转换' pandas '频率字符串为' DateOffset '

Question

我有一个时区感知的pandasDateTimeIndex，我想将其推进一个时间步长，其时间步长由其.freq属性指定。但是，这样做并不尊重时区信息:

import pandas as pd
i = pd.date_range('2020-03-28', freq='D', periods=3, tz='Europe/Amsterdam')
# DatetimeIndex(['2020-03-28 00:00:00+01:00', '2020-03-29 00:00:00+01:00',
#                '2020-03-30 00:00:00+02:00'],
#               dtype='datetime64[ns, Europe/Amsterdam]', freq='D')
i + i.freq
# Not what I want; second timestamp is advanced by 24h instead of 23h and is no longer at midnight:
# DatetimeIndex(['2020-03-29 00:00:00+01:00', '2020-03-30 01:00:00+02:00',
#                '2020-03-31 00:00:00+02:00'],
#               dtype='datetime64[ns, Europe/Amsterdam]', freq='D')

做工作是使用pd.DateOffset:

i + pd.DateOffset(days=1)
# What I want; all timestamps at midnight (I just need to re-set the .freq attribute):
# DatetimeIndex(['2020-03-29 00:00:00+01:00', '2020-03-30 00:00:00+02:00',
#                '2020-03-31 00:00:00+02:00'],
#               dtype='datetime64[ns, Europe/Amsterdam]', freq=None)

然而，由于我事先不知道索引的频率，我想使用i.freq的值来获得正确的DateOffset。有办法做到这一点吗?(除了使用长if... elif... elif...块)

当然也欢迎其他的解决方案。

这是我发现的唯一一个与此相关的问题，但我不能在这里使用它:

i + pd.tseries.frequencies.to_offset(i.freq)
# Not what I want:
# DatetimeIndex(['2020-03-29 00:00:00+01:00', '2020-03-30 01:00:00+02:00',
#                '2020-03-31 00:00:00+02:00'],
#               dtype='datetime64[ns, Europe/Amsterdam]', freq='D')

(实际上，后一项正好返回i.freq)

许多谢谢。

编辑(1)

正如评论中所建议的，在某些情况下使用.shift(1)是有效的，包括我上面所说的情况…

i.shift(1)
# What I want; all timestamps at midnight:
# DatetimeIndex(['2020-03-29 00:00:00+01:00', '2020-03-30 00:00:00+02:00',
#                '2020-03-31 00:00:00+02:00'],
#               dtype='datetime64[ns, Europe/Amsterdam]', freq='D')

…但并非全部如此。实际上，将原始索引中的开始日期提前一天会导致时间戳丢失，其余的时间戳都是错误的:

i2 = pd.date_range('2020-03-29', freq='D', periods=3, tz='Europe/Amsterdam')
# DatetimeIndex(['2020-03-29 00:00:00+01:00', '2020-03-30 00:00:00+02:00',
#               '2020-03-31 00:00:00+02:00'],
#              dtype='datetime64[ns, Europe/Amsterdam]', freq='D')
i2.shift(1)
# Not what I want: timestamps not at midnight, and one got dropped!
# DatetimeIndex(['2020-03-30 01:00:00+02:00', '2020-03-31 01:00:00+02:00'],
#               dtype='datetime64[ns, Europe/Amsterdam]', freq='D')

编辑(2)

正如@MrFruppes的回答所建议的，使用i.freq的.nanos属性作为pd.DateOffset的输入…

i + pd.DateOffset(nanoseconds=i.freq.nanos)
# What I want; all timestamps at midnight (I just need to re-set the .freq attribute):
# DatetimeIndex(['2020-03-29 00:00:00+01:00', '2020-03-30 00:00:00+02:00',
#                '2020-03-31 00:00:00+02:00'],
#               dtype='datetime64[ns, Europe/Amsterdam]', freq=None)

…但是，当我们试图推进到下个月初时，它就中断了:

i3 = pd.date_range('2020-03-01', freq='MS', periods=3, tz='Europe/Amsterdam')
# DatetimeIndex(['2020-03-01 00:00:00+01:00', '2020-04-01 00:00:00+02:00',
#                '2020-05-01 00:00:00+02:00'],
#               dtype='datetime64[ns, Europe/Amsterdam]', freq='MS')
i3 + pd.DateOffset(nanoseconds=i3.freq.nanos)
Traceback (most recent call last):
File "<ipython-input-58-f3a32c654a6e>", line 1, in <module>
i3 + pd.DateOffset(nanoseconds=i3.freq.nanos)
File "pandas_libstslibsoffsets.pyx", line 690, in pandas._libs.tslibs.offsets.BaseOffset.nanos.__get__
ValueError: <MonthBegin> is a non-fixed frequency

Answer 1

如果你有一个固定的频率，你可以使用频率的nanos属性:

import pandas as pd
i = pd.date_range('2020-03-29', freq='D', periods=3, tz='Europe/Amsterdam')
# DatetimeIndex(['2020-03-29 00:00:00+01:00', '2020-03-30 00:00:00+02:00',
#               '2020-03-31 00:00:00+02:00'],
#              dtype='datetime64[ns, Europe/Amsterdam]', freq='D')
i + pd.DateOffset(nanoseconds=i.freq.nanos)
# DatetimeIndex(['2020-03-30 00:00:00+02:00', '2020-03-31 00:00:00+02:00',
#                '2020-04-01 00:00:00+02:00'],
#               dtype='datetime64[ns, Europe/Amsterdam]', freq=None)

Answer 2

pd.DateOffset也不是普遍有效。以下是我目前通过所有单元测试的代码，但我对改进持开放态度:

if i.tz is None:
raise AttributeError("Index is missing timezone information.")
# Get right timestamp for each index value, based on the frequency.
# . This one breaks for 'MS':
# i + pd.DateOffset(nanoseconds=i.freq.nanos)
# . This drops a value at some DST transitions:
# i.shift(1)
# . This one gives wrong value at DST transitions:
# i + i.freq
if i.freq == "15T": # period length always the same
ts_right = i + pd.Timedelta(hours=0.25)
elif i.freq == "H": # period length always the same
ts_right = i + pd.Timedelta(hours=1)
else:               # period length dependent on calendar
if i.freq == "D":
kwargs = {"days": 1}
elif i.freq == "MS":
kwargs = {"months": 1}
elif i.freq == "QS":
kwargs = {"months": 3}
elif i.freq == "AS":
kwargs = {"years": 1}
else:
raise ValueError(f"Invalid frequency: {i.freq}.")
ts_right = i + pd.DateOffset(**kwargs)

(我只实现了与我的用例相关的.freq值。)