遵循this answer的精神，我尝试了以下操作将datetimes的DataFrame列转换为自纪元以来的秒数。
df['date'] = (df['date']+datetime.timedelta(hours=2)-datetime.datetime(1970,1,1))
df['date'].map(lambda td:td.total_seconds())
第二个命令导致以下错误，我不理解。对这里可能发生的事情有任何想法吗？我用apply替换了map，这没有帮助。
---------------------------------------------------------------------------
AttributeError Traceback (most recent call last)
<ipython-input-99-7123e823f995> in <module>()
----> 1 df['date'].map(lambda td:td.total_seconds())
/Users/cpd/.virtualenvs/py27-ipython+pandas/lib/python2.7/site-packages/pandas-0.12.0_937_gb55c790-py2.7-macosx-10.8-x86_64.egg/pandas/core/series.pyc in map(self, arg, na_action)
1932 return self._constructor(new_values, index=self.index).__finalize__(self)
1933 else:
-> 1934 mapped = map_f(values, arg)
1935 return self._constructor(mapped, index=self.index).__finalize__(self)
1936
/Users/cpd/.virtualenvs/py27-ipython+pandas/lib/python2.7/site-packages/pandas-0.12.0_937_gb55c790-py2.7-macosx-10.8-x86_64.egg/pandas/lib.so in pandas.lib.map_infer (pandas/lib.c:43628)()
<ipython-input-99-7123e823f995> in <lambda>(td)
----> 1 df['date'].map(lambda td:td.total_seconds())
AttributeError: 'float' object has no attribute 'total_seconds'
最佳答案
更新:
在0.15.0中，Timedeltas成为了正式的dtype。
因此这成为可能(以及下面的方法)
In [45]: s = Series(pd.timedelta_range('1 day',freq='1S',periods=5))
In [46]: s.dt.components
Out[46]:
days hours minutes seconds milliseconds microseconds nanoseconds
0 1 0 0 0 0 0 0
1 1 0 0 1 0 0 0
2 1 0 0 2 0 0 0
3 1 0 0 3 0 0 0
4 1 0 0 4 0 0 0
In [47]: s.astype('timedelta64[s]')
Out[47]:
0 86400
1 86401
2 86402
3 86403
4 86404
dtype: float64
原始答案:
我看到您掌握了(很快就会发布0.13)，
因此，假设您的numpy> = 1.7。做这个。有关文档，请参见here(这是频率转换)
In [5]: df = DataFrame(dict(date = date_range('20130101',periods=10)))
In [6]: df
Out[6]:
date
0 2013-01-01 00:00:00
1 2013-01-02 00:00:00
2 2013-01-03 00:00:00
3 2013-01-04 00:00:00
4 2013-01-05 00:00:00
5 2013-01-06 00:00:00
6 2013-01-07 00:00:00
7 2013-01-08 00:00:00
8 2013-01-09 00:00:00
9 2013-01-10 00:00:00
In [7]: df['date']+timedelta(hours=2)-datetime.datetime(1970,1,1)
Out[7]:
0 15706 days, 02:00:00
1 15707 days, 02:00:00
2 15708 days, 02:00:00
3 15709 days, 02:00:00
4 15710 days, 02:00:00
5 15711 days, 02:00:00
6 15712 days, 02:00:00
7 15713 days, 02:00:00
8 15714 days, 02:00:00
9 15715 days, 02:00:00
Name: date, dtype: timedelta64[ns]
In [9]: (df['date']+timedelta(hours=2)-datetime.datetime(1970,1,1)) / np.timedelta64(1,'s')
Out[9]:
0 1357005600
1 1357092000
2 1357178400
3 1357264800
4 1357351200
5 1357437600
6 1357524000
7 1357610400
8 1357696800
9 1357783200
Name: date, dtype: float64
包含的值是
np.timedelta64[ns]对象，它们与timedelta对象的方法不同，因此没有total_seconds()。In [10]: s = (df['date']+timedelta(hours=2)-datetime.datetime(1970,1,1))
In [11]: s[0]
Out[11]: numpy.timedelta64(1357005600000000000,'ns')
您可以将它们指定为int类型，然后返回
ns单位。In [12]: s[0].astype(int)
Out[12]: 1357005600000000000
您也可以执行此操作(但只能在单个unit元素上)。
In [18]: s[0].astype('timedelta64[s]')
Out[18]: numpy.timedelta64(1357005600,'s')
关于自大纪元以来的Python Pandas日期时间系列，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/19553464/