<p>如果在hese前面加上实际日期,则可以将它们转换为datetime64列:</p>
<pre><code>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
</code></pre>
<p>现在可以减去这些列:</p>
<pre><code>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]
</code></pre>
<p>要获得微秒数,只需将基础纳秒除以1000:</p>
<pre><code>In [15]: t.astype(np.int64) / 10**3
Out[15]:
0 0
1 100000
2 200000
3 400000
dtype: int64
</code></pre>
<p>正如Jeff指出的,在最新版本的numpy上,你可以除以1微秒:</p>
<pre><code>In [16]: t / np.timedelta64(1,'us')
Out[16]:
0 0
1 100000
2 200000
3 400000
dtype: float64
</code></pre>