查找日期列表之间的所有相交期间（用于客房预订）

Question

我需要编写一个组织房间预订的软件。
在这种情况下，我需要为日历记录标题栏，以重新整理房间。
基本上，那里的房间和一个房间一次只能预订一次，如果发生冲突，则标题必须突出显示。
基本上，它需要看起来像这样：

为此，我需要一个红色的日期范围列表，以显示一次以上的预订在同一时间范围内使用的日期范围。
我可以使用某种Python库吗？
基本上，它需要获取“日期日期列表”并吐出不带日期日期的日期列表。

Answer 1

按开始时间升序，然后按结束时间降序对时间间隔进行排序。

然后遍历列表，将每个开始时间与您看到的最大先前结束时间进行比较。如果开始时间早于该结束时间，则您会发现从this_start到min(max_prev_end, this_end)的冲突。

Answer 2

...基本上，它需要获取一份日期日期列表，并吐出一系列非日期日期的日期列表...

为此设置的完美。下面的示例代码将查找所有未使用的日期，从那里开始构建（ini，结束）元组将是正确的，因为当您找到非连续日期时，每个元组都会结束。签出：

from datetime import timedelta, datetime as dt

ranges = [
    ("2018-1-1", "2018-1-31"),   # 1
    ("2018-1-20", "2018-2-10"),  # 2
    ("2018-2-25", "2018-3-15"),  # 3
    ("2018-2-27", "2018-3-20")   # 4
]

# Range 1 and 2 overlaps staring at day (2018-1-20) to day (2018-1-31)
# You have an open window from (2018-2-10) to (2018-2-25)
# And again overlaps in ranges 3 and 4.


# Lets take those ranges specifications in real dates objects.
date_ranges = []
for ini, end in ranges:
    date_ranges.append(
        (
            dt.strptime(ini, "%Y-%m-%d"), 
            dt.strptime(end, "%Y-%m-%d")
        )
    )

# Now in order to be able to use set theory, you need to
# convert those date ranges into sets of data that 
# match your problem.

# I will use set of days, when every element is a different day,
# I,m assuming a day is the unit for meassuring reservation time.

date_sets = []

for ini, end in date_ranges:
    delta = end - ini
    # We don't convert to sets here becouse
    # we want to keep the order til last minunte.
    # Just in case ;)
    dates_ = [ini + timedelta(n) for n in range(delta.days + 1)]
    date_sets.append(dates_)


# Let get also all days invloved in the full date range,
# we need a reference so we can tell what days are free.


all_dates = []
# Assuming date ranges specification are ordered, which is
# the general case.

# Remember that thing about order
# I told earlier?  :)
ini = date_sets[0][0]
end = date_sets[-1][-1]
delta =  end - ini
dates_ = [ini + timedelta(n) for n in range(delta.days + 1)]
all_dates.append(dates_)


# Now we are ready to work with sets.
all_dates = set(*all_dates)
date_sets = map(set, date_sets)

# Having the sets we can operate on them
# see: https://docs.python.org/2/library/sets.html

# Sice you want the days that are free for rental, 
# we are getting symmetric difference (elements in either s or t but not both)
free_for_rental = all_dates ^ (set.union(*date_sets))


# Printing the result so you can check it.
list_of_free_dates = sorted(list(free_for_rental))

for day in list_of_free_dates:
    print(day)

回复：

2018-02-11 00:00:00
2018-02-12 00:00:00
2018-02-13 00:00:00
2018-02-14 00:00:00
2018-02-15 00:00:00
2018-02-16 00:00:00
2018-02-17 00:00:00
2018-02-18 00:00:00
2018-02-19 00:00:00
2018-02-20 00:00:00
2018-02-21 00:00:00
2018-02-22 00:00:00
2018-02-23 00:00:00
2018-02-24 00:00:00