In Python, given a date, how do I find the preceding weekday? (Weekdays are Mon to Fri. I don't care about holidays)
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4 Answers
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Simply subtract a day from the given date, then check if the date is a weekday. If not, subtract another, until you do have a weekday:
from datetime import date, timedelta
def prev_weekday(adate):
adate -= timedelta(days=1)
while adate.weekday() > 4: # Mon-Fri are 0-4
adate -= timedelta(days=1)
return adate
Demo:
>>> prev_weekday(date.today())
datetime.date(2012, 8, 20)
>>> prev_weekday(date(2012, 8, 20))
datetime.date(2012, 8, 17)
Alternatively, use an offset table; no need to make this a mapping, a tuple will do just fine:
_offsets = (3, 1, 1, 1, 1, 1, 2)
def prev_weekday(adate):
return adate - timedelta(days=_offsets[adate.weekday()])

Martijn Pieters
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or just `return adate - timedelta(days=[3, 1, 1, 1, 1, 1, 2][adate.weekday()])` – Bruno Vermeulen Jul 05 '20 at 09:40
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@BrunoVermeulen: yes, that's what the code in the last example does. – Martijn Pieters Jul 05 '20 at 12:24
2
This is an old question, but anyway, a simpler way to do that, which doesn't require loops
from datetime import datetime, timedelta
today = datetime.today() # Today date
weekday = 6
days = today.isoweekday() - weekday
if days<0:
days += 7
previous_date = today - timedelta(days=days)
print(previous_date)

Andres Perez
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1
See the datetime module, in particular the date()
and weekday()
function. For example:
import datetime
temp = datetime.date(2012,8,21) # Today's date: 2012/08/21
print temp.weekday()
This will output 1
. 0
stands for Monday, 1
for Tuesday, etc., until 6
for Sunday. Finding the preceding weekday is easy from here.

HerrKaputt
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1
in datetime module you can do something like this: a = date.today() - timedelta(days=1)
and then a.weekday()
. Where monday is 0 and sunday is 6.

Swair
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