有没有一种方法可以通过使用Python Lambda降至创建的时间进行排序？
我正在与Sqlalchemy合作查询包含事件列表的数据库。每个事件都有创建的时间和一个ARI（亚马逊资源编号）。我可以从一个函数调用中返回最多记录，我想从最新的第一个开始回到记录。为此，我需要手动传递正确对记录的lambda函数；该参数的文档在这里：
cols (default=None):
A function which takes a DBObject class and returns a tuple of columns (or SQLAlchemy
expressions) used to compare it. This overrides the default sorting behaviour.
Example: key=lambda x: (x.created, x.ari)
如您所见，默认情况下，它按X.创建，然后按X.Ari进行升级。实际上，我需要做与此相反的事情，然后按X.创建，然后按x.ari进行排序，以 descing 顺序进行排序。有什么办法可以用lambda做到这一点？使用STROT（）或排序（）并不是我使用此预先存在的功能的选择。
I'm working with SqlAlchemy to query a database containing a list of events. Each event has a created time and an ari (amazon resource number). There's a maximum number of records I can get back from one function call, and I want to get back the records starting with the most recent first. To do this I need to manually pass in a lambda function that sorts the records properly; documentation for the argument is here:
cols (default=None):
A function which takes a DBObject class and returns a tuple of columns (or SQLAlchemy
expressions) used to compare it. This overrides the default sorting behaviour.
Example: key=lambda x: (x.created, x.ari)
As you can see, by default it sorts by x.created, then x.ari, both in ascending order. I actually need to do the opposite of this and sort by x.created, then x.ari, in descending order. Is there any way I can do this with just a lambda? Using sort() or sorted() is not really an option due to me using this pre-existing function.
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