这是高效的-即使在开始和结束之间有一万天的时间-而且仍然非常灵活(它在sum函数内最多迭代7次)：def intervening_weekdays(start, end, inclusive=True, weekdays=[0, 1, 2, 3, 4]):

if isinstance(start, datetime.datetime):

start = start.date() # make a date from a datetime

if isinstance(end, datetime.datetime):

end = end.date() # make a date from a datetime

if end < start:

# you can opt to return 0 or swap the dates around instead

raise ValueError("start date must be before end date")

if inclusive:

end += datetime.timedelta(days=1) # correct for inclusivity

try:

# collapse duplicate weekdays

weekdays = {weekday % 7 for weekday in weekdays}

except TypeError:

weekdays = [weekdays % 7]

ref = datetime.date.today() # choose a reference date

ref -= datetime.timedelta(days=ref.weekday()) # and normalize its weekday

# sum up all selected weekdays (max 7 iterations)

return sum((ref_plus - start).days // 7 - (ref_plus - end).days // 7

for ref_plus in

(ref + datetime.timedelta(days=weekday) for weekday in weekdays))

这将分别为start和{}获取{}和{}的{}对象。在

另外，您可以在闭合(inclusive=True)和半开(inclusive=False)间隔之间进行选择。在

默认情况下，它计算日期之间的工作日数，但也可以选择任意一组工作日(周末：weekdays=[5, 6])或单个工作日(星期三：weekdays=2)。在