Python Pandas 中对象组内的时间差

Question

我有一个如下所示的数据框：

from    to         datetime              other
-------------------------------------------------
11      1     2016-11-06 22:00:00          -
11      1     2016-11-06 20:00:00          -
11      1     2016-11-06 15:45:00          -
11      12    2016-11-06 15:00:00          -
11      1     2016-11-06 12:00:00          -
11      18    2016-11-05 10:00:00          -
11      12    2016-11-05 10:00:00          -
12      1     2016-10-05 10:00:59          -
12      3     2016-09-06 10:00:34          -

我想按 "from" 和 "to" 列分组，然后按降序对 "datetime" 进行排序，最后想计算这些分组对象之间的时间差当前时间和下一次。例如，在这种情况下，我想要一个如下所示的数据框：

from    to     timediff in minutes                                          others
11      1            120
11      1            255
11      1            225
11      1            0 (preferrably subtract this date from the epoch)
11      12           300
11      12           0
11      18           0
12      1            25
12      3            0

我实在想不通！！有出路吗？任何帮助将不胜感激！提前致谢！

Answer 1

我认为你需要：

groupby with apply sort_values with diff, convert Timedelta to minutes by seconds and floor division 60

fillna 和 sort_index，删除索引

中的级别 2

df = df.groupby(['from','to']).datetime
       .apply(lambda x: x.sort_values().diff().dt.seconds // 60)
       .fillna(0)
       .sort_index()
       .reset_index(level=2, drop=True)
       .reset_index(name='timediff in minutes')

print (df)

   from  to  timediff in minutes 
0    11   1                 120.0
1    11   1                 255.0
2    11   1                 225.0
3    11   1                   0.0
4    11  12                 300.0
5    11  12                   0.0
6    11  18                   0.0
7    12   3                   0.0
8    12   3                   0.0

df = df.join(df.groupby(['from','to'])
               .datetime
               .apply(lambda x: x.sort_values().diff().dt.seconds // 60)
               .fillna(0)
               .reset_index(level=[0,1], drop=True)
               .rename('timediff in minutes'))
print (df)
   from  to            datetime other  timediff in minutes
0    11   1 2016-11-06 22:00:00     -                120.0
1    11   1 2016-11-06 20:00:00     -                255.0
2    11   1 2016-11-06 15:45:00     -                225.0
3    11  12 2016-11-06 15:00:00     -                300.0
4    11   1 2016-11-06 12:00:00     -                  0.0
5    11  18 2016-11-05 10:00:00     -                  0.0
6    11  12 2016-11-05 10:00:00     -                  0.0
7    12   3 2016-10-05 10:00:59     -                  0.0
8    12   3 2016-09-06 10:00:34     -                  0.0

Answer 2

几乎同上，但没有 apply:

result = df.sort_values(['from','to','datetime'])\
           .groupby(['from','to'])['datetime']\
           .diff().dt.seconds.fillna(0)

Answer 3

df.assign(
    timediff=df.sort_values(
        'datetime', ascending=False
    ).groupby(['from', 'to']).datetime.diff(-1).dt.seconds.div(60).fillna(0))

Python Pandas 中对象组内的时间差

Time difference within group by objects in Python Pandas

python

pandas

difference

data-science

pandas-groupby