I'm running Python 3.8.3 and I found something weird about the ISO Week format (%V) :
The first day and the last day of 2019 are both in week 1.
from datetime import date
print(date(2019, 1, 1).strftime('%Y-W%V'))
print(date(2019, 12, 29).strftime('%Y-W%V'))
print(date(2019, 12, 31).strftime('%Y-W%V'))
Output:
2019-W01
2019-W52
2019-W01
Why does it behave like that?
It is fully correct.
As you see in your dates, all of them are in 2019, so it is correct to get 2019 with %Y.
Week number is defined by ISO, and so one week could be considered in previous or in next year.
You need to use %G to get year of the week number (%V).
I'm trying to generate week number string using Python time module, considering week starts on Sunday.
If my interpretation of the official documentation is correct then this can be achieved by the following code:
import time
time.strftime("%U", time.localtime())
>> 37
My question is, is the above output correct? Shouldn't the output be 38 instead, considering the below details:
My timezone is IST (GMT+5:30)
import time
#Year
time.localtime()[0]
>> 2019
#Month
time.localtime()[1]
>> 9
#Day
time.localtime()[2]
>> 18
Yes, the output is correct. Week 1 started on January 6th, as that was the first Sunday in 2019. January 1st through 5th were week 0:
>>> time.strftime('%U', time.strptime("2019-1-1", "%Y-%m-%d"))
'00'
>>> time.strftime('%U', time.strptime("2019-1-6", "%Y-%m-%d"))
'01'
This is covered in the documentation:
All days in a new year preceding the first Sunday are considered to be in week 0.
You are perhaps looking for the ISO week date, but note that in this system the first day of the week is a Monday.
You can get the week number using that system with the datetime.date.isocalendar() method, or by formatting with %V:
>>> time.strftime("%V", time.localtime())
'38'
>>> from datetime import date
>>> date.today().isocalendar() # returns ISO year, week, and weekday
(2019, 38, 2)
>>> date.today().strftime("%V")
'38'
It's correct since you start counting from the first Sunday.
%U - week number of the current year, starting with the first Sunday as the first day of the first week
https://www.tutorialspoint.com/python/time_strftime.htm
It's correct. Since all days in a new year preceding the first Sunday are considered to be in week 0 (01/01 to 01/05), this week is the week 37.
I am trying to get a date based on a number of the week, but there are some annoyances.
The date.weekday() returns the day of the week where 0 in Monday and 6 is Sunday.
The %w directive of date.strftime() and date.strptime() uses the 0 for Sunday and 6 for Saturday.
This causes some really annoying issues when trying to figure out a date given a week number from date.weekday().
Is there a better way of getting a date from a week number?
EDIT:
Added the example.
import datetime
original_date = datetime.date(2014, 8, 24)
week_of_the_date = original_date.isocalendar()[1] # 34
day_of_the_date = original_date.isocalendar()[2] # 7
temp = '{0} {1} {2}'.format(*(2014, week_of_the_date, day_of_the_date-1))
date_from_week = datetime.datetime.strptime(temp, '%Y %W %w')
week_from_new_date = date_from_week.isocalendar()[1] # 35!!
EDIT 2:
I ultimately put the date stuff in the view (using jQuery UI), it has more consistent notions of weeks.
I think the Sunday vs. Monday distinction between weekday and strftime using %W is moot - you could use isoweekday to get those to line up, or %U in strftime if you wanted Sunday as the first day of the week. The real problem is that strftime, based on the underlying C function, determines the first week of the year differently than the ISO definition. With %W the docs say: " All days in a new year preceding the first Monday are considered to be in week 0". ISO calendars count the week containing the first Thursday as the first week, for reasons I do not understand.
Two ways I found to work with ISO weeks, either just getting datetime.date instances back or supporting a variety of operations, are:
this answer with a simple timedelta approach:
What's the best way to find the inverse of datetime.isocalendar()?
this third-party library: https://pypi.python.org/pypi/isoweek/
Date is datetime.date(2013, 12, 30)
I am trying to get week number using
import datetime
datetime.date(2013, 12, 30).isocalendar()[1]
I am getting output as ,
1
Why i am not getting week number of last year , instead i am getting week number of current year?
Whats wrong i am doing here ?
You are doing nothing wrong, 2013/12/30 falls in week 1 of 2014, according to the ISO8601 week numbering standard:
The ISO 8601 definition for week 01 is the week with the year's first Thursday in it.
The Thursday in that week is 2014/01/02.
Other ways to explain the definition, from the same linked WikiPedia article:
It is the first week with a majority (four or more) of its days in January (ISO weeks start on Monday)
Its first day is the Monday nearest to 1 January.
It has 4 January in it. Hence the earliest possible dates are 29 December through 4 January, the latest 4 through 10 January.
It has the year's first working day in it, if Saturdays, Sundays and 1 January are not working days.
