我写了一些代码来确定两个日期时间之间每个时隙的小时数。我的代码非常大/复杂，而我之前预计它会相对简短/容易。是否有人建议编写较短/更快/更好的代码以达到相同的结果？
我想对两个日期时间之间的每个时隙的小时数求和，分别称为开始和 end 。它们可以相隔数天，也可以在同一小时内放置，但绝不相同。时隙是预定义的，并且对于工作日和周末而言是不同的。 开始和结束的时间不必等于时隙的开始或结束时间。
5个不同大小的插槽中有工作日，
周末时段存在于3个不同大小的广告位中，分别是：
当我输入以下开始和结束日期时间时，
我希望得到以下结果：
class MyDateRange:
def __init__(self, name, start_datetime, end_datetime):
self.name = name
self.start_datetime = start_datetime
self.end_datetime = end_datetime
def hours_per_timeslot(date_range):
# time-slots week days
time_slots_week = {
1: {"begin": datetime.time(hour=0), "end": datetime.time(hour=5), "diff": datetime.timedelta(hours=5)},
2: {"begin": datetime.time(hour=5), "end": datetime.time(hour=10), "diff": datetime.timedelta(hours=5)},
3: {"begin": datetime.time(hour=10), "end": datetime.time(hour=14), "diff": datetime.timedelta(hours=4)},
4: {"begin": datetime.time(hour=14), "end": datetime.time(hour=20), "diff": datetime.timedelta(hours=6)},
5: {"begin": datetime.time(hour=20), "end": datetime.time(hour=23, minute=59, second=59, microsecond=999999),
"diff": datetime.timedelta(hours=4)}
}
# time-slots weekend days
time_slots_weekend = {
1: {"begin": datetime.time(hour=0), "end": datetime.time(hour=8), "diff": datetime.timedelta(hours=8)},
2: {"begin": datetime.time(hour=8), "end": datetime.time(hour=19), "diff": datetime.timedelta(hours=11)},
3: {"begin": datetime.time(hour=19), "end": datetime.time(hour=23, minute=59, second=59, microsecond=999999),
"diff": datetime.timedelta(hours=5)}
}
# dictionary to store outcome of week days
dict_week = {
1: datetime.timedelta(),
2: datetime.timedelta(),
3: datetime.timedelta(),
4: datetime.timedelta(),
5: datetime.timedelta(),
}
# dictionary to store outcome of weekend days
dict_weekend = {
1: datetime.timedelta(),
2: datetime.timedelta(),
3: datetime.timedelta(),
}
begin = date_range.start_datetime
end = date_range.end_datetime
delta = end - begin
# calculate the hours per time-slot when begin and en are on the same date.
if begin.date() == end.date():
day = calendar.day_name[begin.weekday()]
# in case it is weekend.
if day in ["Saturday", "Sunday"]:
for k, v in time_slots_weekend.items():
# in case begin and end are in the same time-slot
if v["begin"] <= begin.time() < v["end"] and v["begin"] <= end.time() < v["end"]:
dict_weekend[k] = end - begin
# in case they are not in the same time-slot
elif v["begin"] <= begin.time() < v["end"]:
dict_weekend[k] = datetime.datetime.combine(begin.date(), v["end"]) - begin
while True:
k += 1
v = time_slots_weekend[k]
if v["begin"] <= end.time() < v["end"]:
dict_weekend[k] = end - datetime.datetime.combine(end.date(), v["begin"])
break
else:
dict_weekend[k] = time_slots_week[k]["diff"]
# in case it is week.
else:
for k, v in time_slots_week.items():
if v["begin"] <= begin.time() < v["end"] and v["begin"] <= end.time() < v["end"]:
dict_week[k] = end - begin
elif v["begin"] <= begin.time() < v["end"]:
dict_week[k] = datetime.datetime.combine(begin.date(), v["end"]) - begin
while True:
k += 1
v = time_slots_week[k]
if v["begin"] <= end.time() < v["end"]:
dict_week[k] = end - datetime.datetime.combine(end.date(), v["begin"])
break
else:
dict_week[k] = time_slots_week[k]["diff"]
# calculate the hours per time-slot when begin and end are on different date.
else:
# calculate the hours per time-slot for begin time until end of day
day_begin = calendar.day_name[begin.weekday()]
# in case it is weekend.
if day_begin in ["Saturday", "Sunday"]:
for k, v in time_slots_weekend.items():
if v["begin"] <= end.time() < v["end"]:
dict_weekend[k] += (datetime.datetime.combine(begin.date(), v["end"])-begin)
k += 1
while k <= 3:
dict_weekend[k] += time_slots_weekend[k]["diff"]
k += 1
# in case it is week.
else:
for k, v in time_slots_week.items():
if v["begin"] <= end.time() < v["end"]:
dict_week[k] += datetime.datetime.combine(begin.date(), v["end"]) - begin
k += 1
while k <= 5:
dict_week[k] += time_slots_week[k]["diff"]
k += 1
# calculate the hours per time-slot for beginning of day until end time
day_end = calendar.day_name[end.weekday()]
# in case it is weekend.
if day_end in ["Saturday", "Sunday"]:
for k, v in time_slots_weekend.items():
if v["begin"] <= end.time() < v["end"]:
dict_weekend[k] += end - datetime.datetime.combine(end.date(), v["begin"])
k -= 1
while k > 0:
dict_weekend[k] += time_slots_weekend[k]["diff"]
k -= 1
# in case it is week.
else:
for k, v in time_slots_week.items():
if v["begin"] <= end.time() < v["end"]:
dict_week[k] += end - datetime.datetime.combine(end.date(), v["begin"])
k -= 1
while k > 0:
dict_week[k] += time_slots_week[k]["diff"]
k -= 1
# in case there are days between begin and end,
if delta.days > 1:
counted_days = {}
for i in range(delta.days-1):
day = calendar.day_name[(begin + datetime.timedelta(days=i + 1)).weekday()]
counted_days[day] = counted_days[day] + 1 if day in counted_days else 1
for k1, v1 in counted_days.items():
if k1 in ["Saturday", "Sunday"]:
for k2 in dict_weekend.keys():
dict_weekend[k2] += (time_slots_weekend[k2]["diff"] * v1)
else:
for k2 in dict_week.keys():
dict_week[k2] += (time_slots_week[k2]["diff"] * v1)
# put the results together and convert them into hours.
results = {}
for k, v in dict_week.items():
results["time_slot_week_day_"+str(k)] = v.seconds/60/60
for k, v in dict_weekend.items():
results["time_slot_weekend_day_"+str(k)] = v.seconds/60/60
return results
dr = MyDateRange("test", datetime.datetime(year=2019, month=4, day=21, hour=8), datetime.datetime(year=2019, month=4, day=27, hour=15))
print(hours_per_timeslot(dr))
答案 0 :(得分：0)
使用时间，日期时间模块？然后计算秒数然后将其转换回去，不是吗？还是我错过了您不想使用此功能的原因？
from datetime import datetime, time
def date_diff_in_Seconds(dt2, dt1):
timedelta = dt2 - dt1
return timedelta.days * 24 * 3600 + timedelta.seconds
#Specified date
date1 = datetime.strptime('2018-01-01 01:00:00', '%Y-%m-%d %H:%M:%S')
#Current date
date2 = datetime.now()
print("\n%d seconds" %(date_diff_in_Seconds(date2, date1)))
print()