How to Get Day of Week as Integer with First of Month Changing Values?

user_16804086 0
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Using .weekday() to find the day of the week as an integer (Monday = 0 ... Sunday = 6) for everyday from today until next year (+365 days from today). Problem now is that if the 1st of the month starts mid week then I need to return the day of the week with the 1st day of the month now being = 0.

Ex. If the month starts Wednesday then Wednesday = 0... Sunday = 4 (for that week only).

Annotated Picture of Month Explaining What I Want to Do

Originally had the below code but wrong as the first statement will run 7 days regardless.

import datetime
from datetime import date

for day in range (1,365):
    departure_date = date.today() + datetime.timedelta(days=day)

    if departure_date.weekday() < 7:
        day_of_week = departure_date.day
    else:
        day_of_week = departure_date.weekday()
python-datetime python
3 Answers

The following seems to do the job properly:

import datetime as dt

def custom_weekday(date):
    if date.weekday() > (date.day-1):
        return date.day - 1
    else:
        return date.weekday()
        
for day in range (1,366):
    departure_date = dt.date.today() + dt.timedelta(days=day)
    day_of_week = custom_weekday(date=departure_date)
    print(departure_date, day_of_week, departure_date.weekday())

Your code had two small bugs:

  1. the if condition was wrong
  2. days are represented inconsistently: date.weekday() is 0-based, date.day is 1-based
user_33889620

For any date D.M.Y, get the weekday W of 1.M.Y.

Then you need to adjust weekday value only for the first 7-W days of that month. To adjust, simply subtract the value W.

Example for September 2021: the first date of month (1.9.2021) is a Wednesday, so W is 2. You need to adjust weekdays for dates 1.9.2021 to 5.9.2021 (because 7-2 is 5) in that month by minus 2.

user_53788160

For every date, get the first week of that month. Then, check if the date is within that first week. If it is, use the .day - 1 value (since you are 0-based). Otherwise, use the .weekday().

from datetime import date, datetime, timedelta

for day in range (-5, 40):
    departure_date = date.today() + timedelta(days=day)
    
    first_week = date(departure_date.year, departure_date.month, 1).isocalendar()[1]

    if first_week == departure_date.isocalendar()[1]:
        day_of_week = departure_date.day - 1
    else:
        day_of_week = departure_date.weekday()

    print(departure_date, day_of_week)

2021-08-27 4
2021-08-28 5
2021-08-29 6
2021-08-30 0
2021-08-31 1
2021-09-01 0
2021-09-02 1
2021-09-03 2
2021-09-04 3
2021-09-05 4
2021-09-06 0
2021-09-07 1
2021-09-08 2
2021-09-09 3
2021-09-10 4
2021-09-11 5
2021-09-12 6
2021-09-13 0
2021-09-14 1
2021-09-15 2
2021-09-16 3
2021-09-17 4
2021-09-18 5
2021-09-19 6
2021-09-20 0
2021-09-21 1
2021-09-22 2
2021-09-23 3
2021-09-24 4
2021-09-25 5
2021-09-26 6
2021-09-27 0
2021-09-28 1
2021-09-29 2
2021-09-30 3
2021-10-01 0
2021-10-02 1
2021-10-03 2
2021-10-04 0
2021-10-05 1
2021-10-06 2
2021-10-07 3
2021-10-08 4
2021-10-09 5
2021-10-10 6
user_110438250
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