Python 对日期字符串列表进行排序
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Sort list of date strings
提问by user2480542
I have an arbitrary list of date strings (mm-yyyy) as follows:
我有一个任意的日期字符串列表 (mm-yyyy)，如下所示：
d = ['09-2012', '04-2007', '11-2012', '05-2013', '12-2006', '05-2006', '08-2007'...]
I need this list to be sorted first by the level of years (ascending), then on the level of months (ascending)..so that the logical ordering can be:
我需要先按年级别（升序）对这个列表进行排序，然后按月级别（升序）排序。这样逻辑顺序可以是：
d_ordered = ['05-2006', '12-2006', '04-2007', '08-2007', '09-2012', '11-2012', '05-2013' ...]
How can I achieve this?
我怎样才能做到这一点？
采纳答案by amatellanes
Try this:
尝试这个：
import datetime
d = ['09-2012', '04-2007', '11-2012', '05-2013', '12-2006', '05-2006', '08-2007']
sorted(d, key=lambda x: datetime.datetime.strptime(x, '%m-%Y'))
回答by TerryA
Use sorted()with a key:
使用sorted()一键：
>>> d = ['09-2012', '04-2007', '11-2012', '05-2013', '12-2006', '05-2006', '08-2007']
>>> def sorting(L):
... splitup = L.split('-')
... return splitup[1], splitup[0]
...
>>> sorted(d, key=sorting)
['05-2006', '12-2006', '04-2007', '08-2007', '09-2012', '11-2012', '05-2013']
It's better to use a function here instead of a lambda to prevent calling split()twice (and it looks a bit neater :))
最好在这里使用函数而不是 lambda 来防止调用split()两次（而且看起来更整洁 :)）
Note that this will return the sorted list. If you want to sort it in place, use .sort():
请注意，这将返回排序列表。如果要就地排序，请使用.sort()：
>>> d.sort(key=sorting)
>>> d
['05-2006', '12-2006', '04-2007', '08-2007', '09-2012', '11-2012', '05-2013']
