pandas 在时间范围之外放行 [英] Pandas Drop Rows Outside of Time Range

查看:99
本文介绍了 pandas 在时间范围之外放行的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我正在尝试遍历DataFrame索引中的每一行,并删除所有不在特定时间之间的行.

I am trying to go through every row in a DataFrame index and remove all rows that are not between a certain time.

我一直在寻找解决方案,但是没有一个将日期和时间分开,我要做的就是删除时间范围之外的行.

I have been looking for solutions but none of them separate the Date from the Time, and all I want to do is drop the rows that are outside of a Time range.

推荐答案

您可以使用

You can use the between_time function directly:

ts.between_time(datetime.time(18), datetime.time(9), include_start=False, include_end=False)


原始答案:

您可以使用indexer_between_time Index方法.

例如,要在上午9点至下午6点之间(包括以下时间)包括:

For example, to include those times between 9am and 6pm (inclusive):

ts.ix[ts.index.indexer_between_time(datetime.time(9), datetime.time(18))]

做相反的事情,排除在下午6点至上午9点之间(排他性):

to do the opposite and exclude those times between 6pm and 9am (exclusive):

ts.ix[ts.index.indexer_between_time(datetime.time(18), datetime.time(9),
                                    include_start=False, include_end=False)]

注意:indexer_between_time的参数include_startinclude_end在默认情况下为True,将include_start设置为False表示日期时间的时间段正好是start_time(第一个参数),在这种情况下为6pm.

Note: indexer_between_time's arguments include_start and include_end are by default True, setting include_start to False means that datetimes whose time-part is precisely start_time (the first argument), in this case 6pm, will not be included.

示例:

In [1]: rng = pd.date_range('1/1/2000', periods=24, freq='H')

In [2]: ts = pd.Series(pd.np.random.randn(len(rng)), index=rng)

In [3]: ts.ix[ts.index.indexer_between_time(datetime.time(10), datetime.time(14))] 
Out[3]: 
2000-01-01 10:00:00    1.312561
2000-01-01 11:00:00   -1.308502
2000-01-01 12:00:00   -0.515339
2000-01-01 13:00:00    1.536540
2000-01-01 14:00:00    0.108617

注意:相同的语法(使用 ix )适用于DataFrame:

Note: the same syntax (using ix) works for a DataFrame:

In [4]: df = pd.DataFrame(ts)

In [5]: df.ix[df.index.indexer_between_time(datetime.time(10), datetime.time(14))]
Out[5]: 
                            0
2000-01-03 10:00:00  1.312561
2000-01-03 11:00:00 -1.308502
2000-01-03 12:00:00 -0.515339
2000-01-03 13:00:00  1.536540
2000-01-03 14:00:00  0.108617

这篇关于 pandas 在时间范围之外放行的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

查看全文
登录 关闭
扫码关注1秒登录
发送“验证码”获取 | 15天全站免登陆