If you were looking for the last week number of a given year (52 or 53, depending on the year), I'd use December 28th, which is always guaranteed to be in the last week (because January 4th is always part of the first week of the next year):
def lastweeknumber(year):
return datetime.date(year, 12, 28).isocalendar()[1]
from datetime import date
from datetime import datetime
ndate='10/1/2016'
ndate = datetime.strptime(ndate, '%m/%d/%Y').strftime('%Y,%m,%d')
print('new format:',ndate)
d=ndate.split(',')
wkno = date(int(d[0]),int(d[1]),int(d[2])).isocalendar()[1]
print(wkno)
manually or read a date to a string and get the week number, play around with different formats.
The Python datetime.isocalendar() method returns a tuple (ISO_year, ISO_week_number, ISO_weekday) for the given datetime object. Is there a corresponding inverse function? If not, is there an easy way to compute a date given a year, week number and day of the week?
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)
As of Python 3.6, you can use the new %G, %u and %V directives. See issue 12006 and the updated documentation:
%G
ISO 8601 year with century representing the year that contains the greater part of the ISO week (%V).
%u
ISO 8601 weekday as a decimal number where 1 is Monday.
%V
ISO 8601 week as a decimal number with Monday as the first day of the week. Week 01 is the week containing Jan 4.
Given a string with year, weeknumber and weekday number, it is easy to parse those out to a date with:
from datetime import datetime
datetime.strptime('2002 01 1', '%G %V %u').date()
or as a function with integer inputs:
from datetime import datetime
def date_from_isoweek(iso_year, iso_weeknumber, iso_weekday):
return datetime.strptime(
'{:04d} {:02d} {:d}'.format(iso_year, iso_weeknumber, iso_weekday),
'%G %V %u').date()
import datetime
def iso_to_gregorian(iso_year, iso_week, iso_day):
"Gregorian calendar date for the given ISO year, week and day"
fourth_jan = datetime.date(iso_year, 1, 4)
_, fourth_jan_week, fourth_jan_day = fourth_jan.isocalendar()
return fourth_jan + datetime.timedelta(days=iso_day-fourth_jan_day, weeks=iso_week-fourth_jan_week)
This was adapted from #BenJames's very good answer. You don't have to know the first day of the year. You just have to know an example of a date which is certainly in the same ISO year, and the ISO calendar week and day of that date.
The 4th of Jan is simply one example, because, as Ben pointed out, the 4th of Jan always belongs to the same ISO year and Gregorian year, and is the first day of the year to do so.
Since weeks are all the same length, you can simply subtract the days and weeks between the ISO of the date you want, and the ISO of the date which you know in both forms, and add on that number of days and weeks. (It doesn't matter whether these numbers are positive or negative, so you could choose some other 'fixed day' such as Dec 28th.)
Edit
I corrected this because, as was helpfully pointed by #JoSo, the first day of the Gregorian year which also belongs to the ISO year is Jan 4th not Jan 5th. As the explanation says, it doesn't matter which date is chosen as a reference point, but choosing the Jan 4th makes this choice less 'magic'.
For the next people coming here, a shorter, single-def, version of Ben's good solution:
def iso_to_gregorian(iso_year, iso_week, iso_day):
jan4 = datetime.date(iso_year, 1, 4)
start = jan4 - datetime.timedelta(days=jan4.isoweekday()-1)
return start + datetime.timedelta(weeks=iso_week-1, days=iso_day-1)
Starting in Python 3.6, datetime.strptime() will support the %G, %V and %u directives, so you can simply do datetime.strptime('2015 1 2', '%G %V %u').date(). See: https://hg.python.org/cpython/rev/acdebfbfbdcf
Note that %W is the week # (0-53) which is NOT THE SAME as the ISO week (1-53). There will be edge cases where %W will not work.
I came up with a solution similar to the one posted by Ben James, but using a single function:
import datetime
def getDateFromWeek(year,week,day):
"""Method to retrieve the date from the specified week, year and weekday"""
year_start = datetime.date(year,1,1)
ys_weekday = year_start.weekday()
delta = (week*7)+(day-ys_weekday)
if ys_weekday<4:
delta -= 7
return year_start + datetime.timedelta(days=delta)
I tested it out with boundary values such as the last week of 2020 and first week of 2021, and it worked out pretty well.
EDIT: ignore this, the edge cases are a pain. Go with Ben's solution.
Ok, on closer inspection I noticed that strptime has %W and %w parameters, so the following works:
def fromisocalendar(y,w,d):
return datetime.strptime( "%04dW%02d-%d"%(y,w-1,d), "%YW%W-%w")
A couple of gotchas: The ISO week number starts at 1, while %W starts at 0. The ISO week day starts at 1 (Monday), which is the same as %w, so Sunday would probably have to be 0, not 7...