python pandas中两个datetime.time列之间的微秒差异？

Question

我有一个python熊猫数据帧，它包含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 

我怎样才能生成包含time1和time2之间的微秒差的第三列中，在整数如果可能？

Answer 1

如果你在前面加上HESE与实际日期，你可以将它们转换为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 

现在你可以减去这些列：

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] 

要得到微秒的数量，只是鸿沟由1000的基础纳秒：

In [15]: t.astype(np.int64)/10**3 
Out[15]: 
0   0 
1 100000 
2 200000 
3 400000 
dtype: int64 

正如杰夫指出，在最近的版本numpy的的NS您可以通过1微秒划分：

In [16]: t/np.timedelta64(1,'us') 
Out[16]: 
0   0 
1 100000 
2 200000 
3 400000 
dtype: float64 

Answer 2

最简单的方法就是要做到这一点：

(pd.to_datetime(df['time2']) - pd.to_datetime(df['time1']))/np.timedelta64(1, 'us')'

Answer 3

使用dateutil你可以改变你的时间戳列‘真正’的时间戳：

df.time1 = df.time1.apply(dateutil.parser.parse) df.time2 = df.time2.apply(dateutil.parser.parse)

之后，你要定义这样一个新列：

df['delta'] = df.time2 - df.time1

Answer 4

起初我还以为那里是由于没有绿色蜱在这里没有正确答案。但正如杰夫在评论中指出的那样，我错了。

无论哪种方式是我的贡献。

首先，很明显，使得datetime.time成timedelta

df['delta'] = (pd.to_timedelta(df.time2.astype(str)) - pd.to_timedelta(df.time1.astype(str))) 

      time1   time2   delta 
0 13:00:07.294234 13:00:07.294234  00:00:00 
1 14:00:07.294234 14:00:07.394234 00:00:00.100000 
2 15:00:07.294234 15:00:07.494234 00:00:00.200000 
3 16:00:07.294234 16:00:07.694234 00:00:00.400000 

现在我们有timedelta我们可以简单地通过一微秒除以它获得的微秒数。

df['microsecond_delta'] = df.delta/pd.np.timedelta64(1, 'us') 

      time1   time2   delta microsecond_delta 
0 13:00:07.294234 13:00:07.294234  00:00:00     0 
1 14:00:07.294234 14:00:07.394234 00:00:00.100000    100000 
2 15:00:07.294234 15:00:07.494234 00:00:00.200000    200000 
3 16:00:07.294234 16:00:07.694234 00:00:00.400000    400000 

我必须补充一点，这是非常直观的，但它似乎是唯一的方法。似乎没有办法直接访问毫秒。我通过应用lambda函数类似的尝试：

df.delta.apply(lambda x: x.microseconds) 
AttributeError: 'numpy.timedelta64' object has no attribute 'microseconds' 

同样是seconds，nanoseconds，milliseconds等真的...