python – 如何检查循环范围的重叠(重叠的年度循环周期)

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我试图找到一个优雅的算法,以检查两个年度重复周期是否重叠.这个时期与年份无关,但一年可能总是闰年.

例如,期间A =(3月1日至5月1日)和期间B =(4月1日至9月1日)重叠.
此外,期间A =(10月1日至2月1日)和期间B =(1月1日至3月1日)重叠.

但是,我发现这比我预想的要困难得多.复杂性来自于跨越年底的时期.

我有一个有效的解决方案(参见下面的doOverlap(A,B)方法),但我发现它令人沮丧.

# for the rest of the MWE context code, see further
# WORKING, but a bit convulted
def doesOverlap(A, B):  
    '''returns True if yearly period A and B have overlapping dates'''
    # list to track if day in year is part of a period A
    # (this could probably be done a bit cheaper with a dictionary of tuples, but not relevant for my question)
    yeardayCovered = [False for x in range(366)]  # leap year

    # mark the days of A
    for d in range(A.start, A.start + A.length):
        yeardayCovered[d % 366] = True

    # now check each of the days in B with A
    for d in range(B.start, B.start + B.length):
        if yeardayCovered[d % 366]:
            return True
    return False

我相信应该可以用更少的支票和更优雅的方式来做到这一点.我已经尝试将其中一个开始日设置为零偏移,应用一些模运算符,然后是常规(非循环)范围重叠检查(Algorithm to detect overlapping periods).但我还没有为我的所有测试用例工作.

#NOT WORKING CORRECTLY!!
def doesOverlap(A, B):   
    '''determines if two yearly periods have overlapping dates'''
    Astart = A.start
    Astop = A.stop
    Bstart = B.start
    Bstop = B.stop

    # start day counting at Astart, at 0
    offset = Astart
    Astart = 0
    Astop = (Astop - offset) % 366
    Bstart = (Bstart - offset) % 366
    Bstop = (Bstop - offset) % 366

    # overlap?
    # https://stackoverflow.com/a/13513973
    return (Astart <= Bstop and Bstart <= Astop)

注意：我已经用Python完成了代码,但理想情况下解决方案应该不是特定于Python的(即不使用通常只在Python中可用的函数,而不是在C或C#中)

# MWE (Minimal Working Example)

import datetime
import unittest


class TimePeriod:
    def __init__(self, startDay, startMonth, stopDay, stopMonth):
        self.startDay = startDay
        self.startMonth = startMonth
        self.stopDay = stopDay
        self.stopMonth = stopMonth

    def __repr__(self):
        return "From " + str(self.startDay) + "/" + str(self.startMonth) + " to " + \
            str(self.stopDay) + "/" + str(self.stopMonth)

    def _dayOfYear(self, d, m, y=2012):
        '''2012 = leap year'''
        date1 = datetime.date(year=y, day=d, month=m)
        return date1.timetuple().tm_yday

    @property
    def start(self):
        '''day of year of start of period, zero-based for easier modulo operations! '''
        return self._dayOfYear(self.startDay, self.startMonth) - 1

    @property
    def stop(self):
        '''day of year of stop of period, zero-based for easier modulo operations! '''
        return self._dayOfYear(self.stopDay, self.stopMonth) - 1

    @property
    def length(self):
        '''number of days in the time period'''
        _length = (self.stop - self.start) % 366 + 1
        return _length

def doesOverlap(A, B):
    # code from above goes here


class TestPeriods(unittest.TestCase):
    pass


def test_generator(a, b, c):
    def test(self):
        self.assertEqual(doesOverlap(a, b), c)
    return test

if __name__ == '__main__':

    #some unit tests, probably not complete coverage of all edge cases though
    tests = [["max", TimePeriod(1, 1, 31, 12), TimePeriod(1, 1, 1, 1), True],
             ["BinA", TimePeriod(1, 3, 1, 11), TimePeriod(1, 5, 1, 10), True],
             ["BoverEndA", TimePeriod(1, 1, 1, 2), TimePeriod(10, 1, 3, 3), True],
             ["BafterA", TimePeriod(1, 1, 1, 2), TimePeriod(2, 2, 3, 3), False],
             ["sBoutA", TimePeriod(1, 12, 2, 5), TimePeriod(1, 6, 1, 7), False],
             ["sBoverBeginA", TimePeriod(1, 11, 2, 5), TimePeriod(1, 10, 1, 12), True],
             ["sBinA", TimePeriod(1, 11, 2, 5), TimePeriod(1, 1, 1, 2), True],
             ["sBinA2", TimePeriod(1, 11, 2, 5), TimePeriod(1, 12, 10, 12), True],
             ["sBinA3", TimePeriod(1, 11, 2, 5), TimePeriod(1, 12, 1, 2), True],
             ["sBoverBeginA", TimePeriod(1, 11, 2, 5), TimePeriod(1, 10, 1, 12), True],
             ["Leap", TimePeriod(29, 2, 1, 4), TimePeriod(1, 10, 1, 12), False],
             ["BtouchEndA", TimePeriod(1, 2, 1, 2), TimePeriod(1, 2, 1, 3), True]]

    for i, t in enumerate(tests):
        test_name = 'test_%s' % t[0]
        test = test_generator(t[1], t[2], t[3])
        setattr(TestPeriods, test_name, test)

    # unittest.main()
    suite = unittest.TestLoader().loadTestsFromTestCase(TestPeriods)
    unittest.TextTestRunner(verbosity=2).run(suite)

解决方法:

def overlap(a0, a1, b0, b1):

首先,如果在开始日期之前结束,则可以通过调整结束日期将“年”圈中的间隔“提升”到时间“线”…

    if a1 < a0: a1 += 365
    if b1 < b0: b1 += 365

如果两个规则间隔相交,则存在交叉点

    if a1 > b0 and a0 < b1: return True

或者如果他们确实移动[a0 … a1 [前进一年或后退一年

    if a1+365 > b0 and a0+365 < b1: return True
    if a1-365 > b0 and a0-365 < b1: return True

否则没有交集

    return False

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