[ Python 常見問題 ] Calculate the time difference between two consecutive rows in pandas

Source From Here 
Question 
I have a pandas dataframe as follows 

view plaincopy to clipboardprint?

1. Dev_id     Time  
2. 88345      13:40:31  
3. 87556      13:20:33  
4. 88955      13:05:00  
5. .....      ........  
6. 85678      12:15:28  

The above dataframe has 83000 rows. I want to take time difference between two consecutive rows and keep it in a separate column. The desired result would be 

view plaincopy to clipboardprint?

1. Dev_id    Time          Time_diff(in min)  
2. 88345      13:40:31      20  
3. 87556      13:20:33      15  
4. 88955      13:05:00      15  

I have tried df['Time_diff'] = df['Time'].diff(-1) but getting error as shown below 

TypeError: unsupported operand type(s) for -: 'datetime.time' and 'datetime.time'

How to solve this? 

How-To 
Problem is pandas need datetimes or timedeltas for diff function, so first converting by to_timedelta, then get total_seconds and divide by 60: 

view plaincopy to clipboardprint?

1. >>> import pandas as pd  
2. >>> from datetime import datetime, timedelta  
3. >>> df = pd.DataFrame([[1, datetime.now()], [2, datetime.now()- timedelta(hours = 1)]], columns = ['id', 'time'])  
4. >>> df  
5.    id                       time  
6. 0   1 2019-01-24 09:10:19.732798  
7. 1   2 2019-01-24 08:10:19.732864  
8. >>> df['time_diff'] = df['time'].diff(-1).dt.total_seconds().div(60)  
9. >>> df  
10.    id                       time  time_diff  
11. 0   1 2019-01-24 09:10:19.732798  59.999999  
12. 1   2 2019-01-24 08:10:19.732864        NaN  

If want floor or round per minutes: 

view plaincopy to clipboardprint?

1. >>> df['time'].diff(-1).dt  
2.  0x7ff17c4bad90>  

>>> df['time'].diff(-1).dt.floor('T')  

0   00:59:00  

1        NaT  

>>> df['time_diff'] = df['time'].diff(-1).dt.floor('T').dt.total_seconds().div(60)  

>>> df  

   id                       time  time_diff  

0   1 2019-01-24 09:10:19.732798       59.0  

1   2 2019-01-24 08:10:19.732864        NaN