投稿者: ブログの住人

class date_as_integer:
    """符号無し整数で表現された日付に対する演算子をまとめたクラス
    A class of operators for dates represented by unsigned integers
    """

    @staticmethod
    def add_days(orig_date, duration):
        """8か6桁の符号無し整数（yyyyMMdd か yyMMdd）で表現された
        日付について、指定した日数を加算、または減算した結果を返す関数
        A function implementation that returns the result of
        adding or subtracting the specified number of days for a date
        represented by an 8- or 6-digit unsigned integer (yyyyMMdd or yyMMdd).
        """
        Y, _md = divmod(orig_date, 10000)
        M, D = divmod(_md, 100)
        M -= 1
        D -= 1

        def get_ndays_month(year):

            months_of_year = [
                [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31], # not leap
                [31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]  # leap year
            ]

            return months_of_year[
                year % 4 == 0 and (year % 100 != 0 or year % 400 == 0)
            ]

        def future(ndays, duration, Y, M):
            ndays = get_ndays_month(Y)[M]
            M += 1
            duration -= ndays
            if M > 11:
                M = 0
                Y += 1
            return ndays, duration, Y, M

        def past(ndays, duration, Y, M):
            M -= 1
            ndays = get_ndays_month(Y)[M]
            duration += ndays
            if M < 0:
                M = 11
                Y -= 1
            return ndays, duration, Y, M

        step_into_next_month = future if (duration >= 0) else past

        ndays = get_ndays_month(Y)[M]

        while not ndays > (D + duration) >= 0:

            ndays, duration, Y, M = step_into_next_month(ndays, duration, Y, M)

        D += duration

        return Y * 10000 + (M + 1) * 100 + D + 1


    @staticmethod
    def add_months(orig_date, duration):
        """6か4桁の符号無し整数（yyyyMM か yyMM）で表現された
        日付について、指定した月数を加算、または減算した結果を返す関数
        A function implementation that returns the result of
        adding or subtracting the specified number of months for a date
        represented by an 6- or 4-digit unsigned integer (yyyyMM or yyMM).
        """
        Y0, M0 = divmod(int(orig_date), 100)
        dy, dm = divmod(duration, 12)

        if M0 + dm > 12:
            Y1 = Y0 + dy + 1
            M1 = M0 + dm - 12
        else:
            Y1 = Y0 + dy
            M1 = M0 + dm

        return Y1 * 100 + M1


if __name__ == "__main__":
    print('Up and Down at new year boundary')
    for i in range(2):
        print(date_as_integer.add_days(20211231, i))
    for i in range(2):
        print(date_as_integer.add_days(20220101, -i))

    print('At boundary of leap day every 4 years')
    for i in range(3):
        print(date_as_integer.add_days(20200228, i))
    for i in range(3):
        print(date_as_integer.add_days(20200301, -i))

    print('''At boundary of leap day in a year that cannot divide by 100
(28 days are in February) ''')
    for i in range(2):
        print(date_as_integer.add_days(21000228, i))
    for i in range(2):
        print(date_as_integer.add_days(21000301, -i))

    print('''At boundary of leap day in a year that can divide by 400
(29 days are in February) ''')
    for i in range(3):
        print(date_as_integer.add_days(20000228, i))
    for i in range(3):
        print(date_as_integer.add_days(20000301, -i))

    print('Find the date N months later (or earlier)')
    for i in range(2):
        print(date_as_integer.add_months(200012, i))
    for i in range(2):
        print(date_as_integer.add_months(200101, -i))

    print('38 == 6 + 12 + 12 + 8')
    print(date_as_integer.add_months(200006, 38))
    print(date_as_integer.add_months(200006, -38))