一般要获取当前周数(从周日开始):

from datetime import *
today = datetime.today()
print today.strftime("%U")

假设您需要将一周与当天的年份组合为字符串。

import datetime
year,week = datetime.date.today().isocalendar()[:2]
week_of_the_year = f"{year}-{week}"
print(week_of_the_year)

你可能会在2021-28年

iscalendar()对于某些日期返回错误的年和周数值:

Python 2.7.3 (default, Feb 27 2014, 19:58:35) 
[GCC 4.6.3] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import datetime as dt
>>> myDateTime = dt.datetime.strptime("20141229T000000.000Z",'%Y%m%dT%H%M%S.%fZ')
>>> yr,weekNumber,weekDay = myDateTime.isocalendar()
>>> print "Year is " + str(yr) + ", weekNumber is " + str(weekNumber)
Year is 2015, weekNumber is 1

与Mark Ransom的方法相比:

>>> yr = myDateTime.year
>>> weekNumber = ((myDateTime - dt.datetime(yr,1,1)).days/7) + 1
>>> print "Year is " + str(yr) + ", weekNumber is " + str(weekNumber)
Year is 2014, weekNumber is 52

我们也有类似的问题，于是我们想出了这个逻辑 我已经测试了1年的测试用例，并且全部通过

import datetime


def week_of_month(dt):

    first_day = dt.replace(day=1)
    dom = dt.day
    if first_day.weekday() == 6:
        adjusted_dom = dom
    else:
        adjusted_dom = dom + first_day.weekday()
    if adjusted_dom % 7 == 0 and first_day.weekday() != 6:
       value = adjusted_dom / 7.0 + 1
    elif first_day.weekday() == 6 and adjusted_dom % 7 == 0 and adjusted_dom == 7:
        value = 1
    else:
        value = int(ceil(adjusted_dom / 7.0))

    return int(value)


year = 2020
month = 01
date = 01

date_value = datetime.datetime(year, month, date).date()
no = week_of_month(date_value)

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

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。