Python:鉴于当前UTC时间,您如何确定特定时区内一天的开始和结束时间?

时间:2011-08-28 05:36:40

标签: python datetime timezone utc

我正在玩Google App Engine,我了解到时区固定为UTC。我想确定用户当地时区当天的开始和结束时间。所以基本上,考虑到UTC的当前时间,如何考虑夏令时转换,如何确定当天的开始和结束时间。

我有一些笨重的示例代码。请注意,我意识到,如果我手动指定日期,我也可以指定明天的日期,但它们是示例,我想以编程方式确定它。我的主要问题是,如果我将timedelta添加到带有时区的日期时间然后将其标准化(就像在pytz文档中建议的那样),我会得到一个日期时间,这是在夏令时切换期间的一小时。

代码中没有提到,但最终的目的是将这些时间转换回UTC,这就是为什么重要的是能够识别时区。

#!/usr/bin/python

import datetime
from pytz.gae import pytz

hobart_tz = pytz.timezone('Australia/Hobart')

utc_dt = pytz.utc.localize(datetime.datetime.utcnow())
hobart_dt = utc_dt.astimezone(hobart_tz)

# create a new datetime for the start of the day and add a day to it to get tomorrow.
today_start = datetime.datetime(hobart_dt.year, hobart_dt.month, hobart_dt.day)
today_start = hobart_tz.localize(today_start)
today_end = hobart_tz.normalize(today_start + datetime.timedelta(days=1))
print 'today:', today_start
print ' next:', today_end
print
# gives:
# today: 2011-08-28 00:00:00+10:00
# next: 2011-08-29 00:00:00+10:00

# but say today is a daylight savings changeover.
# after normalisation, we are off by an hour.

dst_finish_2011 = datetime.datetime(2011, 4, 3)  # this would come from hobart_dt
dst_finish_2011 = hobart_tz.localize(dst_finish_2011)
next = hobart_tz.normalize(dst_finish_2011 + datetime.timedelta(days=1))
print '2011-04-03:', dst_finish_2011
print '2011-04-04:', next   # expect 2011-04-04 00:00:00+10:00
print
# gives
# 2011-04-03: 2011-04-03 00:00:00+11:00
# 2011-04-04: 2011-04-03 23:00:00+10:00 (wrong)

dst_start_2011 = datetime.datetime(2011, 10, 2)  # this would come from hobart_dt
dst_start_2011 = hobart_tz.localize(dst_start_2011)
next = hobart_tz.normalize(dst_start_2011 + datetime.timedelta(days=1))
print '2011-10-02:', dst_start_2011
print '2011-10-03:', next   # expect 2011-10-03 00:00:00+11:00
print
# gives
# 2011-10-02: 2011-10-02 00:00:00+10:00
# 2011-10-03: 2011-10-03 01:00:00+11:00 (wrong)

# I guess we could ignore the timezone and localise *after* ?

dst_finish_2011 = datetime.datetime(2011, 4, 3)  # this would come from hobart_dt
next = dst_finish_2011 + datetime.timedelta(days=1)
# now localise
dst_finish_2011 = hobart_tz.localize(dst_finish_2011)
next = hobart_tz.localize(next)
print '2011-04-03:', dst_finish_2011
print '2011-04-04:', next   # expect 2011-04-04 00:00:00+10:00
print
# gives
# 2011-04-03: 2011-04-03 00:00:00+11:00
# 2011-04-04: 2011-04-04 00:00:00+10:00

4 个答案:

答案 0 :(得分:0)

我相信你得到的结果是因为你要加一天而不是86400秒。天与秒之间没有统一的,始终正确的等价。例如,如果pytz强制一天“真正”为86400秒,那么将一天添加到12月31日或6月30日有时会导致结果的秒字段“一秒钟关闭” ,因为在某些年份那些日子已经有86401秒。 (未来可能会有86402甚至86399秒。)

因此,添加一天意味着简单地将日期递增1,如果需要,可以延续到月份和年份,但不会更改时间字段。尝试添加86400秒,看看是否得到了所需的结果。

答案 1 :(得分:0)

在当地时区找出当天(午夜)的开始时间和当天(明天)的结束时间,知道UTC时间:

