Pandas 日期时间索引上限到一天中的特定时间
[英]Pandas datetime index ceil to specific hour in day
问题描述
I have a datetime index which I would like to round up(ceil) to a specific hour in the day.
我有一个日期时间索引，我想将其四舍五入（ceil）到一天中的特定时间。 I am already aware of pandas' offset aliases and how they work, but specifically I would like to tell it to round the datetime to a specific hour in the day(or a specific day in the month).我已经知道 pandas 的偏移别名及其工作方式，但具体来说，我想告诉它将日期时间四舍五入到一天中的特定时间（或一个月中的特定日期）。 For example I would like to have this kind of transformation:例如，我想进行这种转换：
print(results.index)
DatetimeIndex(['2018-12-14 05:00:00+01:00', '2018-12-14 06:00:00+01:00',
'2018-12-14 07:00:00+01:00', '2018-12-14 08:00:00+01:00',
'2018-12-14 09:00:00+01:00', '2018-12-14 10:00:00+01:00',
'2018-12-14 11:00:00+01:00', '2018-12-14 12:00:00+01:00',
'2018-12-14 13:00:00+01:00', '2018-12-14 14:00:00+01:00',
Turns into变成
DatetimeIndex(['2018-12-14 08:00:00+01:00', '2018-12-14 08:00:00+01:00',
'2018-12-14 08:00:00+01:00', '2018-12-14 08:00:00+01:00',
'2018-12-15 08:00:00+01:00', '2018-12-15 08:00:00+01:00',
'2018-12-15 08:00:00+01:00', '2018-12-15 08:00:00+01:00',
'2018-12-15 08:00:00+01:00', '2018-12-15 08:00:00+01:00',
As far as I'm aware, there does not exist such a parameter that we can pass to ceil() that would allow this, since we can only round to nearest hour, day, month(freq='H', 'D', 'M')... Is there an elegant solution to this or would I have to code my own for loop?据我所知，不存在我们可以传递给 ceil() 的参数，因为我们只能四舍五入到最近的小时、天、月（freq='H', 'D' , 'M')... 对此是否有优雅的解决方案，或者我是否必须编写自己的 for 循环代码？
1 个解决方案
解决方案1
2 2020-10-07 12:07:24
One idea is use numpy.where and offsets.DateOffset - here hour without s means set values to 8 , day with s means add one day to original days:一个想法是使用numpy.where和offsets.DateOffset - 这里没有s hour意味着将值设置为8 ，带s day意味着在原始日期基础上增加一天：
d = pd.DatetimeIndex(['2018-12-14 05:00:00+01:00', '2018-12-14 06:00:00+01:00',
'2018-12-14 07:00:00+01:00', '2018-12-14 08:00:00+01:00',
'2018-12-14 09:00:00+01:00', '2018-12-14 10:00:00+01:00',
'2018-12-14 11:00:00+01:00', '2018-12-14 12:00:00+01:00',
'2018-12-14 13:00:00+01:00', '2018-12-14 14:00:00+01:00'])
results = pd.DataFrame(index=d)
out = np.where(results.index.hour <= 8,
results.index + pd.offsets.DateOffset(hour=8),
results.index + pd.offsets.DateOffset(days=1, hour=8))
print (pd.DatetimeIndex(out))
DatetimeIndex(['2018-12-14 08:00:00+01:00', '2018-12-14 08:00:00+01:00',
'2018-12-14 08:00:00+01:00', '2018-12-14 08:00:00+01:00',
'2018-12-15 08:00:00+01:00', '2018-12-15 08:00:00+01:00',
'2018-12-15 08:00:00+01:00', '2018-12-15 08:00:00+01:00',
'2018-12-15 08:00:00+01:00', '2018-12-15 08:00:00+01:00'],
dtype='datetime64[ns, pytz.FixedOffset(60)]', freq=None)
Another idea is use Timedelta s and add day only if condition is True :另一个想法是使用Timedelta s 并仅在条件为True时添加日期：
m = results.index.hour > 8
out = results.index + pd.offsets.DateOffset(hour=8) + pd.Timedelta(days=1) * m
print (out)
DatetimeIndex(['2018-12-14 08:00:00+01:00', '2018-12-14 08:00:00+01:00',
'2018-12-14 08:00:00+01:00', '2018-12-14 08:00:00+01:00',
'2018-12-15 08:00:00+01:00', '2018-12-15 08:00:00+01:00',
'2018-12-15 08:00:00+01:00', '2018-12-15 08:00:00+01:00',
'2018-12-15 08:00:00+01:00', '2018-12-15 08:00:00+01:00'],
dtype='datetime64[ns, pytz.FixedOffset(60)]', freq=None)
m = results.index.hour > 8
out = results.index.floor('d') + pd.Timedelta(hours=8) + pd.Timedelta(days=1) * m
print (out)
DatetimeIndex(['2018-12-14 08:00:00+01:00', '2018-12-14 08:00:00+01:00',
'2018-12-14 08:00:00+01:00', '2018-12-14 08:00:00+01:00',
'2018-12-15 08:00:00+01:00', '2018-12-15 08:00:00+01:00',
'2018-12-15 08:00:00+01:00', '2018-12-15 08:00:00+01:00',
'2018-12-15 08:00:00+01:00', '2018-12-15 08:00:00+01:00'],
dtype='datetime64[ns, pytz.FixedOffset(60)]', freq=None)
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