[PYTHON] 파이썬 - Ceil, datetime, 다음 1/4 분기

파이썬 - Ceil, datetime, 다음 1/4 분기

이 datetime을 상상해 봅시다.

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

나는 시간의 다음 1/4에 그것을 ceil하고 싶다.

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

나는 상상한다.

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

물론 이것은 작동하지 않습니다.

해결법

1. ==============================

   1.이것은 마이크로 초를 고려합니다!

   이것은 마이크로 초를 고려합니다!

   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)
   

   델타의 설명 :

2. ==============================

   2.

   def ceil(dt):
       if dt.minute % 15 or dt.second:
           return dt + datetime.timedelta(minutes = 15 - dt.minute % 15,
                                          seconds = -(dt.second % 60))
       else:
           return dt
   

   이것은 당신에게 :

   >>> ceil(datetime.datetime(2012,10,25, 17,45))
   datetime.datetime(2012, 10, 25, 17, 45)
   >>> ceil(datetime.datetime(2012,10,25, 17,45,1))
   datetime.datetime(2012, 10, 25, 18, 0)
   >>> ceil(datetime.datetime(2012,12,31,23,59,0))
   datetime.datetime(2013,1,1,0,0)
   

3. ==============================

   3.@ 마크 디킨슨은 지금까지 최고의 공식을 제안했다.

   @ 마크 디킨슨은 지금까지 최고의 공식을 제안했다.

   def ceil_dt(dt, delta):
       return dt + (datetime.min - dt) % delta
   

   Python 3에서 임의의 시간 델타 (15 분이 아님)의 경우 :

   #!/usr/bin/env python3
   import math
   from datetime import datetime, timedelta
   
   def ceil_dt(dt, delta):
       return datetime.min + math.ceil((dt - datetime.min) / delta) * delta
   
   print(ceil_dt(datetime(2012, 10, 25, 17, 32, 16), timedelta(minutes=15)))
   # -> 2012-10-25 17:45:00
   

   중간 수레를 피하기 위해 divmod ()를 사용할 수 있습니다 :

   def ceil_dt(dt, delta):
       q, r = divmod(dt - datetime.min, delta)
       return (datetime.min + (q + 1)*delta) if r else dt
   

   예:

   >>> ceil_dt(datetime(2012, 10, 25, 17, 32, 16), timedelta(minutes=15))
   datetime.datetime(2012, 10, 25, 17, 45)
   >>> ceil_dt(datetime.min, datetime.resolution) 
   datetime.datetime(1, 1, 1, 0, 0)
   >>> ceil_dt(datetime.min, 2*datetime.resolution)
   datetime.datetime(1, 1, 1, 0, 0)
   >>> ceil_dt(datetime.max, datetime.resolution)
   datetime.datetime(9999, 12, 31, 23, 59, 59, 999999)
   >>> ceil_dt(datetime.max, 2*datetime.resolution)
   Traceback (most recent call last):
     File "<stdin>", line 1, in <module>
     File "<stdin>", line 3, in ceil_dt
   OverflowError: date value out of range
   >>> ceil_dt(datetime.min+datetime.resolution, datetime.resolution)
   datetime.datetime(1, 1, 1, 0, 0, 0, 1)
   >>> ceil_dt(datetime.min+datetime.resolution, 2*datetime.resolution)
   datetime.datetime(1, 1, 1, 0, 0, 0, 2)
   >>> ceil_dt(datetime.max-datetime.resolution, datetime.resolution)
   datetime.datetime(9999, 12, 31, 23, 59, 59, 999998)
   >>> ceil_dt(datetime.max-datetime.resolution, 2*datetime.resolution)
   datetime.datetime(9999, 12, 31, 23, 59, 59, 999998)
   >>> ceil_dt(datetime.max-2*datetime.resolution, datetime.resolution)
   datetime.datetime(9999, 12, 31, 23, 59, 59, 999997)
   >>> ceil_dt(datetime.max-2*datetime.resolution, 2*datetime.resolution)
   datetime.datetime(9999, 12, 31, 23, 59, 59, 999998)
   >>> ceil_dt(datetime.max-timedelta(1), datetime.resolution)
   datetime.datetime(9999, 12, 30, 23, 59, 59, 999999)
   >>> ceil_dt(datetime.max-timedelta(1), 2*datetime.resolution)
   datetime.datetime(9999, 12, 31, 0, 0)
   >>> ceil_dt(datetime.min, datetime.max-datetime.min)
   datetime.datetime(1, 1, 1, 0, 0)
   >>> ceil_dt(datetime.max, datetime.max-datetime.min)
   datetime.datetime(9999, 12, 31, 23, 59, 59, 999999)
   

4. ==============================

   4.정확한 분을 계산하고 분, 초를 0으로 설정 한 후 datetime 객체에 추가해야합니다.

   정확한 분을 계산하고 분, 초를 0으로 설정 한 후 datetime 객체에 추가해야합니다.

   import datetime
   
   def quarter_datetime(dt):
       minute = (dt.minute//15+1)*15
       return dt.replace(minute=0, second=0)+datetime.timedelta(minutes=minute)
   
   for minute in [12, 22, 35, 52]:
       print quarter_datetime(datetime.datetime(2012, 10, 25, 17, minute, 16))
   

   모든 경우에 적용됩니다.

   2012-10-25 17:15:00
   2012-10-25 17:30:00
   2012-10-25 17:45:00
   2012-10-25 18:00:00
   

from https://stackoverflow.com/questions/13071384/python-ceil-a-datetime-to-next-quarter-of-an-hour by cc-by-sa and MIT license