Most Recent Previous Business Day in Python

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()

Maybe this code could help:

lastBusDay = datetime.datetime.today()
shift = datetime.timedelta(max(1,(lastBusDay.weekday() + 6) % 7 - 3))
lastBusDay = lastBusDay - shift

The idea is that on Mondays yo have to go back 3 days, on Sundays 2, and 1 in any other day.

The statement (lastBusDay.weekday() + 6) % 7 just re-bases the Monday from 0 to 6.

Really don't know if this will be better in terms of performance.

There seem to be several options if you're open to installing extra libraries.

This post describes a way of defining workdays with dateutil.

BusinessHours lets you custom-define your list of holidays, etc., to define when your working hours (and by extension working days) are.

DISCLAMER: I'm the author...

I wrote a package that does exactly this, business dates calculations. You can use custom week specification and holidays.

I had this exact problem while working with financial data and didn't find any of the available solutions particularly easy, so I wrote one.

Hope this is useful for other people.

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)

timeboard package does this.

Suppose your date is 04 Sep 2017. In spite of being a Monday, it was a holiday in the US (the Labor Day). So, the most recent business day was Friday, Sep 1.

>>> import timeboard.calendars.US as US
>>> clnd = US.Weekly8x5()
>>> clnd('04 Sep 2017').rollback().to_timestamp().date()
datetime.date(2017, 9, 1)

In UK, 04 Sep 2017 was the regular business day, so the most recent business day was itself.

>>> import timeboard.calendars.UK as UK
>>> clnd = UK.Weekly8x5()
>>> clnd('04 Sep 2017').rollback().to_timestamp().date()
datetime.date(2017, 9, 4)

DISCLAIMER: I am the author of timeboard.

This will give a generator of working days, of course without holidays, stop is datetime.datetime object. If you need holidays just make additional argument with list of holidays and check with 'IFology' ;-)

def workingdays(stop, start=datetime.date.today()):
    while start != stop:
        if start.weekday() < 5:
            yield start
        start += datetime.timedelta(1)

Later on you can count them like

workdays = workingdays(datetime.datetime(2015, 8, 8))
len(list(workdays))

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')

 def getNthBusinessDay(startDate, businessDaysInBetween):
    currentDate = startDate
    daysToAdd = businessDaysInBetween
    while daysToAdd > 0:
        currentDate += relativedelta(days=1)
        day = currentDate.weekday()
        if day < 5:
            daysToAdd -= 1

    return currentDate 

Why don't you try something like:

lastBusDay = datetime.datetime.today()
if datetime.date.weekday(lastBusDay) not in range(0,5):
    lastBusDay = 5

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

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.