给定两个日期时间(start_date
和end_date
),我想生成这两个日期之间的其他日期时间的列表,新的日期时间由一个变量间隔分隔。例如,2011-10-10和2011-12-12之间每4天或从现在到明天下午19:00之间每8小时。
可能是大致相当于Dateperiod PHP类的东西。
在Python中实现这一点最有效的方法是什么?
发布于 2012-05-21 23:22:49
from datetime import date, datetime, timedelta
def perdelta(start, end, delta):
curr = start
while curr < end:
yield curr
curr += delta
>>> for result in perdelta(date(2011, 10, 10), date(2011, 12, 12), timedelta(days=4)):
... print result
...
2011-10-10
2011-10-14
2011-10-18
2011-10-22
2011-10-26
2011-10-30
2011-11-03
2011-11-07
2011-11-11
2011-11-15
2011-11-19
2011-11-23
2011-11-27
2011-12-01
2011-12-05
2011-12-09
适用于dates和datetime对象。您的第二个示例:
>>> for result in perdelta(datetime.now(),
... datetime.now().replace(hour=19) + timedelta(days=1),
... timedelta(hours=8)):
... print result
...
2012-05-21 17:25:47.668022
2012-05-22 01:25:47.668022
2012-05-22 09:25:47.668022
2012-05-22 17:25:47.668022
发布于 2015-12-08 18:07:49
我真的很喜欢@Martijn和@Óscar López的两个答案。让我在这两个答案之间提出我的组合解决方案。
from datetime import date, datetime, timedelta
def datetime_range(start, end, delta):
current = start
if not isinstance(delta, timedelta):
delta = timedelta(**delta)
while current < end:
yield current
current += delta
start = datetime(2015,1,1)
end = datetime(2015,1,31)
#this unlocks the following interface:
for dt in datetime_range(start, end, {'days': 2, 'hours':12}):
print dt
print dt
2015-01-01 00:00:00
2015-01-03 12:00:00
2015-01-06 00:00:00
2015-01-08 12:00:00
2015-01-11 00:00:00
2015-01-13 12:00:00
2015-01-16 00:00:00
2015-01-18 12:00:00
2015-01-21 00:00:00
2015-01-23 12:00:00
2015-01-26 00:00:00
2015-01-28 12:00:00
发布于 2019-02-01 23:23:06
到目前为止,这里给出的所有解决方案都是特定于start < stop的情况,但您可以很容易地对它们进行调整,以处理stop < start使用操作符模块的情况,如以下代码所示,改编自@MartijnPieters的答案。
import datetime
import operator
def time_range(start: datetime.datetime, stop, step: datetime.timedelta):
"Imitate range function for datetimes instead of ints."
sec = step.total_seconds()
if sec == 0:
raise ValueError("step must not be 0 seconds")
if sec < 0:
compare = operator.gt
else:
compare = operator.lt
x = start
while compare(x, stop):
yield x
x += step # immutable
https://stackoverflow.com/questions/10688006
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