我有一个python pandas数据框，其中包含2列:time1和time2:
time1 time2
13:00:07.294234 13:00:07.294234
14:00:07.294234 14:00:07.394234
15:00:07.294234 15:00:07.494234
16:00:07.294234 16:00:07.694234
如何生成第三列，其中包含
time1和time2之间的微秒差异(如果可能)，以整数表示？ 最佳答案
如果您将实际日期作为前缀，则可以将它们转换为datetime64列:
In [11]: '2014-03-19 ' + df
Out[11]:
time1 time2
0 2014-03-19 13:00:07.294234 2014-03-19 13:00:07.294234
1 2014-03-19 14:00:07.294234 2014-03-19 14:00:07.394234
2 2014-03-19 15:00:07.294234 2014-03-19 15:00:07.494234
3 2014-03-19 16:00:07.294234 2014-03-19 16:00:07.694234
[4 rows x 2 columns]
In [12]: df = ('2014-03-19 ' + df).astype('datetime64[ns]')
Out[12]:
time1 time2
0 2014-03-19 20:00:07.294234 2014-03-19 20:00:07.294234
1 2014-03-19 21:00:07.294234 2014-03-19 21:00:07.394234
2 2014-03-19 22:00:07.294234 2014-03-19 22:00:07.494234
3 2014-03-19 23:00:07.294234 2014-03-19 23:00:07.694234
现在您可以减去这些列:
In [13]: delta = df['time2'] - df['time1']
In [14]: delta
Out[14]:
0 00:00:00
1 00:00:00.100000
2 00:00:00.200000
3 00:00:00.400000
dtype: timedelta64[ns]
要获得微秒数，只需将基础纳秒除以1000:
In [15]: t.astype(np.int64) / 10**3
Out[15]:
0 0
1 100000
2 200000
3 400000
dtype: int64
正如Jeff指出的那样，在最新版本的numpy中，您可以除以1微秒:
In [16]: t / np.timedelta64(1,'us')
Out[16]:
0 0
1 100000
2 200000
3 400000
dtype: float64
关于python - python pandas中两个datetime.time列之间的微秒差异？，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/22513306/