Dataframe組中的Next/Prev opreation-

Question

我想在給定的數據框按特定列分組之後獲得下一個（第二個）條目。如果其中任何一個不存在，那麼它應該根據時間返回nan/nat。考慮下面的例子：

>>> df1 = pd.DataFrame({'School': {0: 'DEF', 1: 'ABC', 2: 'PQR', 3: 'DEF', 4: 'PQR', 5: 'PQR'}, 'OpenTime': {0: '08:00:00.000', 1: '09:00:00.000', 2: '10:00:23.563', 3: '09:30:05.908', 4: '07:15:50.100', 5: '08:15:00.000'}, 'CloseTime': {0: '13:00:00.000', 1: '14:00:00.000', 2: '13:30:00.100', 3: '15:00:00.768', 4: '13:00:00.500', 5: '15:50:32.534'}, 'IsTopper':{0:'1',1:'1',2:'1',3:'1',4:'1',5:'-1'}}) 
>>> df1 
     CloseTime IsTopper  OpenTime School 
0 13:00:00.000  1 08:00:00.000 DEF 
1 14:00:00.000  1 09:00:00.000 ABC 
2 13:30:00.100  1 10:00:23.563 PQR 
3 15:00:00.768  1 09:30:05.908 DEF 
4 13:00:00.500  1 07:15:50.100 PQR 
5 15:50:32.534  -1 08:15:00.000 PQR 

獲取第一值是簡單的，並且可以通過任一來實現以下

>>> df1.groupby(['School', 'IsTopper'])['OpenTime'].first() 

OR

>>> (df1.groupby(['School', 'IsTopper'])).apply(lambda x:x.iloc[0])['OpenTime'] 

獲取使用...iloc[1]將拋出下一個（第二個）值在上述情況下出錯。

最後，我試圖讓下面的輸出在上面的例子中的情況：

 School IsTopper  OpenTime Next_OpenTime 
0  DEF  1  08:00:00.000 09:30:05.908 
1  ABC  1  09:00:00.000  
2  PQR  1  10:00:23.563 07:15:50.100 
3  DEF  1  09:30:05.908  
4  PQR  1  07:15:50.100  
5  PQR  -1  08:15:00.000  

Answer 1

>>> df1['Next_OpenTime'] = (df1.groupby(['School', 'IsTopper']))['OpenTime'].shift(-1) 
>>> df1 
     IsTopper  OpenTime School Next_OpenTime 
0   1 08:00:00.000 DEF 09:30:05.908 
1   1 09:00:00.000 ABC   NaN 
2   1 10:00:23.563 PQR 07:15:50.100 
3   1 09:30:05.908 DEF   NaN 
4   1 07:15:50.100 PQR   NaN 
5   -1 08:15:00.000 PQR   NaN