Ceil a datetime to next quarter of an hour

Question

Let\'s imagine this datetime

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

I\'d l

Answer 1

@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)

Answer 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)

Answer 3

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".

Answer 4

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

Answer 5

The formula proposed here by @Mark Dickinson worked beautifully, but I needed a solution that also handled timezones and Daylight Savings Time (DST).

Using pytz, I arrived at:

import pytz
from datetime import datetime, timedelta

def datetime_ceiling(dt, delta):
    # Preserve original timezone info
    original_tz = dt.tzinfo
    if original_tz:
        # If the original was timezone aware, translate to UTC.
        # This is necessary because datetime math does not take
        # DST into account, so first we normalize the datetime...
        dt = dt.astimezone(pytz.UTC)
        # ... and then make it timezone naive
        dt = dt.replace(tzinfo=None)
    # We only do math on a timezone naive object, which allows
    # us to pass naive objects directly to the function
    dt = dt + ((datetime.min - dt) % delta)
    if original_tz:
        # If the original was tz aware, we make the result aware...
        dt = pytz.UTC.localize(dt)
        # ... then translate it from UTC back its original tz.
        # This translation applies appropriate DST status.
        dt = dt.astimezone(original_tz)
    return dt

A nearly identical floor function can be made by changing one line of code:

def datetime_floor(dt, delta):
    ...
    dt = dt - ((datetime.min - dt) % delta)
    ...

The following datetime is three minutes before the transition from DST back to Standard Time (STD):

datetime.datetime(2020, 11, 1, 1, 57, tzinfo=<DstTzInfo 'US/Eastern' EDT-1 day, 20:00:00 DST>)

Assuming the above as dt, we can round down to the nearest five minute increment using our floor function:

>>> datetime_floor(dt, timedelta(minutes=5))
datetime.datetime(2020, 11, 1, 1, 55, tzinfo=<DstTzInfo 'US/Eastern' EDT-1 day, 20:00:00 DST>)

The timezone and relationship to DST is preserved. (The same would be true for the ceiling function.)

On this date DST will end at 2 am, at which point the time will "roll back" to 1am STD. If we use our ceiling function to round up from 1:57am DST, we should not end up at 2am DST, but rather at 1:00am STD, which is the result we get:

>>> datetime_ceiling(dt, timedelta(minutes=5))
datetime.datetime(2020, 11, 1, 1, 0, tzinfo=<DstTzInfo 'US/Eastern' EST-1 day, 19:00:00 STD>)

Answer 6

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