有许多周编号系统。下面是最常见的系统代码示例:

ISO: First week starts with Monday and must contain the January 4th (or first Thursday of the year). The ISO calendar is already implemented in Python: >>> from datetime import date >>> date(2014, 12, 29).isocalendar()[:2] (2015, 1) North American: First week starts with Sunday and must contain the January 1st. The following code is my modified version of Python's ISO calendar implementation for the North American system: from datetime import date def week_from_date(date_object): date_ordinal = date_object.toordinal() year = date_object.year week = ((date_ordinal - _week1_start_ordinal(year)) // 7) + 1 if week >= 52: if date_ordinal >= _week1_start_ordinal(year + 1): year += 1 week = 1 return year, week def _week1_start_ordinal(year): jan1 = date(year, 1, 1) jan1_ordinal = jan1.toordinal() jan1_weekday = jan1.weekday() week1_start_ordinal = jan1_ordinal - ((jan1_weekday + 1) % 7) return week1_start_ordinal >>> from datetime import date >>> week_from_date(date(2014, 12, 29)) (2015, 1) MMWR (CDC): First week starts with Sunday and must contain the January 4th (or first Wednesday of the year). I created the epiweeks package specifically for this numbering system (also has support for the ISO system). Here is an example: >>> from datetime import date >>> from epiweeks import Week >>> Week.fromdate(date(2014, 12, 29)) (2014, 53)

我将讨论总结为两个步骤:

将原始格式转换为datetime对象。 使用datetime对象或date对象的函数来计算周数。

热身

from datetime import datetime, date, time
d = date(2005, 7, 14)
t = time(12, 30)
dt = datetime.combine(d, t)
print(dt)

第一步

要手动生成一个datetime对象，可以使用datetime.datetime(2017,5,3)或datetime.datetime.now()。

但实际上，我们通常需要解析一个现有的字符串。我们可以使用strptime函数，例如datetime.strptime('2017-5-3'，'%Y-%m-%d')，其中必须指定格式。不同格式代码的详细信息可以在官方文档中找到。

或者，更方便的方法是使用dateparse模块。例如dateparser。parse('16 Jun 2010')， dateparser.parse('12/2/12')或dateparser.parse('2017-5-3')

以上两种方法将返回一个datetime对象。

第二步

使用获得的datetime对象调用strptime(format)。例如,

python

dt = datetime.strptime('2017-01-1','%Y-%m-%d') # return a datetime object. This day is Sunday
print(dt.strftime("%W")) # '00' Monday as the 1st day of the week. All days in a new year preceding the 1st Monday are considered to be in week 0.
print(dt.strftime("%U")) # '01' Sunday as the 1st day of the week. All days in a new year preceding the 1st Sunday are considered to be in week 0.
print(dt.strftime("%V")) # '52' Monday as the 1st day of the week. Week 01 is the week containing Jan 4.

决定使用哪种格式是很棘手的。更好的方法是获取一个日期对象来调用iscalendar()。例如,

python

dt = datetime.strptime('2017-01-1','%Y-%m-%d') # return a datetime object
d = dt.date() # convert to a date object. equivalent to d = date(2017,1,1), but date.strptime() don't have the parse function
year, week, weekday = d.isocalendar() 
print(year, week, weekday) # (2016,52,7) in the ISO standard

实际上，您更有可能使用date. iscalendar()来准备周报，特别是在圣诞节-新年购物季。

有许多周编号系统。下面是最常见的系统代码示例:

ISO: First week starts with Monday and must contain the January 4th (or first Thursday of the year). The ISO calendar is already implemented in Python: >>> from datetime import date >>> date(2014, 12, 29).isocalendar()[:2] (2015, 1) North American: First week starts with Sunday and must contain the January 1st. The following code is my modified version of Python's ISO calendar implementation for the North American system: from datetime import date def week_from_date(date_object): date_ordinal = date_object.toordinal() year = date_object.year week = ((date_ordinal - _week1_start_ordinal(year)) // 7) + 1 if week >= 52: if date_ordinal >= _week1_start_ordinal(year + 1): year += 1 week = 1 return year, week def _week1_start_ordinal(year): jan1 = date(year, 1, 1) jan1_ordinal = jan1.toordinal() jan1_weekday = jan1.weekday() week1_start_ordinal = jan1_ordinal - ((jan1_weekday + 1) % 7) return week1_start_ordinal >>> from datetime import date >>> week_from_date(date(2014, 12, 29)) (2015, 1) MMWR (CDC): First week starts with Sunday and must contain the January 4th (or first Wednesday of the year). I created the epiweeks package specifically for this numbering system (also has support for the ISO system). Here is an example: >>> from datetime import date >>> from epiweeks import Week >>> Week.fromdate(date(2014, 12, 29)) (2014, 53)

对于一年中的瞬时周的整数值，尝试:

import datetime
datetime.datetime.utcnow().isocalendar()[1]

我相信date. iscalendar()将是答案。本文解释了ISO 8601日历背后的数学原理。查看Python文档的datetime页面的date. iscalendar()部分。

>>> dt = datetime.date(2010, 6, 16) 
>>> wk = dt.isocalendar()[1]
24

. iscalendar()返回一个三元组，包含(年，星期数，星期天数)。dt. iscalendar()[0]返回年份，dt. iscalendar()[1]返回周数，dt. iscalendar()[2]返回周数。尽可能的简单。

已经给出了很多答案，但我想补充一下。

如果您需要将周显示为年/周样式(例如，1953 - 2019年第53周，2001 - 2020年第1周等)，您可以这样做:

import datetime

year = datetime.datetime.now()
week_num = datetime.date(year.year, year.month, year.day).strftime("%V")
long_week_num = str(year.year)[0:2] + str(week_num)

它将使用当前的年份和周，写这篇文章当天的long_week_num将是:

>>> 2006