我怎样才能获得两天之间的所有日子？

7 回答

• 2

  >>> def weekdays_between(s, e):
  ...     return [n % 7 for n in range(s, e + (1 if e > s else 8))]
  ... 
  >>> weekdays_between(2, 4)
  [2, 3, 4]
  >>> weekdays_between(5, 1)
  [5, 6, 0, 1]
  

  如果你必须从/转换到实际天数，它会更复杂一些 .

  >>> days = 'Mon Tue Wed Thu Fri Sat Sun'.split()
  >>> days_1 = {d: n for n, d in enumerate(days)}
  >>> def weekdays_between(s, e): 
  ...     s, e = days_1[s], days_1[e]
  ...     return [days[n % 7] for n in range(s, e + (1 if e > s else 8))]
  ... 
  >>> weekdays_between('Wed', 'Fri')
  ['Wed', 'Thu', 'Fri']
  >>> weekdays_between('Sat', 'Tue')
  ['Sat', 'Sun', 'Mon', 'Tue']
  

  回复于 2024-05-05T17:03:17+08:00
• 0

  怎么样（在伪代码中）：

  weekday[] = {"Mon" .. "Sun"}
  for(i = wkday_start; (i % 7) != wkday_end; i = (i+1) % 7)
      printf("%s ", weekday[i]);
  

  它的工作方式类似于循环缓冲区，wkday_start是从（从0开始）开始的索引，wkday_end是结束索引 .

  希望这可以帮助

  乔治 .

  回复于 2024-05-05T17:03:17+08:00
• 0

  在excellent answer from Stephan202的基础上，您可以概括圆形切片的概念 .

  >>> def circular_slice(r, s, e):
  ... return [r[n % len(r)] for n in range(s, e + (1 if e>s else len(r)+1))]
  ...
  >>> circular_slice(range(0,7), 2, 4)
  [2, 3, 4]
  >>> circular_slice(range(0,7), 5, 1)
  [5, 6, 0, 1]
  >>> circular_slice('Mon Tue Wed Thu Fri Sat Sun'.split(), 5, 1)
  ['Sat', 'Sun', 'Mon', 'Tue']
  

  回复于 2024-05-05T17:03:17+08:00
• 1

  提供的解决方案已经回答了问题，但我想提出一些额外的建议 . 我不知道你在做什么，但也许你想要实际日期呢？

  >>> from datetime import timedelta, date
  >>> from dateutil.rrule import rrule, DAILY
  >>> today = date(2009, 10, 13) # A tuesday
  >>> week = today - timedelta(days=6)
  >>> list(rrule(DAILY, byweekday=xrange(5), dtstart=week, until=today))
  [datetime.datetime(2009, 10, 7, 0, 0),
   datetime.datetime(2009, 10, 8, 0, 0),
   datetime.datetime(2009, 10, 9, 0, 0),
   datetime.datetime(2009, 10, 12, 0, 0),
   datetime.datetime(2009, 10, 13, 0, 0)]
  

  这使用了优秀的python-dateutil模块 .

  回复于 2024-05-05T17:03:17+08:00
• 1

  使用日历模块作为日期名称列表：

  import calendar
  
  def intervening_days(day1, day2):
      weektest = list(calendar.day_name)*2
      d1 = weektest.index(day1)
      d2 = weektest.index(day2,d1+1)
      return weektest[d1:d2+1]
  
  print intervening_days("Monday","Sunday")
  print intervening_days("Monday","Tuesday")
  print intervening_days("Thursday","Tuesday")
  print intervening_days("Monday","Monday")
  

  打印：

  ['Monday', 'Tuesday', 'Wednesday', 'Thursday', 'Friday', 'Saturday', 'Sunday']
  ['Monday', 'Tuesday']
  ['Thursday', 'Friday', 'Saturday', 'Sunday', 'Monday', 'Tuesday']
  ['Monday', 'Tuesday', 'Wednesday', 'Thursday', 'Friday', 'Saturday', 'Sunday', 'Monday']
  

  如果您不希望星期一到星期一返回整整一周的天数，请将d2的确定更改为 d2 = weektest.index(day2,d1) .

  回复于 2024-05-05T17:03:17+08:00
• 9

  你问了一个算法，我明白这应该是语言独立的;但是，以下代码使用C＃和LINQ表达式：

  DayOfWeek start = DayOfWeek.Wednesday;
  DayOfWeek end = DayOfWeek.Friday;
  
  IEnumerable<DayOfWeek> interval = 
      Enum.GetValues(typeof(DayOfWeek)).OfType<DayOfWeek>()
          .Where(d => d >= start && d <= end);
  
  Console.WriteLine(
      String.Join(", ", 
          interval.Select(d => d.ToString()).ToArray()));
  

  也许，使用任何语言，您应该将值归因于每一天（ Sunday=0 等）并查找与您所需的间隔匹配的所有值 .

  回复于 2024-05-05T17:03:17+08:00
• 8

  以下代码在星期一 - 星期一返回1 .

  bool isWeekday(int d) {
      return d >= 1 && d <= 5;
  }
  
  int f(int d1, int d2) {
      int res = isWeekday(d1) ? 1 : 0;
      return d1 == d2 ?
             res :
             res + f(d1 % 7 + 1, d2);
  }
  

  回复于 2024-05-05T17:03:17+08:00