pandas 如何创建熊猫时间戳对象？
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How to create a pandas Timestamp object?
提问by Mannaggia
It might sound like a trivial question, but I was not able to find proper documentation on this including help(pd.Timestamp) which does not provide a clear constructor.
这听起来像是一个微不足道的问题，但我无法找到有关此问题的适当文档，包括 help(pd.Timestamp) ，它没有提供明确的构造函数。
I want to create the last time stamp of eg day 2012-12-13, so hour is 23, minute is 59, seconds is 59, nanosecond is 999999999
我想创建例如 2012-12-13 的最后一个时间戳，所以小时是 23，分钟是 59，秒是 59，纳秒是 999999999
alternatively, if I have microsecond precision the last timestamp of a day would be as before except microsecond is 999999
或者，如果我有微秒精度，一天的最后一个时间戳将与以前一样，除了微秒是 999999
I need this to filter out all timestamps until a given (end of) day by indexing the original series by something like df.ix[:last_timestamp]
我需要通过使用 df.ix[:last_timestamp] 之类的东西索引原始系列来过滤掉所有时间戳，直到给定（结束）一天
Thx
谢谢
采纳答案by Jeff
In [1]: Timestamp('20131213 11:59:59.999999999')
Out[1]: Timestamp('2013-12-13 11:59:59.999999', tz=None)
You can also do
你也可以这样做
In [3]: pd.Timestamp('20141213')-pd.Timedelta('1ns')
Out[3]: Timestamp('2014-12-12 23:59:59.999999999')
Sounds like you actually want to use partial string slicing, see here
听起来您实际上想使用部分字符串切片，请参见此处
In [19]: s = Series(1,pd.date_range('20131212',freq='H',periods=25))
In [20]: s
Out[20]:
2013-12-12 00:00:00 1
2013-12-12 01:00:00 1
2013-12-12 02:00:00 1
..
2013-12-12 22:00:00 1
2013-12-12 23:00:00 1
2013-12-13 00:00:00 1
Freq: H, dtype: int64
In [21]: s['2013':'20131212']
Out[21]:
2013-12-12 00:00:00 1
2013-12-12 01:00:00 1
2013-12-12 02:00:00 1
..
2013-12-12 21:00:00 1
2013-12-12 22:00:00 1
2013-12-12 23:00:00 1
Freq: H, dtype: int64
