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Get week number with week start day different than monday - Python

I have a dataset with a date column. I want to get the week number associated with each date. I know I can use:

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

But it gives me the week num with start day = monday. While I need the week to start on a friday.

How do you suggest I go about doing that?

like image 956
Nouha Avatar asked Mar 23 '20 15:03

Nouha


2 Answers

tl;dr

The sections "ISO Standard" and "What you want" is to clarify your need.

You could just copy paste the code in the section "Solution" and see if the result is what you want.


ISO Standard

Definition

  • Weeks start with Monday.
  • Each week's year is the Gregorian year in which the Thursday falls.

Result of Python Standard Library 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)

Calendar Sketch

#                 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 

What You Want

Definition

  • Weeks start with Friday.
  • Each week's year is the Gregorian year in which the Monday falls.

Calendar Sketch

#                 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 

Solution

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)))

To Test

We could simply initialize CustomizedCalendar with original ISO settings, and verify if the outcome is the same with original isocalendar()'s result.

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)
like image 104
AnnieFromTaiwan Avatar answered Sep 24 '22 07:09

AnnieFromTaiwan


Here's the minimal logic:

You just need to add 3 days to a Monday to get to a Thursday. Just add the days to Monday and call the ISO Weeknumber. You'll get the shifted weeknumber.

from datetime import datetime, timedelta

x = datetime(2020, 1, 2) # this is Thursday and week 1 in ISO calendar; should be 1 in custom calendar w/ week starting Thu
y = datetime(2020, 1, 3) # this is Friday and week 1 in ISO calendar; should be 2 in custom calendar
print(x)
print(y)

def weeknum(dt):
    return dt.isocalendar()[1]

def myweeknum(dt):
    offsetdt = dt + timedelta(days=3);  # you add 3 days to Mon to get to Thu 
    return weeknum(offsetdt);

print(weeknum(x));
print(myweeknum(x));

print(weeknum(y));
print(myweeknum(y));

Output:

2020-01-02 00:00:00
2020-01-03 00:00:00
1
1
1
2
like image 36
vvg Avatar answered Sep 21 '22 07:09

vvg