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

Question

I have a python pandas data frame, which contains 2 columns: time1 and 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 

How can I generate a third column which contains the microsecond difference between time1 and time2 , in integer if possible?

Answer 1

If you prepend hese with an actual date you can convert them to datetime64 columns:

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

Now you can subtract these columns:

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]

To get the number of microseconds, just divide the underlying nanoseconds by 1000:

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

As Jeff points out, on recent versions of numpy you can divide by 1 micro second:

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

Answer 2

the easiest way is just to do this:

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

Answer 3

At first I thought there was no correct answers here due to no green ticks. But as pointed out by Jeff in the comments, I was wrong.

Either way here is my contribution.

First, the obvious, making the datetime.time into a 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

Now that we have the timedelta we can simply divide it by one microsecond to get the number of microseconds.

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

I have to add that this is very counter intuitive, but it seems it is the only way. There seem to be no way of accessing the milliseconds directly. I tried via applying lambda functions like:

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

Same is true for seconds , nanoseconds , milliseconds and so on...

Answer 4

Using dateutil you could transform your timestamp columns to 'real' timestamps:

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

After that you want to define a new column like this:

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