How to get start of week for given (iso) weeknumber / year in python

Question

I know that I can use datetime.isocalendar() to get the right number per week. How can I do the inverse, given the week number and year, to get the first day of this week.

Answer 1

If you are limited to stdlib, you can do the following:

 >>> datetime.datetime.strptime('2011, 4, 0', '%Y, %U, %w') datetime.datetime(2011, 1, 23, 0, 0) 

Answer 2

Could not find the standard library, so I had to flip it. The main difficulty lies in finding the first day of the year ISO (which may be last year). You need to add some input checks, etc.

 import datetime def isoWeekToGregorian(isoYear, isoWeek, isoDayOfWeek): #we have to find the date of the iso year t0 = datetime.datetime(isoYear,1,1,0,0,0) t0iso = t0.isocalendar() if (t0iso[1] != 1): t0prime = t0 + datetime.timedelta(7-t0iso[2]) else: t0prime = t0 - datetime.timedelta(t0iso[2]) #we can add our weeks and days... return t0prime + datetime.timedelta(7*(isoWeek-1) + isoDayOfWeek) 

We can create a simple test:

 #TEST: we know 2004 has 53 weeks... t0 = datetime.datetime(2004,12,26,0,0,0) t1 = datetime.datetime(2005,1,10,0,0,0) ndays = (t1-t0).days + 1 for i in range(ndays): d0 = t0 + datetime.timedelta(i) d0iso = d0.isocalendar() if (d0 != isoWeekToGregorian(d0iso[0],d0iso[1],d0iso[2])): print "failed: %s" % d0 else: print d0, d0iso 

What prints correctly:

 2004-12-26 00:00:00 (2004, 52, 7) 2004-12-27 00:00:00 (2004, 53, 1) 2004-12-28 00:00:00 (2004, 53, 2) 2004-12-29 00:00:00 (2004, 53, 3) 2004-12-30 00:00:00 (2004, 53, 4) 2004-12-31 00:00:00 (2004, 53, 5) 2005-01-01 00:00:00 (2004, 53, 6) 2005-01-02 00:00:00 (2004, 53, 7) 2005-01-03 00:00:00 (2005, 1, 1) 2005-01-04 00:00:00 (2005, 1, 2) 2005-01-05 00:00:00 (2005, 1, 3) 2005-01-06 00:00:00 (2005, 1, 4) 2005-01-07 00:00:00 (2005, 1, 5) 2005-01-08 00:00:00 (2005, 1, 6) 2005-01-09 00:00:00 (2005, 1, 7) 2005-01-10 00:00:00 (2005, 2, 1)