递归列表函数

Question

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轴中。它应该扫描列表并应用以下规则：

• 如果在期间内找到无值：将无替换为datetime.date.today日期（）

• 如果一段时间在内开始，而在另一段时间内结束：删除它。

• 如果一段时间在之前开始，但在另一段时间内结束：延长开始日期。

• 如果一段时间在之后结束，但在另一段时间内开始：延长结束日期。

• 如果一段时间在之后开始，而在另一段时间之后结束：保留它，这是一个单独的时间段。

• 如果一段时间在另一段时间之前开始并且在另一段时间之前结束：保留它，这是一个单独的时间段。

举一个输入和所需输出的例子可能要容易得多（假设值是用datetime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)]

Answer 1

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()

可用的格式是here

# 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)]

尝试一下你的任何一个日期列表，你应该观察到同样的行为。你知道吗

HTH公司