我正在尝试创建一个递归的Python函数,它接受一个时间段列表,并将它们合并成一个干净的时间线。它应该扫描列表并应用以下规则:
如果在时间段中找到 None 的值: 用datetime.date.today()替换None
如果一个时间段在另一个时间段内开始和在另一个时间段内结束: 删除它.
如果一个时间段在另一个时间段之前开始但在另一个时间段内结束:延长开始日期。
如果一个时间段在另一个时间段内开始但在另一个时间段之后结束:延长结束日期。
如果一个时间段在另一个时间段之后开始和在另一个时间段之后结束:保留它,这是一个独立的时间段。
如果一个时间段在另一个时间段之前开始和在另一个时间段之前结束:保留它,这是一个独立的时间段。
可能更容易通过示例说明输入和期望的输出(假设值已用datetime格式化):
[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 的帮助,但我仍然无法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)]