Python dataframe：对具有相同第一个字符的列进行分组

我有一个如下所示的数据帧

                            LIT__0001   LIT__002    AAA__0001   AAA__0002   XYZ
2019-10-31 13:40:00-04:00   NaN         0.014786    10          55          1
2019-10-31 13:45:00-04:00   NaN         0.012143    33          11          2
2019-10-31 13:50:00-04:00   NaN         NaN         NaN         NaN         3
2019-10-31 13:55:00-04:00   NaN         0.020000    14          13          4
2019-10-31 14:00:00-04:00   0.010000    NaN         14          NaN         5

                            LIT         AAA         XYZ
2019-10-31 13:40:00-04:00   0.014786    10          1
2019-10-31 13:45:00-04:00   0.012143    11          2
2019-10-31 13:50:00-04:00   NaN         NaN         3
2019-10-31 13:55:00-04:00   0.020000    13          4
2019-10-31 14:00:00-04:00   0.010000    14          5

我需要将其转换为如下所示的数据帧

                            LIT__0001   LIT__002    AAA__0001   AAA__0002   XYZ
2019-10-31 13:40:00-04:00   NaN         0.014786    10          55          1
2019-10-31 13:45:00-04:00   NaN         0.012143    33          11          2
2019-10-31 13:50:00-04:00   NaN         NaN         NaN         NaN         3
2019-10-31 13:55:00-04:00   NaN         0.020000    14          13          4
2019-10-31 14:00:00-04:00   0.010000    NaN         14          NaN         5

                            LIT         AAA         XYZ
2019-10-31 13:40:00-04:00   0.014786    10          1
2019-10-31 13:45:00-04:00   0.012143    11          2
2019-10-31 13:50:00-04:00   NaN         NaN         3
2019-10-31 13:55:00-04:00   0.020000    13          4
2019-10-31 14:00:00-04:00   0.010000    14          5

也就是说，对于在“_u”之前有公共第一个字符的每一列，为每一行取最小值

我的数据帧非常大，因此我非常欣赏更快的解决方案。

用于轴=1的列和用于拆分的lambda函数：

df = df.groupby(lambda x: x.split('__')[0], axis=1, sort=False).min()

或使用：