Python将tzinfo传递给不带pytz的naive datetime

Question

我在Python的日期/时区上挣扎了太久，我想有人可以帮我一下。

基本上我想在UTC中进行转换，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中有一个任意的date对象，我想通过考虑DST来知道它们的不同小时数。

我试过这样的方法：

>>> 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

我已经看到一些帖子告诉我使用localize模块中的pytz，但是不幸的是我不能使用额外的模块，所以不可能使用pyzt

有人知道我如何在不使用pytz的情况下将这个天真的datetime转换成一个带时间分区的对象吗？

Answer 1

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

x_dt.replace(tzinfo=Eastern).utcoffset()返回对应于-4小时的datetime.timedelta(-1, 72000)！（摘自问题的评论）

Answer 2

就其价值而言，@skyl提供的答案或多或少相当于pytz所做的。

这是相关的pytz来源。它只是用tzinfokwarg调用replace对象上的datetime：

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)