计算pandas中连续两行之间的时间差

Question

I have a pandas dataframe as follows

Dev_id     Time
88345      13:40:31
87556      13:20:33
88955      13:05:00
.....      ........
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

Dev_id    Time          Time_diff(in min)
88345      13:40:31      20
87556      13:20:33      15
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

Answer 1

Problem is pandas need datetime s or timedelta s for diff function, so first converting by to_timedelta , then get total_seconds and divide by 60 :

df['Time_diff'] = pd.to_timedelta(df['Time'].astype(str)).diff(-1).dt.total_seconds().div(60)
#alternative
#df['Time_diff'] = pd.to_datetime(df['Time'].astype(str)).diff(-1).dt.total_seconds().div(60)
print (df)
   Dev_id      Time  Time_diff
0   88345  13:40:31  19.966667
1   87556  13:20:33  15.550000
2   88955  13:05:00  49.533333
3   85678  12:15:28        NaN

If want floor or round per minutes:

df['Time_diff'] = (pd.to_timedelta(df['Time'].astype(str))
                     .diff(-1)
                     .dt.floor('T')
                     .dt.total_seconds()
                     .div(60))
print (df)
   Dev_id      Time  Time_diff
0   88345  13:40:31       19.0
1   87556  13:20:33       15.0
2   88955  13:05:00       49.0
3   85678  12:15:28        NaN

Answer 2

您应该首先将 df['Time'] 列转换/转换为pd.Timedelta然后进行减法