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Dec 14th, 2018
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Python 0.77 KB | None | 0 0
  1. weekdays = ["man", "tir", "ons", "tor", "fre", "lør", "søn"]
  2.  
  3. def ily(year): # ily = is_leap_year()
  4.     return True if year % 400 == 0 else False if year % 100 == 0 else True if year % 4 == 0 else False
  5.  
  6. def weekday_newyear(year):
  7.     if year == 1900:
  8.         return 0
  9.    
  10.     else:
  11.         if ily(year):
  12.             return ((year - 1900) + (year - 1900) // 4 - 1) % 7
  13.  
  14.         else:
  15.             return ((year - 1900) + (year - 1900) // 4) % 7
  16.            
  17. def is_workday(weekday):
  18.     return False if weekday in (5, 6) else True
  19.  
  20. def workday_in_year(year):
  21.     weekday = weekday_newyear(year)
  22.    
  23.     if not ily(year):
  24.         return 261 if is_workday(weekday) else 260
  25.    
  26.     else:
  27.         return 262 if weekday in range(4) else 261 if weekday == 4 else 260
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