Most recent previous business day in Python
PythonDatetimePython Problem Overview
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?
Python Solutions
Solution 1 - Python
Use pandas!
import datetime
# BDay is business day, not birthday...
from pandas.tseries.offsets import BDay
today = datetime.datetime.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)
Solution 2 - 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()
Solution 3 - Python
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.
Solution 4 - Python
There seem to be several options if you're open to installing extra libraries.
This post describes a way of defining workdays with dateutil.
http://coding.derkeiler.com/Archive/Python/comp.lang.python/2004-09/3758.html
BusinessHours lets you custom-define your list of holidays, etc., to define when your working hours (and by extension working days) are.
Solution 5 - Python
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.
Solution 6 - Python
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)
Solution 7 - Python
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.
Solution 8 - Python
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))
Solution 9 - Python
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')
Solution 10 - Python
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
Solution 11 - Python
Why don't you try something like:
lastBusDay = datetime.datetime.today()
if datetime.date.weekday(lastBusDay) not in range(0,5):
lastBusDay = 5
Solution 12 - Python
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 13 - Python
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.