How to round the minute of a datetime object python

0 votes
asked Aug 12, 2010 by lucas-manco

I have a datetime object produced using strptime().

>>> tm
datetime.datetime(2010, 6, 10, 3, 56, 23)

What I need to do is round the minute to the closest 10th minute. What I have been doing up to this point was taking the minute value and using round() on it.

min = round(tm.minute, -1)

However, as with the above example, it gives an invalid time when the minute value is greater than 56. i.e.: 3:60

What is a better way to do this? Does datetime support this?

8 Answers

0 votes
answered Aug 12, 2010 by mykhal

if you don't want to use condition, you can use modulo operator:

minutes = int(round(tm.minute, -1)) % 60

UPDATE

did you want something like this?

def timeround10(dt):
    a, b = divmod(round(dt.minute, -1), 60)
    return '%i:%02i' % ((dt.hour + a) % 24, b)

timeround10(datetime.datetime(2010, 1, 1, 0, 56, 0)) # 0:56
# -> 1:00

timeround10(datetime.datetime(2010, 1, 1, 23, 56, 0)) # 23:56
# -> 0:00

.. if you want result as string. for obtaining datetime result, it's better to use timedelta - see other responses ;)

0 votes
answered Aug 12, 2010 by omnifarious

This will get the 'floor' of a datetime object stored in tm rounded to the 10 minute mark before tm.

tm = tm - datetime.timedelta(minutes=tm.minute % 10,
                             seconds=tm.second,
                             microseconds=tm.microsecond)

If you want classic rounding to the nearest 10 minute mark, do this:

discard = datetime.timedelta(minutes=tm.minute % 10,
                             seconds=tm.second,
                             microseconds=tm.microsecond)
tm -= discard
if discard >= datetime.timedelta(minutes=5):
    tm += datetime.timedelta(minutes=10)

or this:

tm += datetime.timedelta(minutes=5)
tm -= datetime.timedelta(minutes=tm.minute % 10,
                         seconds=tm.second,
                         microseconds=tm.microsecond)
0 votes
answered Aug 1, 2012 by le-droid

General function to round a datetime at any time laps in seconds:

def roundTime(dt=None, roundTo=60):
   """Round a datetime object to any time laps in seconds
   dt : datetime.datetime object, default now.
   roundTo : Closest number of seconds to round to, default 1 minute.
   Author: Thierry Husson 2012 - Use it as you want but don't blame me.
   """
   if dt == None : dt = datetime.datetime.now()
   seconds = (dt.replace(tzinfo=None) - dt.min).seconds
   rounding = (seconds+roundTo/2) // roundTo * roundTo
   return dt + datetime.timedelta(0,rounding-seconds,-dt.microsecond)

Samples with 1 hour rounding & 30 minutes rounding:

print roundTime(datetime.datetime(2012,12,31,23,44,59,1234),roundTo=60*60)
2013-01-01 00:00:00

print roundTime(datetime.datetime(2012,12,31,23,44,59,1234),roundTo=30*60)
2012-12-31 23:30:00
0 votes
answered Jan 8, 2015 by waybehind

Based on Stijn Nevens and modified for Django use to round current time to the nearest 15 minute.

from datetime import date, timedelta, datetime, time

    def roundTime(dt=None, dateDelta=timedelta(minutes=1)):

        roundTo = dateDelta.total_seconds()

        if dt == None : dt = datetime.now()
        seconds = (dt - dt.min).seconds
        # // is a floor division, not a comment on following line:
        rounding = (seconds+roundTo/2) // roundTo * roundTo
        return dt + timedelta(0,rounding-seconds,-dt.microsecond)

    dt = roundTime(datetime.now(),timedelta(minutes=15)).strftime('%H:%M:%S')

 dt = 11:45:00

if you need full date and time just remove the .strftime('%H:%M:%S')

