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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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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.
To convert Python datetime to string, use the strftime() function. The strftime() method is a built-in Python method that returns the string representing date and time using date, time, or datetime object.
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()]) If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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