 'D5',
2783: 'D5'},
('Label7', ''): {2363: 'A S',
2375: 'B S',
2387: 'C C',
2783: 'W I'},
('count', 'LARGE'): {2363: 777.0,
2375: 777.0,
2387: 777.0,
2783: 777.0},
          ('50%', 'LARGE'): {2363: pd.Timedelta('0 days 00:00:20'),
2375: pd.Timedelta('0 days 00:15:53'),
2387: pd.Timedelta('0 days 00:16:00'),
2783: pd.Timedelta('0 days 00:01:04')}
         }
        )

Answer 1

您可以只保留level1等于“”或“LARGE”的列，使用boolen索引：

df.loc[:, df.columns.get_level_values(1).isin(['', 'LARGE'])]

Answer 2

一种模式filter

df.filter(like='L')