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Python-Ceil a datetime para el próximo cuarto de hora

Question

Imaginemos esta fecha y hora

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

Me gustaría ceñirlo al siguiente cuarto de hora para obtener

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

Me imagino algo como

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

Pero claro que esto no funciona.

Answer 1

¡Éste tiene en cuenta los microsegundos!

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)

Explicación del delta :

• 900 segundos son 15 minutos (un cuarto de hora sin segundos de salto, lo cual no creo que datetime maneje ... )
• nsecs / 900 es el número de trozos de cuarto de hora que han transcurrido. Tomando el ceil de este número se redondea la cantidad de trozos de cuarto de hora.
• Multiplique el número de segmentos de cuarto de hora por 900 para calcular cuántos segundos han transcurrido desde el comienzo de la hora después del "redondeo".

Answer 2

Solo necesita calcular los minutos correctos y agregarlos en el objeto datetime después de configurar los minutos, segundos a cero

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))

Funciona para todos los casos:

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

Answer 3

@Mark Dickinson sugirió la mejor fórmula hasta ahora:

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

En Python 3, para un delta de tiempo arbitrario (no solo 15 minutos):

#!/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

Para evitar flotadores intermedios, divmod() podría ser usado:

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

Ejemplo:

>>> 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)

Answer 4

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

Esto te da:

>>> 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)