简化处理当前使用 datetime 和 isodate 的时间和持续时间的代码？

Question

下面的代码应该是：

将输入字符串解析为某种持续时间
验证duration不为空，不为负，不超过10年

输入字符串的示例是：

duration_string = "P10W"
duration_string = "P1Y"

这是代码

    duration = isodate.parse_duration(duration_string)

    if isinstance(duration, datetime.timedelta):
        if not duration > datetime.timedelta(0):
            raise Exception('duration invalid')
        if duration > datetime.timedelta(3660):
            raise Exception('duration cannot be longer than 10 years')
    elif isinstance(duration, isodate.Duration):
        if not duration > 0:
            raise Exception('duration invalid')
        if duration > isodate.duration.Duration(0, 0, 0, years=10, months=0):
            log.debug("duration %s isodate %s" % (duration, isodate.duration.Duration(0, 0, 0, years=10, months=0)))
            raise Exception('duration cannot be longer than 10 years')

有比我制作的怪物更简单的方法吗？

除了需要简化之外，行duration > isodate.duration.Duration(0, 0, 0, years=10, months=0)行不通。

我正在使用 Python 2.7

Answer 1

好的，所以如果您绝对必须使用 isodate 持续时间解析，请保留 isodate 库。然而，我要提到的是 isodate 库不完整，有许多糟糕的设计决策，而且总体来说很糟糕。

但是如果你必须使用他们的解析工具，这可能是一个好方法。

import isodate
import functools

@functools.total_ordering  # if we implement < ==, will implement <=, >, >=
class Duration(isodate.Duration):
    # inherit from isodate.Duration -- gives us ==

    # constants 
    seconds_in_day = 60**2 * 24
    approx_days_in_month = 30
    approx_days_in_year = 365

    def approx_total_seconds(self):
        """approx total seconds in duration"""
        # self.months and self.years are stored as `Decimal`s for some reason...
        return self.tdelta.total_seconds() \
               + float(self.months) * self.approx_days_in_month *  self.seconds_in_day \
               + float(self.years) * self.approx_days_in_year * self.seconds_in_day

    def __lt__(self, other):
        """defines self < other"""
        if not isinstance(other, Duration):
            return NotImplemented
        return self.approx_total_seconds() < other.approx_total_seconds()

    @classmethod
    def parse_duration(cls, datestring):
        """a version of isodate.parse_duration that returns out class"""

        iso_dur = isodate.parse_duration(datestring)

        # iso_date.parse_duration can return either a Duration or a timedelta...
        if isinstance(iso_dur, isodate.Duration):
            return cls(seconds=iso_dur.tdelta.total_seconds(),
                       months=iso_dur.months, years=iso_dur.years)
        else:
            return cls(seconds=iso_dur.total_seconds())


ten_weeks = Duration.parse_duration('P10W')
one_year = Duration.parse_duration('P1Y')

print(ten_weeks.approx_total_seconds())
print(one_year.approx_total_seconds())

print(ten_weeks < one_year)
print(ten_weeks > one_year)

产出

6048000.0
31536000.0
True
False

如果您不需要 isodate 解析（我怀疑您不需要），您可以这样做

@functools.TotalOrdering
class ApproxTimeDelta:

    approx_days_in_week = 7
    approx_days_in_month = 30
    approx_days_in_year = 365

    def __init__(self, days, weeks, months, years):
        self.days = days + \
                    weeks * self.approx_days_in_week + \
                    months * self.approx_days_in_month + \
                    years * self.approx_days_in_year

    def __eq__(self, other):
        return self.days == other.days

    def __lt__(self, other):
        return self.days < other.days

并将 years/months/weeks/days 作为整数传递并像以前一样进行比较。

Answer 2

这是我最终使用的替代解决方案：

    if isinstance(duration, datetime.timedelta):
        if not duration > 0:
            raise Exception('duration invalid')
        if duration > 3650:
            raise Exception('maximum duration is 3650 days')
    elif isinstance(duration, isodate.Duration):
        if duration.years > 10:
            raise Exception('maximum duration is 10 years')
        if duration.months > 120:
            raise Exception('maximum duration is 120 months')

简化处理当前使用 datetime 和 isodate 的时间和持续时间的代码？

Simplify code dealing with time and duration that currently uses datetime and isodate?

python

time

duration

python-2.7