获取周开始日不同于周一的周数 - Python

Question

我有一个带有日期列的数据集。我想获取与每个日期关联的周数。我知道我可以使用：

x['date'].isocalendar()[1]

但它给了我开始日 = 星期一的周数。虽然我需要一周从星期五开始。

你建议我怎么做？

Answer 1

tl;博士

“ ISO 标准”和“您想要的”部分是为了阐明您的需求。

您可以将代码复制粘贴到“解决方案”部分，看看结果是否是您想要的。

ISO标准

定义

• 周从星期一开始。
• 每周的一年是星期四所在的公历年。

Python 标准库的结果 datetime

>>> datetime(2020, 1, 1).isocalendar()
(2020, 1, 3)  # The 3rd day of the 1st week in 2020
>>> datetime(2019, 12, 31).isocalendar()
(2020, 1, 2)  # The 2nd day of the 1st week in 2020
>>> datetime(2019, 1, 1).isocalendar()
(2019, 1, 2)
>>> datetime(2017, 1, 1).isocalendar()
(2016, 52, 7)
>>> datetime(2016, 12, 26).isocalendar()
(2016, 52, 1)
>>> datetime(2015, 12, 31).isocalendar()
(2015, 53, 4)
>>> datetime(2016, 1, 1).isocalendar()
(2015, 53, 5)

日历草图

#                 Mo Tu Wd Th Fr Sa Sn
# [2019-52w] DEC/ 23 24 25 26 27 28 29 /DEC
# [2020-1w]  DEC/ 30 31  1  2  3  4  5 /JAN

# [2019-1w]  DEC/ 31  1  2  3  4  5  6 /JAN

# [2016-52w] DEC/ 26 27 28 29 30 31  1 /JAN

# [2015-53w] DEC/ 28 29 30 31  1  2  3 /JAN
# [2016-1w]  JAN/  4  5  6  7  8  9 10 /JAN 

你想要什么

定义

• 周从星期五开始。
• 每周的一年是星期一所在的公历年。

日历草图

#                 Fr Sa Sn. Mo Tu Wd Th 
# [2019-51w] DEC/ 20 21 22. 23 24 25 26  /DEC
# [2019-52w] DEC/ 27 28 29. 30 31  1  2  /JAN
# [2020-1w]  JAN/  3  4  5.  6  7  8  9  /JAN

# [2018-53w] DEC/ 28 29 30. 31  1  2  3  /JAN
# [2019-1w]  JAN/  4  5  6.  7  8  9 10  /JAN

# [2016-52w] DEC/ 23 24 25. 26 27 28 29  /DEC
# [2017-1w]  DEC/ 30 31  1.  2  3  4  5  /JAN

# [2015-52w] DEC/ 25 26 27. 28 29 30 31  /DEC
# [2016-1w]  JAN/  1  2  3.  4  5  6  7  /JAN 

解决方案

from datetime import datetime, timedelta
from enum import IntEnum

WEEKDAY = IntEnum('WEEKDAY', 'MON TUE WED THU FRI SAT SUN', start=1)

class CustomizedCalendar:

    def __init__(self, start_weekday, indicator_weekday=None):
        self.start_weekday = start_weekday
        self.indicator_delta = 3 if not (indicator_weekday) else (indicator_weekday - start_weekday) % 7

    def get_week_start(self, date):
        delta = date.isoweekday() - self.start_weekday
        return date - timedelta(days=delta % 7)

    def get_week_indicator(self, date):
        week_start = self.get_week_start(date)
        return week_start + timedelta(days=self.indicator_delta)

    def get_first_week(self, year):
        indicator_date = self.get_week_indicator(datetime(year, 1, 1))
        if indicator_date.year == year:  # The date "year.1.1" is on 1st week.
            return self.get_week_start(datetime(year, 1, 1))
        else:  # The date "year.1.1" is on the last week of "year-1".
            return self.get_week_start(datetime(year, 1, 8))
    
    def calculate(self, date):
        year = self.get_week_indicator(date).year
        first_date_of_first_week = self.get_first_week(year)
        diff_days = (date - first_date_of_first_week).days
        return year, (diff_days // 7 + 1), (diff_days % 7 + 1)

if __name__ == '__main__':
    # Use like this:
    my_calendar = CustomizedCalendar(start_weekday=WEEKDAY.FRI, indicator_weekday=WEEKDAY.MON)
    print(my_calendar.calculate(datetime(2020, 1, 2)))

去测试

我们可以简单地CustomizedCalendar使用原始 ISO 设置进行初始化，并验证结果是否与 originalisocalendar()的结果相同。

my_calendar = CustomizedCalendar(start_weekday=WEEKDAY.MON)
s = datetime(2019, 12, 19)
for delta in range(20):
    print my_calendar.calculate(s) == s.isocalendar()
    s += timedelta(days=1)