Question

我一直在Python的日期/时区上挣扎太久，并且认为有人可以帮我一把。

基本上我想用UTC进行转换并考虑DST的变化。

我从一个Python教程中创建了以下tzinfo类（我知道这不是100％准确但不需要）：

from datetime import tzinfo, timedelta, datetime

ZERO = timedelta(0)
HOUR = timedelta(hours=1)

def first_sunday_on_or_after(dt):
    days_to_go = 6 - dt.weekday()
    if days_to_go:
        dt += timedelta(days_to_go)
    return dt

DSTSTART_2007 = datetime(1, 3, 8, 2)
DSTEND_2007 = datetime(1, 11, 1, 1)
DSTSTART_1987_2006 = datetime(1, 4, 1, 2)
DSTEND_1987_2006 = datetime(1, 10, 25, 1)
DSTSTART_1967_1986 = datetime(1, 4, 24, 2)
DSTEND_1967_1986 = DSTEND_1987_2006

class USTimeZone(tzinfo):

    def __init__(self, hours, reprname, stdname, dstname):
        self.stdoffset = timedelta(hours=hours)
        self.reprname = reprname
        self.stdname = stdname
        self.dstname = dstname

    def __repr__(self):
        return self.reprname

    def tzname(self, dt):
        if self.dst(dt):
            return self.dstname
        else:
            return self.stdname

    def utcoffset(self, dt):
        return self.stdoffset + self.dst(dt)

    def dst(self, dt):
        if dt is None or dt.tzinfo is None:
            # An exception may be sensible here, in one or both cases.
            # It depends on how you want to treat them.  The default
            # fromutc() implementation (called by the default astimezone()
            # implementation) passes a datetime with dt.tzinfo is self.
            return ZERO
        assert dt.tzinfo is self

        # Find start and end times for US DST. For years before 1967, return
        # ZERO for no DST.
        if 2006 < dt.year:
            dststart, dstend = DSTSTART_2007, DSTEND_2007
        elif 1986 < dt.year < 2007:
            dststart, dstend = DSTSTART_1987_2006, DSTEND_1987_2006
        elif 1966 < dt.year < 1987:
            dststart, dstend = DSTSTART_1967_1986, DSTEND_1967_1986
        else:
            return ZERO

        start = first_sunday_on_or_after(dststart.replace(year=dt.year))
        end = first_sunday_on_or_after(dstend.replace(year=dt.year))

        # Can't compare naive to aware objects, so strip the timezone from
        # dt first.
        if start <= dt.replace(tzinfo=None) < end:
            return HOUR
        else:
            return ZERO

另一方面，我在EST中有一个任意date个对象，我想通过考虑DST来了解它们的不同小时数。

我尝试过这样的事情：

>>> Eastern = ustimezone.USTimeZone(-5, "Eastern",  "EST", "EDT")
>>> x = datetime.date.today() # I actually get an arbitrary date but this is for the example
>>> x_dt = datetime.datetime.combine(x, datetime.time())
>>> x_dt_tz = x_dt.astimezone(Eastern)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: astimezone() cannot be applied to a naive datetime

我看过几个帖子告诉他们使用localize模块中的pytz，但遗憾的是我无法使用其他模块，因此无法使用pyzt

有谁知道我如何将这个天真的日期时间变成一个时间点对象而不使用pytz ？

Answer 1

对于它的价值，@skyl提供的答案或多或少等同于pytz。

以下是相关的pytz来源。它只需使用replace kwarg在datetime对象上调用tzinfo：

def localize(self, dt, is_dst=False):
    '''Convert naive time to local time'''
    if dt.tzinfo is not None:
        raise ValueError('Not naive datetime (tzinfo is already set)')
    return dt.replace(tzinfo=self)

Answer 2

使用x_dt.replace(tzinfo=Eastern)（found from this Google Groups thread）。

x_dt.replace(tzinfo=Eastern).utcoffset()返回datetime.timedelta(-1, 72000)，相当于-4小时！（来自问题的评论）

Python将tzinfo传递给没有pytz的天真日期时间

2 个答案: