Most recent previous business day in Python

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Question :

Most recent previous business day in Python

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 =
 if == 5:      #if it's Saturday
     lastBusDay = lastBusDay - datetime.timedelta(days = 1) #then make it Friday
 elif == 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?

Answer #1:

Use pandas!

import datetime
# BDay is business day, not birthday...
from pandas.tseries.offsets import BDay

today =
print(today - BDay(4))

Since today is Thursday, Sept 26, that will give you an output of:

datetime.datetime(2013, 9, 20, 14, 8, 4, 89761)
Answered By: Kyle Hannon

Answer #2:

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.

Answered By: Alison R.

Answer #3:

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 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:

Answered By: lakenen

Answer #4:

Maybe this code could help:

lastBusDay =
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.

Answered By: TMC

Answer #5:

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.

Answered By: antoniobotelho

Answer #6:

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(), 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(), 9, 4)

DISCLAIMER: I am the author of timeboard.

Answered By: Maxim Mamaev

Answer #7:

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
        return previous_working_day(most_recent, holidays)

check_day =, 12, 28)

which produce:, 12, 24)
Answered By: ZdPo Ster

Answer #8:

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,
    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))
Answered By: mastier

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