This is working. Is it ok?

import calendar

from datetime import datetime





def second_fourth_saturday(year):

    holidays = {}

    for month in range(1, 13):

        cal = calendar.monthcalendar(year, month)

        if cal[0][calendar.SATURDAY]:

            holidays[month] = (

                cal[1][calendar.SATURDAY],

                cal[3][calendar.SATURDAY]

            )

        else:

            holidays[month] = (

                cal[2][calendar.SATURDAY],

                cal[4][calendar.SATURDAY]

            )

    return holidays





if __name__ == "__main__":

    today = datetime.today().day

    tomonth = datetime.today().month

    if today in second_fourth_saturday(2019)[tomonth]:

        print("Enjoy")

    else:

        print("Start Working")

A few suggestions

• 
  

  collections.defaultdict

  
  
  

  
  

  holidays = {}
  
  for month in range(1, 13):
  
      holidays[month] = 
  
      ...
  
  

  
  

  
  
  

  There is a module for dictionaries starting with a basic datatype

  
  
  

  from collections import defaultdict
  
  holidays = defaultdict(tuple)
  
  

  
  
  

  Secondly you first init it as a list, and afterwards you make it a tuple this is odd. Pick one and stick with it

  
  


• 
  

  You don't have to calculate all the months only the specific month

  
  
  

  Since you already know which month it is now, just calculate only the month you are interested in

  
  
  

  To do this you would have to give the month as a second parameter

  
  
  

  def second_fourth_saturday(year, month):
  
      cal = calendar.monthcalendar(year, month)
  
      ...
  
  

  
  


• 
  

  Don't Repeat Yourself

  
  
  

  
  

  if cal[0][calendar.SATURDAY]:
  
      holidays[month] = (
  
          cal[1][calendar.SATURDAY],
  
          cal[3][calendar.SATURDAY]
  
      )
  
  else:
  
      holidays[month] = (
  
          cal[2][calendar.SATURDAY],
  
          cal[4][calendar.SATURDAY]
  
      )
  
  

  
  

  
  
  

  If you calculate the weeks beforehand you don't have to repeat yourself

  
  
  

  second_fourth_saturday = (1, 3) if cal[0][calendar.SATURDAY] else (2, 4)
  
  

  
  


• 
  

  Return what is asked

  
  
  

  Instead of return a dict of month with second/fourth saturdays, I think it would be more clear if the function returns a boolean value if the day is a second or fourth saturday

  
  

Code

from calendar import monthcalendar, SATURDAY

from datetime import datetime



def second_fourth_saturday(date):

    month_calender = monthcalendar(date.year, date.month)

    second_fourth_saturday = (1, 3) if month_calender[0][SATURDAY] else (2, 4)

    return any(date.day == month_calender[i][SATURDAY] for i in second_fourth_saturday)



if __name__ == "__main__":

    is_second_fourth_saturday = second_fourth_saturday(datetime.today())

    print("Enjoy" if is_second_fourth_saturday else "Start working")

As suggested by @HagenvonEitzen in the comments:

Today is the first Saturday of the current month iff it is Saturday
and the day is between 1 and 7, inclusive. It is the second Saturday
iff it is Saturday and the day is between 8 and 14, inc[l]usive. It is
the third Saturday iff it is Saturday and the day is between 15 and
21, inc[l]usive. It is the fourth Saturday iff it is Saturday and the
day is between 22 and 28, inc[l]usive.

So, just test against that:

from calendar import SATURDAY



def second_fourth_saturday(date):

    if date.weekday() != SATURDAY:

        return False

    day = date.day

    return day in range(8, 14 + 1) or day in range(22, 28 + 1)

This uses the fact that in tests for range are constant time in Python 3 (in Python 2 they were not).

You could also hardcode those numbers:

POSSIBLE_DAYS = set(range(8, 14 + 1)) | set(range(22, 28 + 1))



def second_fourth_saturday(date):

    return date.weekday() == SATURDAY and date.day in POSSIBLE_DAYS