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/
Related
I'm trying to write a bit of code to check if a document has been updated this week, and if not to read in the data and update it. I need to be able to check if the last modified date/time of the document occurred in this week or not (Monday-Sunday).
I know this code gives me the last modified time of the file as a float of secconds since the epoch:
os.path.getmtime('path')
And I know I can use time.ctime to get that as a string date:
time.ctime(os.path.getmtime('path'))
But I'm not sure how to check if that date was in the current week. I also don't know if its easier to convert to a datetime object rather than ctime for this?
you can use datetime.isocalendar and compare the week attribute, basicallly
import os
from datetime import datetime
t_file = datetime.fromtimestamp(os.path.getmtime(filepath))
t_now = datetime.now()
print(t_file.isocalendar().week == t_now.isocalendar().week)
# or print(t_file.isocalendar()[1]== t_now.isocalendar()[1])
# to compare the year as well, use e.g.
print(t_file.isocalendar()[:2] == t_now.isocalendar()[:2])
The ISO year consists of 52 or 53 full weeks, and where a week starts on a Monday and ends on a Sunday. The first week of an ISO year is the first (Gregorian) calendar week of a year containing a Thursday. This is called week number 1, and the ISO year of that Thursday is the same as its Gregorian year.
I have the following if statement:
if date.weekday() != 0:
How can I change this if statement to check every other Monday (one day every other week)?
Thanks in advance for your help!
You post doesn't mention what language you're using. That would be useful to know!
At any rate, a function like weekday probably returns a number from 0-6, which isn't going to give you enough info.
Most languages have a function that turns a date into a millisecond value, typically milliseconds since 1970-01-01 00:00 UTC.
If you take such a value and divide it by 86400000 (number of milliseconds in a day), you get a day number. Day 0 (January 1, 1970) happens to be a Thursday.
In JavaScript, this day number can be obtained from a Date object like this:
let dayNum = Math.floor(someDate.getTime() / 86400000)
From this, dayNum % 7 gives you a number from 0-6 for Thursday through Wednesday. You can add an offset to change what 0 means. (dayNum + 4) % 7 produces 0-6 for Sunday through Saturday.
You want a 14-day cycle for every other week, so dayNum % 14 is what you need, and you then just have to decide which value from 0 to 13 represents your target dates.
Take any sample date that is one of your desired dates, compute the remainder value you get for that date, and then test for that value to be able to match any qualifying date.
Two caveats:
For dates before 1970, the % operator might produces negative numbers. Python is one of the few languages that doesn't do this.
The timezone offset from UTC might be important to your task. And if that's important, Daylight Saving Time changes and/or historical changes in UTC offset can complicate matters.
Python update:
Let's say this coming Monday (July 5, 2021) is one of the every-other-week days that you want. This:
date(2021, 7, 5).toordinal() % 14
...produces a value of 8. Every other Monday before or after will also be 8. The toordinal() method goes straight to a day number, without having to fuss with milliseconds, 86400000, or having to worry about timezone offsets.
Recently I was bothered with the function that would return starting day of the week for a given string concatenation of year and week and surprisingly couldn't find it anywhere.
I saw similar ones (ie. question1, question2), but I am only given the year and week, without days.
Let's have an example input and a desired output:
func('202005') -> Monday 27.01.2020
I found a lot of counting, but not anything elegant, a smooth one-liner. Is there any?
During investigation I combined some similar approaches with python docs regading datetime, so leaving my proposal here that I run with for any bypassers:
# python3.7
import datetime
def get_week_start_date(yearweek: str) -> datetime:
"""
Return date of starting the week (Monday)
i.e: 201944 returns datetime.datetime(2019, 11, 4, 0, 0)
%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.
Arguments:
yearweek {str} -- year to be calculated
Returns:
datetime -- Date of starting the week in passed year, week
"""
return datetime.datetime.strptime(f"{yearweek}-1", "%G%V-%u")
time.strptime accepts %U in format string, which mean 2-digit number of week, however it requires day of week to work and starts counting weeks at 0, that is
import time
yearweek = '202005'
yearweekmonday = yearweek+'1'
print(time.strptime(yearweekmonday, '%Y%U%w'))
Output: time.struct_time(tm_year=2020, tm_mon=2, tm_mday=3, tm_hour=0, tm_min=0, tm_sec=0, tm_wday=0, tm_yday=34, tm_isdst=-1)
As weeks numbers starts with 0, 202005 result in 2020-02-03 rather than 2020-01-27, but after you conver time.struct_time to datatime object from that you should be able to easily manipulate using timedelta.
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.
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.