递归列表功能

Question

我试图创建一个接受时间的列表，并将它们整合到一个干净的时间表递归Python函数。它应该在列表中扫描并应用这些规则：

• 如果值没有是在周期中发现：更换无与datetime.date.today（）

• 如果一段内和开始内结束另一个时期：删除。

• 如果一段时间之前开始内但两端另一段：延长起始日期。

• 如果一段结束但内开始另一个周期之后：延伸结束日期。

• 如果和经过一段时间开始结束后另一段：保持它，它是一个独立的时期。

• 如果和前一段时间开始之前结束另一个时期：保持它，它是一个独立的时期。

这也许是更容易得到的输入的示例，并且所期望的输出（假定值被格式化与日期时间）：

[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. 

由于@khredos，我已写入的下面，但它仍然不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() 

种

可用的格式都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) 

输出应该类似于

[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)] 

与你有日期的任何名单试试吧，你应该遵守相同的行为。

HTH