python-2.7 - Python2 : Retrieve Sunday - Saturday Week Start/End Dates For Given Date Range

Question

有很多帖子都解决了类似的问题，但没有一个帖子具有与我的问题相同的限制。

我正在编写一个脚本，可以从数据中心获取任意周数的数据。它获取的周数取决于外部用户向我的脚本提供的日期范围。数据中心的一周从周日到周六。 Python 的一周从周一到周日。

我需要能够获取日期范围内每个日期之前的周日和之后的周六的日期。更复杂的是，一周开始日期和周末结束日期都不能超出要求的范围。这可以防止我简单地从范围内的每个日期中减去一天。

一些示例场景:

示例1)

requested_date_range = [datetime(2016,7,1,0,0),datetime(2016,8,5,0,0)]
what I get from the various Python utilities (dateutil, datetime_periods, etc):

[
[datetime(2016,6,27,0,0),datetime(2016,7,3,0,0)],
[datetime(2016,7,4,0,0),datetime(2016,7,10,0,0)],
[datetime(2016,7,11,0,0),datetime(2016,7,17,0,0)],
[datetime(2016,7,18,0,0),datetime(2016,7,24,0,0)],
[datetime(2016,7,25,0,0),datetime(2016,7,31,0,0)],
[datetime(2016,8,1,0,0),datetime(2016,8,7,0,0)]
]

what I actually need:
[
[datetime(2016,7,1,0,0),datetime(2016,7,2,0,0)], #"week" starts on first day of requested range and ends on the following Saturday
[datetime(2016,7,3,0,0),datetime(2016,7,9,0,0)], #Sunday through Saturday
[datetime(2016,7,10,0,0),datetime(2016,7,16,0,0)], #Sunday through Saturday
[datetime(2016,7,17,0,0),datetime(2016,7,23,0,0)], #Sunday through Saturday
[datetime(2016,7,24,0,0),datetime(2016,7,30,0,0)], #Sunday through Saturday
[datetime(2016,7,31,0,0),datetime(2016,8,5,0,0)] #"week" starts on Sunday and ends on last day of requested range
] 

示例2)

requested_date_range = [datetime(2016,7,3,0,0),datetime(2016,8,7,0,0)]
what I get from the various Python utilities (dateutil, datetime_periods, etc):
[
[datetime(2016,6,27,0,0),datetime(2016,7,3,0,0)],
[datetime(2016,7,4,0,0),datetime(2016,7,10,0,0)],
[datetime(2016,7,11,0,0),datetime(2016,7,17,0,0)],
[datetime(2016,7,18,0,0),datetime(2016,7,24,0,0)],
[datetime(2016,7,25,0,0),datetime(2016,7,31,0,0)],
[datetime(2016,8,1,0,0),datetime(2016,8,7,0,0)]
]
what I actually need: 
[
[datetime(2016,7,3,0,0),datetime(2016,7,9,0,0)], #"week" starts on first day of requested range
[datetime(2016,7,10,0,0),datetime(2016,7,16,0,0)], #Sunday through Saturday
[datetime(2016,7,17,0,0),datetime(2016,7,23,0,0)], #Sunday through Saturday
[datetime(2016,7,24,0,0),datetime(2016,7,30,0,0)], #Sunday through Saturday
[datetime(2016,7,31,0,0),datetime(2016,8,6,0,0)], #Sunday through Saturday
[datetime(2016,8,7,0,0),datetime(2016,8,7,0,0)]  #"week" ends up being only one day long because the max requested date falls on a Sunday
]

