How to get last Friday?

Question

The code below should return last Friday, 16:00:00. But it returns Friday of previous week. How to fix that?

now = datetime.datetime.now() test = (now - datetime.timedelta(days=now.weekday()) + timedelta(days=4, weeks=-1)) test = test.replace(hour=16,minute=0,second=0,microsecond=0)

Upd. I use the following approach now - is it the best one?

now = datetime.datetime.now() if datetime.datetime.now().weekday() > 4: test = (now - datetime.timedelta(days=now.weekday()) + timedelta(days=4)) else: test = (now - datetime.timedelta(days=now.weekday()) + timedelta(days=4, weeks=-1)) test = test.replace(hour=16,minute=0,second=0,microsecond=0)

Upd2. Just to give an example. Let's assume that today is Oct 5, 2012. In case current time is equal to or less than 16:00 it should return Sep 28, 2012, otherwise - Oct 5, 2012.

@LA_ How do you want it to work if the current day is Friday? — bohney
– bohney, Commented Oct 2, 2012 at 8:52
@bohney, depends on the time. If 16:00 already passed, then it should be today. Otherwise - Friday of prev week. — LA_
– LA_, Commented Oct 2, 2012 at 9:03

jfs · Accepted Answer · 2016-02-25 16:58:54Z

47

The dateutil library is great for things like this:

>>> from datetime import datetime >>> from dateutil.relativedelta import relativedelta, FR >>> datetime.now() + relativedelta(weekday=FR(-1)) datetime.datetime(2012, 9, 28, 9, 42, 48, 156867)

edited Feb 25, 2016 at 16:58

jfs

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answered Oct 2, 2012 at 8:43

Jon Clements

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6 Comments

bohney Over a year ago

This solution is good. You only need to add the .replace call for the time of day.

Jon Clements Over a year ago

@LA_ That's a shame - wasn't aware that GAE was a constraint

LA_ Over a year ago

I think I can install it separately. But would prefer to do it without 3rd party libraries. Anyway, thanks for your help.

American curl Over a year ago

How can I get last Friday by this code if the current day is Friday?

eugene Over a year ago

@Americancurl you can do min(mon-1day-relativedelta(weekday=MO(-1)), mon-relativedelta(weekday=MO(-1))

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moooeeeep · Accepted Answer · 2012-10-02 12:01:46Z

As in the linked question, you need to use datetime.date objects instead of datetime.datetime. To get a datetime.datetime in the end, you can use datetime.datetime.combine():

import datetime current_time = datetime.datetime.now() # get friday, one week ago, at 16 o'clock last_friday = (current_time.date() - datetime.timedelta(days=current_time.weekday()) + datetime.timedelta(days=4, weeks=-1)) last_friday_at_16 = datetime.datetime.combine(last_friday, datetime.time(16)) # if today is also friday, and after 16 o'clock, change to the current date one_week = datetime.timedelta(weeks=1) if current_time - last_friday_at_16 >= one_week: last_friday_at_16 += one_week

Looks like your code if today is Oct 6, 2012 will return Sep 28, when Oct 5 is expected.
Thanks. I have to clearly explain what I need next time... Your code on Oct 5, 2012 will always return Sep 28, 2012. But in case current time is equal to or less than 16:00 it should return Sep 28, 2012, otherwise - Oct 5, 2012.

Bertrand Bordage · Accepted Answer · 2016-08-27 12:17:15Z

Simplest solution without dependency:

from datetime import datetime, timedelta def get_last_friday(): now = datetime.now() closest_friday = now + timedelta(days=(4 - now.weekday())) return (closest_friday if closest_friday < now else closest_friday - timedelta(days=7))

bohney · Accepted Answer · 2012-10-02 08:50:35Z

This was borrowed from Jon Clements, but is the full solution:

>>> from datetime import datetime >>> from dateutil.relativedelta import relativedelta, FR >>> lastFriday = datetime.now() + relativedelta(weekday=FR(-1)) >>> lastFriday.replace(hour=16,minute=0,second=0,microsecond=0) datetime.datetime(2012, 9, 28, 16, 0, 0, 0)

Community · Accepted Answer · 2017-05-23 12:24:34Z

The principle is the same as in your other question.

Get the friday of the current week and, if we are later, subtract one week.

import datetime from datetime import timedelta now = datetime.datetime.now() today = now.replace(hour=16,minute=0,second=0,microsecond=0) sow = (today - datetime.timedelta(days=now.weekday())) this_friday = sow + timedelta(days=4) if now > this_friday: test = this_friday else: test = this_friday + timedelta(weeks=-1)

in this answer, I suppose that from Fr. 16:00:01 on, you want the datetime of this week.

Amro Hassaan · Accepted Answer · 2014-09-04 01:21:28Z

Could be lame but to me simplest. Get the last day of the current month and start checking in a loop (which wouldn't cost anything since max loops before finding last friday is 7) for friday. if last day is not friday decrement and the check the day before.

import calendar from datetime import datetime, date def main(): year = datetime.today().year month = datetime.today().month x = calendar.monthrange(year,month) lastday = x[1] while True: z = calendar.weekday(year, month, lastday) if z != 4: lastday -= 1 else: print(date(year,month,lastday)) break if __name__ == "__main__": main()

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