Most recent previous business day in Python

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I need to subtract business days from the current date.

I currently have some code which needs always to be running on the most recent business day. So that may be today if we're Monday thru Friday, but if it's Saturday or Sunday then I need to set it back to the Friday before the weekend. I currently have some pretty clunky code to do this:

 lastBusDay = datetime.datetime.today()
 if datetime.date.weekday(lastBusDay) == 5:      #if it's Saturday
     lastBusDay = lastBusDay - datetime.timedelta(days = 1) #then make it Friday
 elif datetime.date.weekday(lastBusDay) == 6:      #if it's Sunday
     lastBusDay = lastBusDay - datetime.timedelta(days = 2); #then make it Friday

Is there a better way?

Can I tell timedelta to work in weekdays rather than calendar days for example?

datetime python
13 Answers

If you want to skip US holidays as well as weekends, this worked for me (using pandas 0.23.3):

import pandas as pd
from pandas.tseries.holiday import USFederalHolidayCalendar
from pandas.tseries.offsets import CustomBusinessDay
US_BUSINESS_DAY = CustomBusinessDay(calendar=USFederalHolidayCalendar())
july_5 = pd.datetime(2018, 7, 5)
result = july_5 - 2 * US_BUSINESS_DAY # 2018-7-2

To convert to a python date object I did this:

result.to_pydatetime().date()
user_4949540
user_598700

If somebody is looking for solution respecting holidays (without any huge library like pandas), try this function:

import holidays
import datetime


def previous_working_day(check_day_, holidays=holidays.US()):
    offset = max(1, (check_day_.weekday() + 6) % 7 - 3)
    most_recent = check_day_ - datetime.timedelta(offset)
    if most_recent not in holidays:
        return most_recent
    else:
        return previous_working_day(most_recent, holidays)

check_day = datetime.date(2020, 12, 28)
previous_working_day(check_day)

which produces:

datetime.date(2020, 12, 24)
user_94246170

For the pandas usecase, I found the following to be quite useful and compact, although not completely readable:

Get most recent previous business day:

In [2]: datetime.datetime(2019, 11, 30) + BDay(1) - BDay(1)  # Saturday
Out[2]: Timestamp('2019-11-29 00:00:00')


In [3]: datetime.datetime(2019, 11, 29) + BDay(1) - BDay(1)  # Friday
Out[3]: Timestamp('2019-11-29 00:00:00')

In the other direction, simply use:

In [4]: datetime.datetime(2019, 11, 30) + BDay(0)  # Saturday
Out[4]: Timestamp('2019-12-02 00:00:00')

In [5]: datetime.datetime(2019, 11, 29) + BDay(0)  # Friday
Out[5]: Timestamp('2019-11-29 00:00:00')
user_8534420

another simplify version

lastBusDay = datetime.datetime.today()
wk_day = datetime.date.weekday(lastBusDay)
if wk_day > 4:      #if it's Saturday or Sunday
    lastBusDay = lastBusDay - datetime.timedelta(days = wk_day-4) #then make it Friday
user_39699420

Solution irrespective of different jurisdictions having different holidays:

If you need to find the right id within a table, you can use this snippet. The Table model is a sqlalchemy model and the dates to search from are in the field day.

def last_relevant_date(db: Session, given_date: date) -> int:
    available_days = (db.query(Table.id, Table.day)
                      .order_by(desc(Table.day))
                      .limit(100).all())
    close_dates = pd.DataFrame(available_days)
    close_dates['delta'] = close_dates['day'] - given_date
    past_dates = (close_dates
                  .loc[close_dates['delta'] < pd.Timedelta(0, unit='d')])
    table_id = int(past_dates.loc[past_dates['delta'].idxmax()]['id'])
    return table_id

This is not a solution that I would recommend when you have to convert in bulk. It is rather generic and expensive as you are not using joins. Moreover, it assumes that you have a relevant day that is one of the 100 most recent days in the model Table. So it tackles data input that may have different dates.

user_74924830
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