Python 3.8 added the fromisocalendar() method:
>>> datetime.fromisocalendar(2011, 22, 1)
datetime.datetime(2011, 5, 30, 0, 0)
Python 3.6 added the %G
, %V
and %u
directives:
>>> datetime.strptime('2011 22 1', '%G %V %u')
datetime.datetime(2011, 5, 30, 0, 0)
Original answer
I recently had to solve this problem myself, and came up with this solution:
import datetime
def iso_year_start(iso_year):
"The gregorian calendar date of the first day of the given ISO year"
fourth_jan = datetime.date(iso_year, 1, 4)
delta = datetime.timedelta(fourth_jan.isoweekday()-1)
return fourth_jan - delta
def iso_to_gregorian(iso_year, iso_week, iso_day):
"Gregorian calendar date for the given ISO year, week and day"
year_start = iso_year_start(iso_year)
return year_start + datetime.timedelta(days=iso_day-1, weeks=iso_week-1)
A few test cases:
>>> iso = datetime.date(2005, 1, 1).isocalendar()
>>> iso
(2004, 53, 6)
>>> iso_to_gregorian(*iso)
datetime.date(2005, 1, 1)
>>> iso = datetime.date(2010, 1, 4).isocalendar()
>>> iso
(2010, 1, 1)
>>> iso_to_gregorian(*iso)
datetime.date(2010, 1, 4)
>>> iso = datetime.date(2010, 1, 3).isocalendar()
>>> iso
(2009, 53, 7)
>>> iso_to_gregorian(*iso)
datetime.date(2010, 1, 3)