How would one go about finding the date of the next Saturday in Python? Preferably using datetime
and in the format '2013-05-25'?

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1Google around a little before you post here. Check this out. See if its useful: http://stackoverflow.com/questions/2436840/how-to-calculate-next-friday-at-3am – Anil May 27 '13 at 09:16
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Thanks, saw that before but I don't need the time and was wondering whether there was a more concise way. – akkatracker May 27 '13 at 09:18
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2This link helps, my apologies for wasting time. If anyone has a better solution feel free to answer. http://stackoverflow.com/questions/8801084/how-to-calculate-next-friday-in-python – akkatracker May 27 '13 at 09:21
6 Answers
>>> from datetime import datetime, timedelta
>>> d = datetime.strptime('2013-05-27', '%Y-%m-%d') # Monday
>>> t = timedelta((12 - d.weekday()) % 7)
>>> d + t
datetime.datetime(2013, 6, 1, 0, 0)
>>> (d + t).strftime('%Y-%m-%d')
'2013-06-01'
I use (12 - d.weekday()) % 7
to compute the delta in days between given day and next Saturday because weekday
is between 0 (Monday) and 6 (Sunday), so Saturday is 5. But:
- 5 and 12 are the same modulo 7 (yes, we have 7 days in a week :-) )
- so
12 - d.weekday()
is between 6 and 12 where5 - d.weekday()
would be between 5 and -1 - so this allows me not to handle the negative case (-1 for Sunday).
Here is a very simple version (no check) for any weekday:
>>> def get_next_weekday(startdate, weekday):
"""
@startdate: given date, in format '2013-05-25'
@weekday: week day as a integer, between 0 (Monday) to 6 (Sunday)
"""
d = datetime.strptime(startdate, '%Y-%m-%d')
t = timedelta((7 + weekday - d.weekday()) % 7)
return (d + t).strftime('%Y-%m-%d')
>>> get_next_weekday('2013-05-27', 5) # 5 = Saturday
'2013-06-01'

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2There's no need to add the modulo base (7) to avoid negative numbers. ["The modulo operator always yields a result with the same sign as its second operand."](https://docs.python.org/3/reference/expressions.html#:~:text=the%20modulo%20operator%20always%20yields%20a%20result%20with%20the%20same%20sign%20as%20its%20second%20operand) – Mattwmaster58 Dec 30 '21 at 19:17
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1
I found this pendulum pretty useful. Just one line
In [4]: pendulum.now().next(pendulum.SATURDAY).strftime('%Y-%m-%d')
Out[4]: '2019-04-27'
See below for more details:
In [1]: import pendulum
In [2]: pendulum.now()
Out[2]: DateTime(2019, 4, 24, 17, 28, 13, 776007, tzinfo=Timezone('America/Los_Angeles'))
In [3]: pendulum.now().next(pendulum.SATURDAY)
Out[3]: DateTime(2019, 4, 27, 0, 0, 0, tzinfo=Timezone('America/Los_Angeles'))
In [4]: pendulum.now().next(pendulum.SATURDAY).strftime('%Y-%m-%d')
Out[4]: '2019-04-27'

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This is best. There are many SO posts like this and its a bog of confusing implementations. – Merlin Jan 25 '23 at 19:37
You need two main packages,
import datetime
import calendar
Once you have those, you can simply get the desired date for the week by the following code,
today = datetime.date.today() #reference point.
saturday = today + datetime.timedelta((calendar.SATURDAY-today.weekday()) % 7 )
saturday
Bonus Following the content, if you type
saturday.weekday()
It will result 5
.
Therefore, you can also use 5
in place of calendar.SATURDAY
and you will get same result.
saturday = today + datetime.timedelta((5-today.weekday()) % 7 )

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if you just want the date from today (inspired from Emanuelle )
def get_next_weekday(weekday_number):
"""
@weekday: week day as a integer, between 0 (Monday) to 6 (Sunday)
"""
assert 0 <= weekday_number <= 6
today_date = datetime.today()
next_week_day = timedelta((7 + weekday_number - today_date.weekday()) % 7)
return (today_date + next_week_day).strftime('%d/%m/%Y')

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Again, based on Emmanuel's example, but making 0-6 conform to your week:
ScheduleShift = -1 # make Saturday end of week
EndofWeekDay = lambda do : do + datetime.timedelta( ( ScheduleShift + (13 - do.weekday() ) %7 ) )
Which can be called with:
EndofWeekDay( datetime.date.today() )
returning a datetime.date object

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Just wanted to share a code. With this, you will get a list of dates when Saturday days will be in the next 10 days.
from datetime import datetime, timedelta
target_day = 'Saturday'
now_ = datetime.today().date()
how_many_days = 10
next_date = [now_ + timedelta(days=x) for x in range(how_many_days) if (now_ + timedelta(days=x)).strftime("%A") == target_day]
print(next_date)

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