pandas 日期时间索引偏移量
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DatetimeIndex offset with
提问by Mikhail Elizarev
I have a dataframe, generated with the following code:
我有一个数据框，使用以下代码生成：
time_index = pd.date_range(start=datetime(2013, 1, 1, 3),
end=datetime(2013, 1, 2, 2, 59),
freq='5T')
grid_columns = [u'in_brd', u'in_alt', u'out_brd', u'out_alt']
grid_proto = pd.DataFrame(index=time_index, columns=grid_columns)
I've also added some data to this dataframe.
我还在这个数据框中添加了一些数据。
When I'm trying to navigate through the index with intoffsets in basic dataframe I get everthing correct:
当我尝试int在基本数据帧中使用偏移量浏览索引时，我得到了正确的结果：
In[152]: grid_proto.index[0] + 1
Out[152]: Timestamp('2013-01-01 03:05:00', tz=None)
But if I'm trying to work with some kind of slice, I get an error:
但是，如果我尝试使用某种切片，则会收到错误消息：
In[153]: z = grid_proto[pd.notnull(x.in_brd)]
In[154]: z.index[0] + 1
Traceback (most recent call last):
File "<ipython-input-151-3ce8a4e5e2d6>", line 1, in <module>
z.index[0] + 1
File "tslib.pyx", line 664, in pandas.tslib._Timestamp.__add__ (pandas\tslib.c:12372)
ValueError: Cannot add integral value to Timestamp without offset.
I understand that this is because in first case I work with a link to DatetimeIndexelements instead of scalar. And in second case I get exactly scalar Timestampvalue of first index element. Am I right?
我知道这是因为在第一种情况下，我使用指向DatetimeIndex元素的链接而不是标量。在第二种情况下，我得到Timestamp第一个索引元素的准确标量值。我对吗？
How to deal with this offset correctly? (I need to navigate through such slice)
如何正确处理这个偏移？（我需要浏览这样的切片）
回答by joris
The reason is that in the first case you have a regular DatetimeIndex with a frequency of 5 minutes. So the integer 1 will be interpreted as one unit of the frequency (5 mins).
While in the second case, because of the slicing, you don't have a regular timeseries anymore, and the DatetimeIndex has no frequency anymore (z.index.freqwill give None, while grid_proto.index.freqwill give 5 mins).
原因是在第一种情况下，您有一个频率为 5 分钟的常规 DatetimeIndex。因此整数 1 将被解释为频率的一个单位（5 分钟）。
而在第二种情况下，由于切片，您不再有常规时间序列，并且 DatetimeIndex 不再有频率（z.index.freq将给出 None，而grid_proto.index.freq将给出 5 分钟）。
To solve this, you can just explicitely add 5 mins:
为了解决这个问题，你可以明确地添加 5 分钟：
In [22]: import datetime as dt
In [23]: z.index[0] + dt.timedelta(minutes=5)
Out[23]: Timestamp('2013-01-01 03:05:00', tz=None)
or alternatively you can add pd.DateOffset(minutes=5)(this will give the same result).
或者，您可以添加pd.DateOffset(minutes=5)（这将产生相同的结果）。