#!/usr/bin/env python
from datetime import datetime, time, timedelta
import pytz # $ pip install pytz
from tzlocal import get_localzone # $ pip install tzlocal

tz = get_localzone() # get the local timezone as pytz.timezone
now = datetime.now(pytz.utc) # some UTC time
dt = now.astimezone(tz) # the same time in the local timezone
today = dt.date() # today in the local timezone (naive date object)
midnight = datetime.combine(today, time()) # midnight in the local timezone
aware_midnight = tz.localize(midnight, is_dst=None) # raise exception
                                                    # for ambiguous or
                                                    # non-existing
                                                    # times
tomorrow = midnight + timedelta(1)
aware_tomorrow = tz.localize(tomorrow, is_dst=None)

def print_time(aware_dt, fmt="%Y-%m-%d %H:%M:%S %Z%z"):
    print(aware_dt.strftime(fmt))
    utc_dt = aware_dt.astimezone(pytz.utc) # the same time in UTC
    print(utc_dt.strftime(fmt))

print_time(aware_midnight)
print_time(aware_tomorrow)

输出

2014-09-01 00:00:00 EST+1000
2014-08-31 14:00:00 UTC+0000
2014-09-02 00:00:00 EST+1000
2014-09-01 14:00:00 UTC+0000

另见,

答案 2 :(得分:-1)

经过一些实验和思考后,我相信我有一个解决方案。你指出我以前的答案是不正确的;天数= 1的timedelta对象与秒= 86400基本相同(除非涉及闰秒)。

我建议的一种方式是,在不考虑时间的情况下增加日期是使用datetime.date对象而不是datetime.datetime对象:

>>> oneday = datetime.timedelta(days=1)
>>> d = datetime.date(2011,4,3)
>>> str(d + oneday)
'2011-04-04'

然后可以添加一天中的时间以形成一个完整的datetime.datetime对象,您可以在该对象中了解当前时间字段与原始值不变的情况。

另一种我觉得可以安全使用的方法是临时使用“天真”的日期。这样,添加timedelta时无需应用时区政策。

>>> hob = pytz.timezone('Australia/Hobart')
>>> dstlast = datetime.datetime(2011,4,3)
>>> str(dstlast)
'2011-04-03 00:00:00'
>>> dstlasthob = hob.localize(dstlast)
>>> str(dstlasthob)
'2011-04-03 00:00:00+11:00'
>>> oneday = datetime.timedelta(days=1)
>>> str(hob.normalize(dstlasthob + oneday))
'2011-04-03 23:00:00+10:00'
>>> nextday = hob.localize(dstlasthob.replace(tzinfo=None) + oneday)
>>> str(nextday)
'2011-04-04 00:00:00+10:00'

我测试了这个方法,其中包含闰秒的日期(一个例子是2008-12-31),结果的时间是00:00:00。这可能实际上是错的,我不确定,但这是你想要的: - )

答案 3 :(得分:-1)

以下代码试图获得午夜的时间;如果时区调整使其失败,则会使用新的区域偏移重新调整到午夜。

def DayStartEnd(localized_dt):
    tz = localized_dt.tzinfo
    start = tz.normalize(datetime.datetime(localized_dt.year,localized_dt.month,localized_dt.day,0,0,0,0,tz))
    after_midnight = start.hour*60*60 + start.minute*60 + start.second
    if start.day != localized_dt.day:
        start += datetime.timedelta(seconds = 24*60*60 - after_midnight)
    elif after_midnight != 0:
        start -= datetime.timedelta(seconds = after_midnight)
    end = tz.normalize(start + datetime.timedelta(hours=24))
    after_midnight = end.hour*60*60 + end.minute*60 + end.second
    if end.day == localized_dt.day:
        end += datetime.timedelta(seconds = 24*60*60 - after_midnight)
    elif after_midnight != 0:
        end -= datetime.timedelta(seconds = after_midnight)
    return start,end

>>> hobart_tz = pytz.timezone('Australia/Hobart')
>>> dst_finish_2011 = datetime.datetime(2011, 4, 3)
>>> dst_finish_2011 = hobart_tz.localize(dst_finish_2011)
>>> start,end = DayStartEnd(dst_finish_2011)
>>> print start,end
2011-04-03 00:00:00+11:00 2011-04-04 00:00:00+10:00
>>> dst_start_2011 = datetime.datetime(2011, 10, 2)
>>> dst_start_2011 = hobart_tz.localize(dst_start_2011)
>>> start,end = DayStartEnd(dst_start_2011)
>>> print start,end
2011-10-02 00:00:00+10:00 2011-10-03 00:00:00+11:00