Get the last thursday of the current month using python

You should pass 2 to TH instead of 1, as 1 doesn’t change anything. Modify your code to:

while (nthu + relativedelta(weekday=TH(2))).month == cmon:
    nthu += relativedelta(weekday=TH(2))

print nthu.strftime('%d-%b-%Y').upper()
# prints 26-MAY-2016

Note that I modified the loop’s condition in order to break in the last occurrence of the day on the month, otherwise it’ll break in the next month (in this case, June).

from datetime import datetime , timedelta

todayte = datetime.today()
cmon = todayte.month

nthu = todayte
while todayte.month == cmon:
    todayte += timedelta(days=1)
    if todayte.weekday()==3: #this is Thursday 
        nthu = todayte
print nthu

You can also use calendar package.
Access the calendar in the form of monthcalendar. and notice that, the Friday is the last day of a week.

import calendar
import datetime
now = datetime.datetime.now()
last_sunday = max(week[-1] for week in calendar.monthcalendar(now.year,
                                                              now.month))
print('{}-{}-{:2}'.format(now.year, calendar.month_abbr[now.month],
                              last_sunday))

I think this will be fastest perhaps:

end_of_month = datetime.datetime.today() + relativedelta(day=31)
last_thursday = end_of_month + relativedelta(weekday=TH(-1))

Based on Adam Smith’s answer on How can I get the 3rd Friday of a month in Python?, you can get the date of the last Thursday of the current month as follows:

import calendar
import datetime

def get_thursday(cal,year,month,thursday_number):
    '''
    For example, get_thursday(cal, 2017,8,0) returns (2017,8,3) 
    because the first thursday of August 2017 is 2017-08-03
    '''
    monthcal = cal.monthdatescalendar(year, month)
    selected_thursday = [day for week in monthcal for day in week if 
                    day.weekday() == calendar.THURSDAY and 
                    day.month == month][thursday_number]
    return selected_thursday

def main():
    '''
    Show the use of get_thursday()
    '''
    cal = calendar.Calendar(firstweekday=calendar.MONDAY)
    today = datetime.datetime.today()
    year = today.year
    month = today.month
    date = get_thursday(cal,year,month,-1) # -1 because we want the last Thursday 
    print('date: {0}'.format(date)) # date: 2017-08-31        

if __name__ == "__main__":
    main()

This code can be used in python 3.x for finding last Thursday of current month.

import datetime
dt = datetime.datetime.today()
def lastThurs(dt):
    currDate, currMth, currYr = dt, dt.month, dt.year
    for i in range(31):
        if currDate.month == currMth and currDate.year == currYr and currDate.weekday() == 3:
            #print('dt:'+ str(currDate))
            lastThuDate = currDate
        currDate += datetime.timedelta(1)

    return lastThuDate

import datetime
def get_thursday(_month,_year):
    for _i in range(1,32):
        if _i > 9:
            _dateStr = str(_i)
        else:
            _dateStr = '0' + str(_i)
        _date = str(_year) + '-' + str(_month) + '-' + _dateStr
        try:
            a = datetime.datetime.strptime(_date, "%Y-%m-%d").strftime('%a')
        except:
             continue
        if a == 'Thu':
            _lastThurs = _date
    return _lastThurs

x = get_thursday('05','2017')
print(x)

You can do something like this:

import pandas as pd
from dateutil.relativedelta import relativedelta, TH

expiry_type = 0
today = pd.datetime.today()
expiry_dates = []

if expiry_type == 0:
    # Weekly expiry
    for i in range(1,13):
         expiry_dates.append((today + relativedelta(weekday=TH(i))).date())
else:
    # Monthly expiry
    for i in range(1,13):
        x = (today + relativedelta(weekday=TH(i))).date()
        y = (today + relativedelta(weekday=TH(i+1))).date()
        if x.month != y.month :
            if x.day > y.day :
                expiry_dates.append(x)

print(expiry_dates)

It is straightforward fast and easy to understand, we take the first of every month and then subtract it by 3 coz 3 is Thursday weekday number and then multiply it by either 4 and check if it is in the same month if it is then that is last Thursday otherwise we multiply it with 3 and we get our last Thursday

import datetime as dt

from datetime import timedelta

#start is the first of every month

start = dt.datetime.fromisoformat('2022-08-01')

if start.month == (start + timedelta((3 - start.weekday()) + 4*7)).month:

  exp = start + timedelta((3 - start.weekday()) + 4*7)

else:

  exp = start + timedelta((3 - start.weekday()) + 3*7)