Rounding time in Python

问题

What would be an elegant, efficient and Pythonic way to perform a h/m/s rounding operation on time related types in Python with control over the rounding resolution?

My guess is that it would require a time modulo operation. Illustrative examples:

20:11:13 % (10 seconds) => (3 seconds)
20:11:13 % (10 minutes) => (1 minutes and 13 seconds)

Relevant time related types I can think of:

datetime.datetime \ datetime.time
struct_time

回答1:

For a datetime.datetime rounding, see this function: https://stackoverflow.com/a/10854034/1431079

Sample of use:

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

回答2:

How about use datetime.timedeltas:

import time
import datetime as dt

hms=dt.timedelta(hours=20,minutes=11,seconds=13)

resolution=dt.timedelta(seconds=10)
print(dt.timedelta(seconds=hms.seconds%resolution.seconds))
# 0:00:03

resolution=dt.timedelta(minutes=10)
print(dt.timedelta(seconds=hms.seconds%resolution.seconds))
# 0:01:13

回答3:

This will round up time data to a resolution as asked in the question:

import datetime as dt
current = dt.datetime.now()
current_td = dt.timedelta(hours=current.hour, minutes=current.minute, seconds=current.second, microseconds=current.microsecond)

# to seconds resolution
to_sec = dt.timedelta(seconds=round(current_td.total_seconds()))
print dt.datetime.combine(current,dt.time(0))+to_sec

# to minute resolution
to_min = dt.timedelta(minutes=round(current_td.total_seconds()/60))
print dt.datetime.combine(current,dt.time(0))+to_min

# to hour resolution
to_hour = dt.timedelta(hours=round(current_td.total_seconds()/3600))
print dt.datetime.combine(current,dt.time(0))+to_hour

回答4:

I think I'd convert the time in seconds, and use standard modulo operation from that point.

20:11:13 = 20*3600 + 11*60 + 13 = 72673 seconds

72673 % 10 = 3

72673 % (10*60) = 73

This is the easiest solution I can think about.

回答5:

You can convert both times to seconds, do the modulo operati

from datetime import time

def time2seconds(t):
    return t.hour*60*60+t.minute*60+t.second

def seconds2time(t):
    n, seconds = divmod(t, 60)
    hours, minutes = divmod(n, 60)
    return time(hours, minutes, seconds)

def timemod(a, k):
    a = time2seconds(a)
    k = time2seconds(k)
    res = a % k
    return seconds2time(res)

print(timemod(time(20, 11, 13), time(0,0,10)))
print(timemod(time(20, 11, 13), time(0,10,0)))

Outputs:

00:00:03
00:01:13

回答6:

I use following code snippet to round to the next hour:

import datetime as dt

tNow  = dt.datetime.now()
# round to the next full hour
tNow -= dt.timedelta(minutes = tNow.minute, seconds = tNow.second, microseconds =  tNow.microsecond)
tNow += dt.timedelta(hours = 1)

回答7:

Here is a lossy* version of hourly rounding:

dt = datetime.datetime
now = dt.utcnow()
rounded = dt.utcfromtimestamp(round(now.timestamp() / 3600, 0) * 3600)

Same principle can be applied to different time spans.

*The above method assumes UTC is used, as any timezone information will be destroyed in conversion to timestamp.

回答8:

def round_dt_to_seconds(dt):
    datetime.timedelta(seconds=dt.seconds)

来源：https://stackoverflow.com/questions/6806467/rounding-time-in-python

标签

python

datetime

time

modulo

rounding