Question

我有三个列表定义何时应该执行任务：

分钟：0-59之间的整数列表，表示执行应该发生的一小时的分钟数;
小时：0-23之间的整数列表，表示执行一天的小时数
day_of_week ：0-6的整数列表，其中星期日= 0，星期六= 6，表示应执行一周的日期。

有没有一种简单的方法来计算在Python下次执行之前的timedelta是什么时候？

谢谢！

修改例如，如果我们有以下列表：

day_of_week = [0]
hour = [1]
minute = [0, 30]

任务应该每周运行两次，每周日1:00和1:30。我想根据当前时间计算timedelta直到下一次出现。

Answer 1

使用dateutil（编辑以解决OP的更新问题）：

import datetime
import random
import dateutil.relativedelta as dr
import itertools

day_of_week = [1,3,5,6]
hour = [1,10,15,17,20]
minute = [4,34,51,58]

now=datetime.datetime.now()
deltas=[]

for min,hr,dow in itertools.product(minute,hour,day_of_week):
    # dateutil convention: Monday = 0, Sunday = 6.
    next_dt=now+dr.relativedelta(minute=min,hour=hr,weekday=dow)
    delta=next_dt-now
    deltas.append(delta)

deltas.sort()

这是下一次的时间：

print(deltas[0])
# 4 days, 14:22:00

这是相应的日期时间：

print(now+deltas[0])
# 2010-09-02 01:04:23.258204

请注意，dateutil使用约定Monday = 0，Sunday = 6。

Answer 2

如果有人感兴趣，这是我使用~unutbu建议开发的代码。它的主要优点是它可以很好地扩展。

import datetime
import dateutil.relativedelta as dr

def next_ocurrance(minutes, hours, days_of_week):
    # days_of_week convention: Sunday = 0, Saturday = 6
    # dateutil convention: Monday = 0, Sunday = 6

    now = datetime.datetime.now()
    weekday = now.isoweekday()
    execute_this_hour = weekday in days_of_week \
                        and now.hour in hours \
                        and now.minute < max(minutes)

    if execute_this_hour:
        next_minute = min([minute for minute in minutes
                           if minute > now.minute])
        return now + dr.relativedelta(minute=next_minute,
                                      second=0,
                                      microsecond=0)
    else:
        next_minute = min(minutes)

    execute_today = weekday in day_of_week \
                    and (now.hour < max(hours) or execute_this_hour)

    if execute_today:
        next_hour = min([hour for hour in hours if hour > now.hour])
        return now + dr.relativedelta(hour=next_hour,
                                      minute=next_minute,
                                      second=0,
                                      microsecond=0)
    else:
        next_hour = min(hours)
        next_day = min([day for day in days_of_week if day > weekday] \
                       or days_of_week)

        return now + dr.relativedelta(weekday=(next_day - 1) % 7,
                                      hour=next_hour,
                                      minute=next_minute,
                                      second=0,
                                      microsecond=0)
if __name__=='__main__':
    day_of_week = [4]
    hour = [1, 10, 12, 13]
    minute = [4, 14, 34, 51, 58]
    print next_ocurrance(minute, hour, day_of_week)

如何计算timedelta直到下次执行预定事件

2 个答案: