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How to know if today is second or fourth Saturday


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2












$begingroup$


This is working. Is it ok?



import calendar
from datetime import datetime


def second_fourth_saturday(year):
holidays = {}
for month in range(1, 13):
holidays[month] = []
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")









share|improve this question











$endgroup$












  • $begingroup$
    Do you mean that your program takes the current day and returns whether that current day happens to be the second or fourth Saturday of the current month?
    $endgroup$
    – okcapp
    4 hours ago










  • $begingroup$
    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, inckusive. It is the third Saturday iff it is Saturday and the day is between 15 and 21, inckusive. It is the fourth Saturday iff it is Saturday and the day is between 22 and 28, inckusive.
    $endgroup$
    – Hagen von Eitzen
    55 secs ago
















2












$begingroup$


This is working. Is it ok?



import calendar
from datetime import datetime


def second_fourth_saturday(year):
holidays = {}
for month in range(1, 13):
holidays[month] = []
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")









share|improve this question











$endgroup$












  • $begingroup$
    Do you mean that your program takes the current day and returns whether that current day happens to be the second or fourth Saturday of the current month?
    $endgroup$
    – okcapp
    4 hours ago










  • $begingroup$
    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, inckusive. It is the third Saturday iff it is Saturday and the day is between 15 and 21, inckusive. It is the fourth Saturday iff it is Saturday and the day is between 22 and 28, inckusive.
    $endgroup$
    – Hagen von Eitzen
    55 secs ago














2












2








2





$begingroup$


This is working. Is it ok?



import calendar
from datetime import datetime


def second_fourth_saturday(year):
holidays = {}
for month in range(1, 13):
holidays[month] = []
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")









share|improve this question











$endgroup$




This is working. Is it ok?



import calendar
from datetime import datetime


def second_fourth_saturday(year):
holidays = {}
for month in range(1, 13):
holidays[month] = []
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")






python python-3.x






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 4 hours ago







Rahul Patel

















asked 5 hours ago









Rahul PatelRahul Patel

232413




232413












  • $begingroup$
    Do you mean that your program takes the current day and returns whether that current day happens to be the second or fourth Saturday of the current month?
    $endgroup$
    – okcapp
    4 hours ago










  • $begingroup$
    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, inckusive. It is the third Saturday iff it is Saturday and the day is between 15 and 21, inckusive. It is the fourth Saturday iff it is Saturday and the day is between 22 and 28, inckusive.
    $endgroup$
    – Hagen von Eitzen
    55 secs ago


















  • $begingroup$
    Do you mean that your program takes the current day and returns whether that current day happens to be the second or fourth Saturday of the current month?
    $endgroup$
    – okcapp
    4 hours ago










  • $begingroup$
    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, inckusive. It is the third Saturday iff it is Saturday and the day is between 15 and 21, inckusive. It is the fourth Saturday iff it is Saturday and the day is between 22 and 28, inckusive.
    $endgroup$
    – Hagen von Eitzen
    55 secs ago
















$begingroup$
Do you mean that your program takes the current day and returns whether that current day happens to be the second or fourth Saturday of the current month?
$endgroup$
– okcapp
4 hours ago




$begingroup$
Do you mean that your program takes the current day and returns whether that current day happens to be the second or fourth Saturday of the current month?
$endgroup$
– okcapp
4 hours ago












$begingroup$
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, inckusive. It is the third Saturday iff it is Saturday and the day is between 15 and 21, inckusive. It is the fourth Saturday iff it is Saturday and the day is between 22 and 28, inckusive.
$endgroup$
– Hagen von Eitzen
55 secs ago




$begingroup$
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, inckusive. It is the third Saturday iff it is Saturday and the day is between 15 and 21, inckusive. It is the fourth Saturday iff it is Saturday and the day is between 22 and 28, inckusive.
$endgroup$
– Hagen von Eitzen
55 secs ago










1 Answer
1






active

oldest

votes


















4












$begingroup$

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")





share|improve this answer









$endgroup$













  • $begingroup$
    Really useful. Thanks.
    $endgroup$
    – Rahul Patel
    1 hour ago











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

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")





share|improve this answer









$endgroup$













  • $begingroup$
    Really useful. Thanks.
    $endgroup$
    – Rahul Patel
    1 hour ago
















4












$begingroup$

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")





share|improve this answer









$endgroup$













  • $begingroup$
    Really useful. Thanks.
    $endgroup$
    – Rahul Patel
    1 hour ago














4












4








4





$begingroup$

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")





share|improve this answer









$endgroup$



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")






share|improve this answer












share|improve this answer



share|improve this answer










answered 2 hours ago









LudisposedLudisposed

8,19222161




8,19222161












  • $begingroup$
    Really useful. Thanks.
    $endgroup$
    – Rahul Patel
    1 hour ago


















  • $begingroup$
    Really useful. Thanks.
    $endgroup$
    – Rahul Patel
    1 hour ago
















$begingroup$
Really useful. Thanks.
$endgroup$
– Rahul Patel
1 hour ago




$begingroup$
Really useful. Thanks.
$endgroup$
– Rahul Patel
1 hour ago


















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