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How to know if today is second or fourth Saturday
Trying to use global variables for referencing directories in Python 2.5Finding the date from “2nd Friday of X month”-style inputTDD Hackerrank: Library FinesOrdering an un-ambiguous scrambled dateStart and end times of recurring eventsReading a repetitive file with repetitive codePython library for calculating next and previous time of the cron-like scheduled taskWeb scraping news articles with Beautiful SoupCleaning up date strings in PythonProgram to check if a date is valid or not
$begingroup$
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")
python python-3.x
asked 5 hours ago
Rahul PatelRahul Patel
232413
$endgroup$
add a comment |
$begingroup$
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")
python python-3.x
asked 5 hours ago
Rahul PatelRahul Patel
232413
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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")
python python-3.x
asked 5 hours ago
Rahul PatelRahul Patel
232413
$endgroup$
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
python python-3.x
asked 5 hours ago
Rahul PatelRahul Patel
232413
asked 5 hours ago
Rahul PatelRahul Patel
232413
edited 4 hours ago
Rahul Patel
asked 5 hours ago
Rahul PatelRahul Patel
232413
asked 5 hours ago
Rahul PatelRahul Patel
232413
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
add a comment |
$begingroup$
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
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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")
answered 2 hours ago
LudisposedLudisposed
8,19222161
$endgroup$
$begingroup$
Really useful. Thanks.
$endgroup$
– Rahul Patel
1 hour ago
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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")
answered 2 hours ago
LudisposedLudisposed
8,19222161
$endgroup$
$begingroup$
Really useful. Thanks.
$endgroup$
– Rahul Patel
1 hour ago
add a comment |
$begingroup$
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")
answered 2 hours ago
LudisposedLudisposed
8,19222161
$endgroup$
$begingroup$
Really useful. Thanks.
$endgroup$
– Rahul Patel
1 hour ago
add a comment |
$begingroup$
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")
answered 2 hours ago
LudisposedLudisposed
8,19222161
$endgroup$
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")
answered 2 hours ago
LudisposedLudisposed
8,19222161
answered 2 hours ago
LudisposedLudisposed
8,19222161
answered 2 hours ago
LudisposedLudisposed
8,19222161
answered 2 hours ago
LudisposedLudisposed
8,19222161
8,19222161
$begingroup$
Really useful. Thanks.
$endgroup$
– Rahul Patel
1 hour ago
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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