将“pandas”频率字符串转换为“DateOffset”

Question

我有一个时区感知 pandas DateTimeIndex，我想提前一个时间步长，时间步长由其 .freq 属性指定。但是，这样做不尊重时区信息：

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')

的工作是使用 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)

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

当然也欢迎其他解决方案。

This 是我发现的唯一与此相关的其他问题，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')

（其实后一项returns正好i.freq。）

非常感谢。

编辑 (1)

正如评论中所建议的那样，在某些情况下使用 .shift(1) 是可行的，包括我在上面提到的情况...

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')

...但不是全部。事实上，将我原始索引中的开始日期提前一天会导致时间戳被删除，其余的都是错误的：

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 在回答中所建议的那样，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)

...但是当我们尝试提前到下月初时它会中断：

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\_libs\tslibs\offsets.pyx", line 690, in pandas._libs.tslibs.offsets.BaseOffset.nanos.__get__

ValueError: <MonthBegin> is a non-fixed frequency

Answer 1

如果你有固定的频率，可以使用nanos属性的频率。例如：

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 也不是普遍适用的。这是我目前拥有的，它通过了我所有的单元测试，但我愿意改进：

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。）

将“pandas”频率字符串转换为“DateOffset”

Convert `pandas` frequency string to `DateOffset`

python

datetime

date

dst

pandas

编辑 (1)

编辑 (2)