将日期时间推迟到下一刻钟

Question

让我们想象一下 datetime

>>> import datetime
>>> dt = datetime.datetime(2012, 10, 25, 17, 32, 16)

我想将其推迟到下一个季度，以获取

datetime.datetime(2012, 10, 25, 17, 45)

我想像

>>> quarter = datetime.timedelta(minutes=15)
>>> import math
>>> ceiled_dt = math.ceil(dt / quarter) * quarter

但是，当然，这不起作用

Answer 1

这需要微秒！

import math

def ceil_dt(dt):
    # how many secs have passed this hour
    nsecs = dt.minute*60 + dt.second + dt.microsecond*1e-6  
    # number of seconds to next quarter hour mark
    # Non-analytic (brute force is fun) way:  
    #   delta = next(x for x in xrange(0,3601,900) if x>=nsecs) - nsecs
    # analytic way:
    delta = math.ceil(nsecs / 900) * 900 - nsecs
    #time + number of seconds to quarter hour mark.
    return dt + datetime.timedelta(seconds=delta)

t1 = datetime.datetime(2017, 3, 6, 7, 0)
assert ceil_dt(t1) == t1

t2 = datetime.datetime(2017, 3, 6, 7, 1)
assert ceil_dt(t2) == datetime.datetime(2017, 3, 6, 7, 15)

t3 = datetime.datetime(2017, 3, 6, 7, 15)
assert ceil_dt(t3) == t3

t4 = datetime.datetime(2017, 3, 6, 7, 16)
assert ceil_dt(t4) == datetime.datetime(2017, 3, 6, 7, 30)

t5 = datetime.datetime(2017, 3, 6, 7, 30)
assert ceil_dt(t5) == t5

t6 = datetime.datetime(2017, 3, 6, 7, 31)
assert ceil_dt(t6) == datetime.datetime(2017, 3, 6, 7, 45)

t7 = datetime.datetime(2017, 3, 6, 7, 45)
assert ceil_dt(t7) == t7

t8 = datetime.datetime(2017, 3, 6, 7, 46)
assert ceil_dt(t8) == datetime.datetime(2017, 3, 6, 8, 0)

说明delta：

• 900秒是15分钟（四分之一小时 不包含leap秒，我认为日期时间无法处理… ）
• nsecs / 900是已发生的四分之一小时块的数量。ceil以此取整四分之一小时的块数。
• 将四分之一小时的块数乘以900，以计算自“四舍五入”后的小时开始以来已经发生了多少秒。