Python2：在指定日期范围内检索星期日 - 星期六星期开始/结束日期

Question

有很多帖子可以解决类似的问题，但没有一个与我的问题有相同的限制。

我在写一个脚本，可以从数据中心获取任意数量的数据。它提取哪个星期取决于外部用户提供给我的脚本的日期范围。数据中心的周从周日到周六。 Python的周从星期一到星期日运行。

我需要能够获取日期范围内每个日期之后的星期日和星期六的日期。使问题复杂化的是，周开始日期和周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减去一天。

一些示例场景：

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 
] 

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

你应该能够做到这一点很容易地使用dateutil.relativedelta。下面的例子功能：

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，我先取范围开始，并添加relativedelta(weekday=SU)给它，它给我的第一个星期日上还是原来的日期。然后我在当前日期或之后连续添加relativedelta(weekday=SU(+2))到“当前周”以获得第二天星期天（自“我的星期开始”始终是星期日，始终是下个星期日）。

对于我生成的每个日期，我只是将relativedelta(weekday=SA)添加到它来生成即将到来的星期六，并且如果我在日期范围之外，我会将最后日期“剪辑”为日期范围。

使用你的例子：

>>> 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))] 

取决于你的口味，你也可以使用rruleset完成类似的东西：

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)) 

需要注意的是，如果你想第一个偷懒，只是将dts.append(x)的所有实例替换为yield x。如果你想第二个懒惰，只需在return语句中删除围绕zip_longest的list()包装。

Answer 2

这里有一个稍微不详细，尽管简洁的答案。

import datetime as dt 

if __name__ == "__main__": 
    weekend_index = (6, 5) # Sunday, Saturday 
    requested_range = (dt.datetime(2016, 7, 9, 0, 0), dt.datetime(2016, 8, 11, 0, 0)) 

    start, end = requested_range 
    sun, sat = weekend_index 
    cur = start 
    my_range = [] 

    while cur < end: 
      cr = [] 
      cr.append(cur) 
      cur = end if end < cur+dt.timedelta(days=6) else (cur+dt.timedelta(days=(sun if cur.weekday() == sun else (sat-cur.weekday())))) 
      cr.append(cur) 
      cur += dt.timedelta(days=1) 
      my_range.append(cr) 

print(my_range) # Returns: 
# [[datetime.datetime(2016, 7, 9, 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, 11, 0, 0)]]