从最佳答案我使用,这避免了必须执行转换到秒仅datetime对象修改为适于版本并使得调用代码更易读:
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)
样品用时1小时四舍五入&15分钟四舍五入:
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
啊,但是那么这里的问题是,必须时刻提高以及 – 2010-08-12 01:10:35
@Lucas曼乔 - 我的解决办法也工作正常,我觉得更有意义。 – Omnifarious 2010-08-12 06:03:03