您可以通过添加timedelta 对象来对datetime 对象进行算术运算。
如果每个周期的间隔不是总数的精确除数,您可能需要准确确定所需的行为,但在这种情况下,此示例将给出最终的短周期。
import datetime
tstart = datetime.datetime(2020,8,24,9,30)
tend = datetime.datetime(2020,8,24,11,30)
interval = datetime.timedelta(minutes=30)
periods = []
period_start = tstart
while period_start < tend:
period_end = min(period_start + interval, tend)
periods.append((period_start, period_end))
period_start = period_end
print(periods)
这给出(插入换行符以提高可读性):
[(datetime.datetime(2020, 8, 24, 9, 30), datetime.datetime(2020, 8, 24, 10, 0)),
(datetime.datetime(2020, 8, 24, 10, 0), datetime.datetime(2020, 8, 24, 10, 30)),
(datetime.datetime(2020, 8, 24, 10, 30), datetime.datetime(2020, 8, 24, 11, 0)),
(datetime.datetime(2020, 8, 24, 11, 0), datetime.datetime(2020, 8, 24, 11, 30))]
对于你想要的字符串输出格式,你可以这样做:
def format_time(dt):
return dt.strftime("%Y-%m-%d %H:%M")
print(['{} - {}'.format(format_time(start), format_time(end))
for start, end in periods])
给予:
['2020-08-24 09:30 - 2020-08-24 10:00',
'2020-08-24 10:00 - 2020-08-24 10:30',
'2020-08-24 10:30 - 2020-08-24 11:00',
'2020-08-24 11:00 - 2020-08-24 11:30']