Round time to nearest hour python

Question

I was trying to round off time to the nearest hour in python in a dataframe.

Suppose if a timestamp is 2017-11-18 0:16 it should come as 2017-11-18 0:00 and 2017-11-18 1:56 should round off as 2017-11-18 2:00

Please (re-)read "[ask]", then [edit] your question to include a [mcve] that demonstrates what you've tried and what didn't work. — Kevin J. Chase, Feb 22 '18 at 22:55
@user123, did one of the below solutions help? if so, feel free to accept one (tick on left). — jpp, Feb 28 '18 at 11:34

score 42 · Accepted Answer · answered Feb 22 '18 at 22:45

42

I experimented a bit with jpp but ended up with a different solution as adding one hour at hour 23 crashed the thing.

from datetime import datetime, timedelta

now = datetime.now()

def hour_rounder(t):
    # Rounds to nearest hour by adding a timedelta hour if minute >= 30
    return (t.replace(second=0, microsecond=0, minute=0, hour=t.hour)
               +timedelta(hours=t.minute//30))

print(now)
print(hour_rounder(now))

Returns:

2018-02-22 23:42:43.352133
2018-02-23 00:00:00

answered Feb 22 '18 at 22:45

Anton vBR

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hour=t.hour necessary? – kmecpp Jul 30 '22 at 16:38
@kmecpp hi there. Yes, as this function rounds the hour we need t.hour to retrieve the hour from our datetime. – Anton vBR Aug 02 '22 at 19:12
But we're doing t.replace() on the original datetime. Wont it already have that hour? – kmecpp Aug 02 '22 at 21:17
@kmecpp True! But this follow the Python Zen. Explicit is better than implicit. So if we would exclude that someone might ask the opposite. – Anton vBR Aug 04 '22 at 07:44

score 18 · Answer 2 · edited Feb 22 '18 at 22:48

18

import pandas as pd

pd.Timestamp.now().round('60min').to_pydatetime()

Returns:

datetime.datetime(2018, 2, 23, 0, 0)

edited Feb 22 '18 at 22:48

Anton vBR

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answered Feb 22 '18 at 22:20

user123

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score 12 · Answer 3 · edited Feb 22 '18 at 22:45

12

This is one way.

from datetime import datetime

now = datetime.now()

def rounder(t):
    if t.minute >= 30:
        return t.replace(second=0, microsecond=0, minute=0, hour=t.hour+1)
    else:
        return t.replace(second=0, microsecond=0, minute=0)

now           # 2018-02-22 22:03:53.831589
rounder(now)  # 2018-02-22 22:00:00.000000

edited Feb 22 '18 at 22:45

Anton vBR

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answered Feb 22 '18 at 22:04

jpp

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5

Sorry for experimenting with your solution. I rolled it back. Basically I found that it would crash when hour 23 turned hour 24. I could have changed your solution but ended up posting my own. I gave you an upvote though. – Anton vBR Feb 22 '18 at 22:47

score 5 · Answer 4 · answered Dec 01 '18 at 14:07

5

There is a general function to round a datetime at any time lapse in seconds here

Sample:

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

answered Dec 01 '18 at 14:07

Le Droid

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score 0 · Answer 5 · answered Mar 27 '23 at 01:16

Here is one way to do it (based on another solution provided here)

def round_to_closest_hour(dttm):
    add_hour = True if dttm.minute >= 30 else False
    dttm = dttm.replace(second=0, microsecond=0, minute=0)
    if add_hour:
        dttm += timedelta(hours=1)
    return dttm

Basically, check if an hour is needed to be added or not (depending on whether it's past 30 minutes mark). Then reset minutes, seconds, microseconds and add an hour if required.

score 0 · Answer 6 · answered Jul 04 '23 at 17:23

0

For those who need to change a whole column:

df = df['Timestamp'].dt.round('60min')

You can also use ceil() and floor() functions to replace round, depending on the rules you want to approximate.

answered Jul 04 '23 at 17:23

Songhua Hu

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Round time to nearest hour python

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