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检查月份的日期是否在两个月之间

python-datetime python

-1

Hani Gotc · 技术社区 · 7 年前

(month,day) 在两个月之间 不管是哪一年 。我在建模方面遇到了问题 .我试着在13岁之前代表1月,但我被卡住了。

返回值

3. =6月25日至8月22日
2.
=9月1日至3月31日或11月1日至12月16日或1月2日至3月31日
0

源代码

from datetime import date, datetime, timedelta

def get_season(date):
        month = date.month
        day   = date.day
        
        if month == 1:
            month = 13
        
        if (6,25) <= (month, day) <= (8,22):
            return 3
        
        elif (4, 1) <= (month, day) <= (6,24) or (8, 22) < (month, day) <= (10, 31) or (12,17) <= (month, day) <= (13,1):
            return 2
        
        elif  (9, 1) <= (month, day) <= (3, 31) or  (11,1) <= (month, day) <= (12, 16) or (13,2) <= (month, day) <= (3, 31):
            return 1
        else:
            return 0
        
if __name__ == "__main__":
    
    date = datetime.strptime("2016-02-01",'%Y-%m-%d').date()
    
    if get_season(date) == 0:
      print "WRONG DOES NOT EXIST THERE!!!!!!!!!!!!!!"

1 回复 | 直到 4 年前

Adam Smith jeffpkamp 7 年前

datetime.date 对象,并将其作为一年重新写入。

from datetime import date

def get_season(d):

    d = date(year=1900, month=d.month, day=d.day)

    if date(1900, 6, 25) <= d <= date(1900, 8, 22):
        return 3

    elif date(1900, 4, 1) <= d <= date(1900, 6, 24) or \
         date(1900, 8, 22) <= d <= date(1900, 10, 31) or \
         date(1900, 12, 17) <= d <= date(1900 12, 31) or \
         date(1900, 1, 1) == d:
        return 2

    elif date(1900, 9, 1) <= d <= date(1900, 3, 31) or \
         date(1900, 11, 1) <= d <= date(1900, 12, 16) or \
         date(1900, 1, 2) <= d <= date(1900, 3, 31):
        return 1

这也使得创建边界对变得更加容易。

def get_season(d):
    d = date(year=1900, month=d.month, day=d.day)

    boundarydict = {1: [(date(1900, 9, 1), date(1900, 3, 31)),
                        (date(1900, 11, 1), date(1900, 12, 16)),
                        (date(1900, 1, 2), date(1900, 3, 31))],
                    2: [(date(1900, 4, 1), date(1900, 6, 24)),
                        (date(1900, 8, 22), date(1900, 10, 31)),
                        (date(1900, 12, 17), date(1900, 12, 31)),
                        (date(1900, 1, 1), date(1900, 1, 1))], # note this one!
                    3: [(date(1900, 6, 25), date(1900, 8, 22))]}

    for retval, boundaries in boundarydict.values():
        if any(a <= d <= b for a, b in boundaries):
            return retval

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