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, Oct 02 '12 at 08:52
@bohney, depends on the time. If 16:00 already passed, then it should be today. Otherwise - Friday of prev week. — LA_, Oct 02 '12 at 09:03

score 44 · Answer 1 · edited Feb 25 '16 at 16:58

44

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 '16 at 16:58

jfs

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answered Oct 02 '12 at 08:43

Jon Clements

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This solution is good. You only need to add the .replace call for the time of day. – bohney Oct 02 '12 at 08:47
@LA_ That's a shame - wasn't aware that GAE was a constraint – Jon Clements Oct 02 '12 at 11:25
I think I can install it separately. But would prefer to do it without 3rd party libraries. Anyway, thanks for your help. – LA_ Oct 02 '12 at 11:37
2

How can I get last Friday by this code if the current day is Friday? – American curl Feb 02 '18 at 02:16
@Americancurl you can do min(mon-1day-relativedelta(weekday=MO(-1)), mon-relativedelta(weekday=MO(-1)) – eugene Jan 17 '20 at 15:18
datetime.now() + relativedelta(days=-1, weekday=FR(-1)) will get you last Friday even if the current date is Friday. – Sibs Dec 21 '20 at 20:35

moooeeeep · Accepted Answer · 2012-10-02T12:01:46.463

17

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

edited Oct 02 '12 at 12:01

answered Oct 02 '12 at 08:43

moooeeeep

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Looks like your code if today is Oct 6, 2012 will return Sep 28, when Oct 5 is expected. – LA_ Oct 02 '12 at 08:47
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. – LA_ Oct 02 '12 at 10:20

score 4 · Answer 3 · answered Oct 02 '12 at 08:50

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)

score 4 · Answer 4 · answered Aug 27 '16 at 12:17

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))

score 1 · Answer 5 · edited May 23 '17 at 12:24

1

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)

edited May 23 '17 at 12:24

Community

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answered Oct 02 '12 at 08:50

glglgl

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in this answer, I suppose that from Fr. 16:00:01 on, you want the datetime of this week. – glglgl Oct 02 '12 at 08:51

score 0 · Answer 6 · answered Sep 04 '14 at 01:21

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()

How to get last Friday?

6 Answers6

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