Python - Ceil a datetime to next quarter of an hour

问题

Let's imagine this datetime

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

I'd like to ceil it to the next quarter of hour, in order to get

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

I imagine something like

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

But of course this does not work

回答1:

This one takes microseconds into account!

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)

Explanation of delta:

900 seconds is 15 minutes (a quarter of an hour sans leap seconds which I don't think datetime handles...)
nsecs / 900 is the number of quarter hour chunks that have transpired. Taking the ceil of this rounds up the number of quarter hour chunks.
Multiply the number of quarter hour chunks by 900 to figure out how many seconds have transpired in since the start of the hour after "rounding".

回答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

This gives you:

>>> 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)

回答3:

@Mark Dickinson suggested the best formula so far:

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

In Python 3, for an arbitrary time delta (not just 15 minutes):

#!/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

To avoid intermediate floats, divmod() could be used:

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

Example:

>>> 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)

回答4:

You just need to calculate correct minutes and add them in datetime object after setting minutes, seconds to zero

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))

It works for all cases:

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

回答5:

Here is my code working with any periods:

def floorDT(dt, secperiod):
    tstmp = dt.timestamp()
    return datetime.datetime.fromtimestamp(
        math.floor(tstmp/secperiod)*secperiod).astimezone().astimezone(datetime.timezone.utc)


def ceilDT(dt, secperiod):
    tstmp = dt.timestamp()
    return datetime.datetime.fromtimestamp(
        math.ceil(tstmp/secperiod)*secperiod).astimezone().astimezone(datetime.timezone.utc)

Note: we must use astimezone().astimezone() trick else it uses local timezone during converting from timestamp

来源：https://stackoverflow.com/questions/13071384/python-ceil-a-datetime-to-next-quarter-of-an-hour

标签

python

math

datetime

ceil