您可以直接从datetime作为字符串获取周数。

>>> import datetime
>>> datetime.date(2010, 6, 16).strftime("%V")
'24'

你也可以得到不同的“类型”的周数的年份改变strftime参数:

%U - Week number of the year (Sunday as the first day of the week) as a zero padded decimal number. All days in a new year preceding the first Sunday are considered to be in week 0. Examples: 00, 01, …, 53 %W - Week number of the year (Monday as the first day of the week) as a decimal number. All days in a new year preceding the first Monday are considered to be in week 0. Examples: 00, 01, …, 53 [...] (Added in Python 3.6, backported to some distribution's Python 2.7's) Several additional directives not required by the C89 standard are included for convenience. These parameters all correspond to ISO 8601 date values. These may not be available on all platforms when used with the strftime() method. [...] %V - ISO 8601 week as a decimal number with Monday as the first day of the week. Week 01 is the week containing Jan 4. Examples: 01, 02, …, 53 from: datetime — Basic date and time types — Python 3.7.3 documentation

我是从这里知道的。它在Python 2.7.6中对我有效

对于一年中的瞬时周的整数值，尝试:

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

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

对于pandas用户，如果你想获得一列周数:

df['weekofyear'] = df['Date'].dt.week

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)

别人建议的ISO周是很好的，但可能不适合你的需求。它假设每周从星期一开始，这导致了年初和年底的一些有趣的异常情况。

如果你宁愿使用一个定义，说第一周总是1月1日到1月7日，而不管星期几，可以使用这样的推导:

>>> testdate=datetime.datetime(2010,6,16)
>>> print(((testdate - datetime.datetime(testdate.year,1,1)).days // 7) + 1)
24