迭代 Pandas dataframe 上的唯一日期

Question

I have a pandas dataframe like this

id        date      time    dif
01  2020-04-02  09:44:00
02  2020-04-02  09:50:23
03  2020-04-02  09:54:56
04  2020-04-03  10:24:42
05  2020-04-03  10:32:12
06  2020-04-03  11:12:21
...

What I'm tryng to do is calculate time difference, in minutes, between each row and its previous one per day. So the result should be like this

id        date      time    dif
01  2020-04-02  09:44:00      6
02  2020-04-02  09:50:23      4
03  2020-04-02  09:54:56
04  2020-04-03  10:24:42      7
05  2020-04-03  10:32:12     40
06  2020-04-03  11:12:21
...

My first thought was to create a list with the unique values of the column date and tried this:

import pandas a dp
import numpy as np

...

dates = df.date.unique()

for d in dates:
  df['dif'] = round(df['time'].diff(-1).dt.total_seconds().div(60),0) * -1

But I think it isn't so easy...

Answer 1

Use DataFrameGroupBy.diff with Series.dt.total_seconds and Series.round :

df['time'] = pd.to_timedelta(df['time'])

df['dif'] = df.groupby('date')['time'].diff(-1).dt.total_seconds().div(60).round().mul(-1)

Or use DataFrameGroupBy.shift with subtracting:

df['dif'] = (df.groupby('date')['time'].shift(-1)
               .sub(df['time'])
               .dt.total_seconds()
               .div(60)
               .round())
print (df)
   id        date     time   dif
0   1  2020-04-02 09:44:00   6.0
1   2  2020-04-02 09:50:23   5.0
2   3  2020-04-02 09:54:56   NaN
3   4  2020-04-03 10:24:42   8.0
4   5  2020-04-03 10:32:12  40.0
5   6  2020-04-03 11:12:21   NaN