如何使用python获得本周的内容?

当心!如果要定义自己的周数,可以使用提供的生成器表达式 in your first question 顺便说一下,得到了一个很棒的答案)。如果要遵循周数的ISO约定,则需要小心:

一年中的第一个日历周是 包括第一个的那个 那年的星期四和 日历年的最后一个日历周 是在 下一日历周的第一个日历周 历年。

比如说, 2010年1月1日和2日不是2010年的第一周 但2009年第53周。

python提供了一个使用iso日历查找周数的模块:

示例代码:

h[1] >>> import datetime
h[1] >>> Jan1st = datetime.date(2010,1,1)
h[1] >>> Year,WeekNum,DOW = Jan1st.isocalendar() # DOW = day of week
h[1] >>> print Year,WeekNum,DOW
2009 53 5

请再次注意,2010年1月1日与2009年第53周的对应关系。

使用前面答案中提供的生成器:

from datetime import date, timedelta


def allsundays(year):
    """This code was provided in the previous answer! It's not mine!"""
    d = date(year, 1, 1)                    # January 1st                                                          
    d += timedelta(days = 6 - d.weekday())  # First Sunday                                                         
    while d.year == year:
        yield d
        d += timedelta(days = 7)

Dict = {}
for wn,d in enumerate(allsundays(2010)):
    # This is my only contribution!
    Dict[wn+1] = [(d + timedelta(days=k)).isoformat() for k in range(0,7) ]

print Dict

dict包含您请求的词典。

你如何确定周数?在这里,我用一个函数来标识一周中的某一天,这个函数得到一周中的星期日(您在示例中使用的),然后返回它加上接下来的6天。

import datetime

one_day = datetime.timedelta(days=1)

def get_week(date):
  """Return the full week (Sunday first) of the week containing the given date.

  'date' may be a datetime or date instance (the same type is returned).
  """
  day_idx = (date.weekday() + 1) % 7  # turn sunday into 0, monday into 1, etc.
  sunday = date - datetime.timedelta(days=day_idx)
  date = sunday
  for n in xrange(7):
    yield date
    date += one_day

print list(get_week(datetime.datetime.now().date()))
# [datetime.date(2010, 1, 3), datetime.date(2010, 1, 4),
#  datetime.date(2010, 1, 5), datetime.date(2010, 1, 6),
#  datetime.date(2010, 1, 7), datetime.date(2010, 1, 8),
#  datetime.date(2010, 1, 9)]
print [d.isoformat() for d in get_week(datetime.datetime.now().date())]
# ['2010-01-03', '2010-01-04', '2010-01-05', '2010-01-06', '2010-01-07',
#  '2010-01-08', '2010-01-09']

您可以使用datetime模块。您可以指定格式和所有内容。链接如下: http://docs.python.org/library/datetime.html

查看datetime.datetime(params)和datetime.timedelta(params)。希望一切顺利;—)

例子:

import datetime

numweeks = 5
start_date = datetime.datetime(year=2010,month=1,day=4)    

weeks = {}

offset = datetime.timedelta(days=0)
for week in range(numweeks):
   this_week = []
   for day in range(7):
        date = start_date + offset
        date = date.strftime( some_format_string )
        this_week.append( date )
        offset += datetime.timedelta(days=1)
   weeks[week] = this_week 

如果您同意ISO标准:

>>> import collections
>>> dd = collections.defaultdict(list)
>>> jan1 = datetime.date(2010, 1, 1)
>>> oneday = datetime.timedelta(days=1)
>>> allyear = [jan1 + k*oneday for k in range(365 + 6)]
>>> for d in allyear:
...   y, w, wd = d.isocalendar()
...   if y == 2010: dd[w].append(d.strftime('%m/%d/%Y'))
... 

这产生的结果与您要查找的结果稍有不同(根据ISO标准,周从星期一开始,而不是星期日……),例如:

>>> dd[1]
['01/04/2010', '01/05/2010', '01/06/2010', '01/07/01/2010', '01/08/2010', '01/09/2010', '01/10/2010']

但是你可以通过模拟一个适当的“一个接一个”的错误来调整它!-)

这个 calendar 模块允许您将任何工作日设置为“每周的第一天”,但没有简单的方法来获取所有的周(当一周被拆分为两个月时,不需要重复),因此我认为直接工作 datetime 可能是个更好的主意。

在读过这个问题之后,我开发了一个三行的方法:

from datetime import timedelta

def get_week_dates(base_date, start_day, end_day=None):
    """
    Return entire week of dates based on given date limited by start_day and end_day.
    If end_day is None, return only start_day.

    >>> from datetime import date
    >>> get_week_dates(date(2015,1,16), 3, 5)
    [datetime.date(2015, 1, 14), datetime.date(2015, 1, 15), datetime.date(2015, 1, 16)]

    >>> get_week_dates(date(2015,1,15), 2, 5)
    [datetime.date(2015, 1, 13), datetime.date(2015, 1, 14), datetime.date(2015, 1, 15), datetime.date(2015, 1, 16)]
    """
    monday = base_date - timedelta(days=base_date.isoweekday() - 1)
    week_dates = [monday + timedelta(days=i) for i in range(7)]
    return week_dates[start_day - 1:end_day or start_day]

使用 get_week_dates(date.today(), 1, 7) 获取当前周日期。

下面是一些代码:

import datetime

now = datetime.datetime.now()

now_day_1 = now - datetime.timedelta(days=now.weekday())

dates = {}

for n_week in range(3):
    dates[n_week] = [(now_day_1 + datetime.timedelta(days=d+n_week*7)).strftime("%m/%d/%Y") for d in range(7)]

print dates

印刷品:

{
 0: ['01/04/2010', '01/05/2010', '01/06/2010', '01/07/2010', '01/08/2010', '01/09/2010', '01/10/2010'], 
 1: ['01/11/2010', '01/12/2010', '01/13/2010', '01/14/2010', '01/15/2010', '01/16/2010', '01/17/2010'], 
 2: ['01/18/2010', '01/19/2010', '01/20/2010', '01/21/2010', '01/22/2010', '01/23/2010', '01/24/2010']
}

current_week = datetime.datetime.now().isocalendar()[1]