我正在尝试创建一个递归Python函数，该函数接受句点列表并将它们合并到一个干净的时间线中。它应该扫描列表并应用以下规则：
-
如果在期间内找到None的值：将None替换为datetime.date.today（）
-
如果期间在内开始，结束于另一个期间：删除它。
-
如果期间在之前开始，但结束于另一个期间：延长开始日期。
-
如果期间在之后结束，但开始于另一个期间：延长结束日期
-
如果一个时期在之后开始，而在留它，这是一个单独的时期
-
如果一个周期在之前开始，而在另一个周期之前结束：保留它，则它是一个单独的周期
举一个输入和期望输出的例子可能要容易得多（假设值是用日期时间格式化的）:
[I] = [(01/2011, 02/2015), (04/2012, 08/2014), (09/2014, 03/2015), (05/2015, 06/2016)]
[O] = [(01/2011, 03/2015), (05/2015, 06/2016)]
# Notice how the output has produced a set of minimum length whilst covering all periods.
[I] = [(07/2011, 02/2015), (04/2012, 08/2014), (09/2014, 04/2015), (06/2015, None)]
[O] = [(07/2011, 04/2015), (06/2015, date.today())]
# Also, notice how the output has amended None, so it can compare dates.
感谢@khredos，我写了以下内容，但它仍然没有输出所需的最小字符串：
from datetime import datetime
# Here is an example list of time periods
periods = [('01/2011', '02/2015'), ('04/2012', '08/2014'), ('09/2014', '03/2015'), ('05/2015', '06/2016')]
# this lambda function converts a string of the format you have specified to a
# datetime object. If the string is None or empty, it uses today's date
cvt = lambda ds: datetime.strptime(ds, '%m/%Y') if ds else datetime.today()
# Now convert your original list to an iterator that contains datetime objects
periods = list(map(lambda s_e : (cvt(s_e[0]), cvt(s_e[1])), periods))
# Next get the start dates into one list and the end dates into another
starts, ends = zip(*periods)
# Finally get the timeline by sorting the two lists
timeline = sorted(starts + ends)
# Output: [datetime.datetime(2011, 1, 1, 0, 0), datetime.datetime(2012, 4, 1, 0, 0), datetime.datetime(2014, 8, 1, 0, 0), datetime.datetime(2014, 9, 1, 0, 0), datetime.datetime(2015, 2, 1, 0, 0), datetime.datetime(2015, 3, 1, 0, 0), datetime.datetime(2015, 5, 1, 0, 0), datetime.datetime(2016, 6, 1, 0, 0)]
from datetime import datetime
# Here is an example list of time periods
periods = [('01/2011', '02/2015'), ('04/2012', '08/2014'), ('09/2014', '03/2015'), ('05/2015', '06/2016')]
# this lambda function converts a string of the format you have specified to a
# datetime object. If the string is None or empty, it uses today's date
cvt = lambda ds: datetime.strptime(ds, '%m/%Y') if ds else datetime.today()
可用的格式在这里
# Now convert your original list to an iterator that contains datetime objects
periods = list(map(lambda s_e : (cvt(s_e[0]), cvt(s_e[1])), periods))
# Next get the start dates into one list and the end dates into another
starts, ends = zip(*periods)
# Finally get the timeline by sorting the two lists
timeline = sorted(starts + ends)
输出应类似于
[datetime.datetime(2011, 1, 1, 0, 0), datetime.datetime(2012, 4, 1, 0, 0), datetime.datetime(2014, 8, 1, 0, 0), datetime.datetime(2014, 9, 1, 0, 0), datetime.datetime(2015, 2, 1, 0, 0), datetime.datetime(2015, 3, 1, 0, 0), datetime.datetime(2015, 5, 1, 0, 0), datetime.datetime(2016, 6, 1, 0, 0)]
试着把它和你的约会清单放在一起，你应该观察到同样的行为。
HTH