Answer 1

您应该能够使用dateutil.relativedelta轻松完成此操作。下面是一个示例函数:

from dateutil.relativedelta import relativedelta
from dateutil.relativedelta import MO, TU, WE, TH, FR, SA, SU

def week_range(range_start, range_end):
    dts = []
    WEEK_START = relativedelta(weekday=SU(+2))
    WEEK_END = relativedelta(weekday=SA)

    c_wstart = range_start + relativedelta(weekday=SU(+1))
    c_wend = c_wstart + WEEK_END

    if range_start < c_wstart:
        dts.append((range_start, range_start + WEEK_END))

    while True:
        if c_wend > range_end:
            c_wend = range_end

        dts.append((c_wstart, c_wend))

        if c_wend >= range_end:
            break

        c_wstart = c_wstart + WEEK_START
        c_wend = c_wstart + WEEK_END

        if c_wstart > range_end:
            break

    return dts

在上面的函数中，我首先获取范围开头并向其添加relativedelta(weekday=SU)，这给了我原始日期或之后的第一个星期日。然后，我连续将 relativedelta(weekday=SU(+2)) 添加到“当前周”，以获得当前日期或之后的第二星期日(这是因为我的“周开始”始终是星期日，始终是下一个星期日)。

对于我生成的每个日期，我只需添加 relativedelta(weekday=SA) 即可生成即将到来的星期六，如果我超出了日期范围，我会“剪辑”最后一个日期为日期范围。

使用您的示例:

>>> week_range(datetime(2016, 7, 1), datetime(2016, 8, 5))
[(datetime.datetime(2016, 7, 1, 0, 0), datetime.datetime(2016, 7, 2, 0, 0)),
 (datetime.datetime(2016, 7, 3, 0, 0), datetime.datetime(2016, 7, 9, 0, 0)),
 (datetime.datetime(2016, 7, 10, 0, 0), datetime.datetime(2016, 7, 16, 0, 0)),
 (datetime.datetime(2016, 7, 17, 0, 0), datetime.datetime(2016, 7, 23, 0, 0)),
 (datetime.datetime(2016, 7, 24, 0, 0), datetime.datetime(2016, 7, 30, 0, 0)),
 (datetime.datetime(2016, 7, 31, 0, 0), datetime.datetime(2016, 8, 5, 0, 0))]
>>> week_range(datetime(2016, 7, 3), datetime(2016, 8, 7))
[(datetime.datetime(2016, 7, 3, 0, 0), datetime.datetime(2016, 7, 9, 0, 0)),
 (datetime.datetime(2016, 7, 10, 0, 0), datetime.datetime(2016, 7, 16, 0, 0)),
 (datetime.datetime(2016, 7, 17, 0, 0), datetime.datetime(2016, 7, 23, 0, 0)),
 (datetime.datetime(2016, 7, 24, 0, 0), datetime.datetime(2016, 7, 30, 0, 0)),
 (datetime.datetime(2016, 7, 31, 0, 0), datetime.datetime(2016, 8, 6, 0, 0)),
 (datetime.datetime(2016, 8, 7, 0, 0), datetime.datetime(2016, 8, 7, 0, 0))]

根据您的喜好，您还可以使用 rruleset 完成类似的操作:

from dateutil.rrule import rrule, rruleset
from dateutil.rrule import WEEKLY, SU, SA
from datetime import timedelta

from itertools import zip_longest, chain
def week_range_rrule(range_start, range_end, weekday_start=SU, weekday_end=SA):
    # Beginning of the week rule
    rr1 = rrule(WEEKLY, byweekday=weekday_start,
                dtstart=range_start, until=range_end)

    # End of the week rule - adding 1 second to the range end because
    # "until" isn't inclusive
    rr2 = rrule(WEEKLY, byweekday=weekday_end,
                dtstart=range_start+relativedelta(SA),
                until=range_end+timedelta(seconds=1))

    # Combine these into a rule set
    rrs = rruleset()
    rrs.rrule(rr1)
    rrs.rrule(rr2)

    # Explicitly add range start and end to the rules, in case they don't
    # fall on neat week boundaries
    rrs.rdate(range_start)
    rrs.rdate(range_end)

    if next(iter(rr2)) == range_start:
        rrs = chain((range_start, ), rrs)

    # Modified version of the "grouper" recipe from itertools
    args = [iter(rrs)] * 2

    return list(zip_longest(*args, fillvalue=range_end))

请注意，如果您希望第一个是惰性的，只需将 dts.append(x) 的所有实例替换为 yield x 即可。如果您希望第二个是惰性的，只需删除 return 语句中 zip_longest 周围的 list() 包装器即可。