0 votes
answered Aug 13, 2015 by mza

I used Stijn Nevens code to great effect (thank you Stijn) and had a little add-on to share. Rounding up, down and rounding to nearest.

def round_time(dt=None, date_delta=timedelta(minutes=1), to='average'):
    """
    Round a datetime object to a multiple of a timedelta
    dt : datetime.datetime object, default now.
    dateDelta : timedelta object, we round to a multiple of this, default 1 minute.
    from:  http://stackoverflow.com/questions/3463930/how-to-round-the-minute-of-a-datetime-object-python
    """
    round_to = date_delta.total_seconds()

    if dt is None:
        dt = datetime.now()
    seconds = (dt - dt.min).seconds

    if to == 'up':
        # // is a floor division, not a comment on following line (like in javascript):
        rounding = (seconds + round_to) // round_to * round_to
    elif to == 'down':
        rounding = seconds // round_to * round_to
    else:
        rounding = (seconds + round_to / 2) // round_to * round_to

return dt + timedelta(0, rounding - seconds, -dt.microsecond)
0 votes
answered Aug 23, 2015 by stijn-nevens

From the best answer I modified to an adapted version using only datetime objects, this avoids having to do the conversion to seconds and makes the calling code more readable:

def roundTime(dt=None, dateDelta=datetime.timedelta(minutes=1)):
    """Round a datetime object to a multiple of a timedelta
    dt : datetime.datetime object, default now.
    dateDelta : timedelta object, we round to a multiple of this, default 1 minute.
    Author: Thierry Husson 2012 - Use it as you want but don't blame me.
            Stijn Nevens 2014 - Changed to use only datetime objects as variables
    """
    roundTo = dateDelta.total_seconds()

    if dt == None : dt = datetime.datetime.now()
    seconds = (dt - dt.min).seconds
    # // is a floor division, not a comment on following line:
    rounding = (seconds+roundTo/2) // roundTo * roundTo
    return dt + datetime.timedelta(0,rounding-seconds,-dt.microsecond)

Samples with 1 hour rounding & 15 minutes rounding:

print roundTime(datetime.datetime(2012,12,31,23,44,59),datetime.timedelta(hour=1))
2013-01-01 00:00:00

print roundTime(datetime.datetime(2012,12,31,23,44,49),datetime.timedelta(minutes=15))
2012-12-31 23:30:00
0 votes
answered Aug 28, 2015 by julienv
def get_rounded_datetime(self, dt, freq, nearest_type='inf'):

    if freq.lower() == '1h':
        round_to = 3600
    elif freq.lower() == '3h':
        round_to = 3 * 3600
    elif freq.lower() == '6h':
        round_to = 6 * 3600
    else:
        raise NotImplementedError("Freq %s is not handled yet" % freq)

    # // is a floor division, not a comment on following line:
    seconds_from_midnight = dt.hour * 3600 + dt.minute * 60 + dt.second
    if nearest_type == 'inf':
        rounded_sec = int(seconds_from_midnight / round_to) * round_to
    elif nearest_type == 'sup':
        rounded_sec = (int(seconds_from_midnight / round_to) + 1) * round_to
    else:
        raise IllegalArgumentException("nearest_type should be  'inf' or 'sup'")

    dt_midnight = datetime.datetime(dt.year, dt.month, dt.day)

    return dt_midnight + datetime.timedelta(0, rounded_sec)
0 votes
answered Sep 15, 2017 by james

Not the best for speed when the exception is caught, however this would work.

def _minute10(dt=datetime.utcnow()):
    try:
        return dt.replace(minute=round(dt.minute, -1))
    except ValueError:
        return dt.replace(minute=0) + timedelta(hours=1)

Timings

%timeit _minute10(datetime(2016, 12, 31, 23, 55))
100000 loops, best of 3: 5.12 µs per loop

%timeit _minute10(datetime(2016, 12, 31, 23, 31))
100000 loops, best of 3: 2.21 µs per loop
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