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简化处理当前使用datetime和isodate的时间和持续时间的代码?

duration time python-2.7 python

user1283776 · 技术社区 · 6 年前

下面的代码应该是:

验证持续时间不为空,不为负,不超过10年

输入字符串示例如下:

duration_string = "P10W"
duration_string = "P1Y"

这是密码

    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')

有没有比我制造的怪物更简单的方法呢?

除了需要简化,这条线 duration > isodate.duration.Duration(0, 0, 0, years=10, months=0)

我使用的是python2.7

2 回复 | 直到 6 年前

FHTMitchell 6 年前

好的,所以如果你必须使用isodate持续时间解析,保持 isodate

但是,如果您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)

6048000.0
31536000.0
True
False

如果你不需要isodate解析(我猜你不需要),你可以这样做

@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

user1283776 6 年前

下面是我最终使用的另一种解决方案:

    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')