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import datetime as dt
dt.datetime.strptime('2019-01-01', '%Y-%m-%d').isocalendar()[1]
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import datetime
my_date = datetime.date.today() # if date is 01/01/2018
year, week_num, day_of_week = my_date.isocalendar()
print("Week #" + str(week_num) + " of year " + str(year))
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>>> from datetime import datetime
>>> datetime.today().strftime('%A')
'Wednesday'
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>>> import datetime
>>> datetime.datetime.today()
datetime.datetime(2012, 3, 23, 23, 24, 55, 173504)
>>> datetime.datetime.today().weekday()
4
source: https://stackoverflow.com/questions/9847213/how-do-i-get-the-day-of-week-given-a-date
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>>> import datetime
>>> datetime.datetime.today()
datetime.datetime(2012, 3, 23, 23, 24, 55, 173504)
>>> datetime.datetime.today().weekday()
4
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>>> import datetime
>>> datetime.datetime.now() + datetime.timedelta(days=7)
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dt = '2001-10-18'
year, month, day = (int(x) for x in dt.split('-'))
answer = datetime.date(year, month, day).weekday()
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import datetime
from dateutil.relativedelta import relativedelta
week = 25
year = 2021
date = datetime.date(year, 1, 1) + relativedelta(weeks=+week)
print(date)