Python datetime isocalendar 给出了错误的元组

Question

这是我的 python 代码

import datetime
a = datetime.datetime(2012, 1, 1)
print(a.isocalendar())

输出为

(2011, 52, 7)

输出的格式是一个元组，其中包含年、周数和工作日，分别按给定日期实例的顺序排列。

为什么会这样？不应该是(2012, 1, 1)吗，如果不是我怎么能把它转换成绝对周数？

Answer 1

根据 documentation of isocalendar:

Return a 3-tuple, (ISO year, ISO week number, ISO weekday).

The ISO calendar is a widely used variant of the Gregorian calendar. See https://www.staff.science.uu.nl/~gent0113/calendar/isocalendar.htm for a good explanation.

The ISO year consists of 52 or 53 full weeks, and where a week starts on a Monday and ends on a Sunday. The first week of an ISO year is the first (Gregorian) calendar week of a year containing a Thursday. This is called week number 1, and the ISO year of that Thursday is the same as its Gregorian year.

For example, 2004 begins on a Thursday, so the first week of ISO year 2004 begins on Monday, 29 Dec 2003 and ends on Sunday, 4 Jan 2004, so that date(2003, 12, 29).isocalendar() == (2004, 1, 1) and date(2004, 1, 4).isocalendar() == (2004, 1, 7).

在您的情况下，2012 年 1 月 1 日是星期日，因此根据 ISO 日历规则，2012 年的第一周是 1 月 2 日 - 1 月 8 日（因为第一个星期四是 1 月 5 日），而 1 月 1 日属于2011 年的最后一周。

Answer 2

您似乎需要北美周数，而不是 ISO 周数。

>>> a = datetime.datetime(2012, 1, 1)
>>> a.strftime("%Y %U")
'2012 01'

Python datetime isocalendar 给出了错误的元组

Python datetime isocalendar giving wrong tuple

python

datetime

iso