Sort list while pushing None values to the end

Question

I have a homogeneous list of objects with None, but it can contain any type of values. Example:

>>> l = [1, 3, 2, 5, 4, None, 7]
>>> sorted(l)
[None, 1, 2, 3, 4, 5, 7]
>>> sorted(l, reverse=True)
[7, 5, 4, 3, 2, 1, None]

Is there a way without reinventing the wheel to get the list sorted the usual python way, but with None values at the end of the list, like that:

[1, 2, 3, 4, 5, 7, None]

I feel like here can be some trick with "key" parameter

Answer 1

>>> l = [1, 3, 2, 5, 4, None, 7]
>>> sorted(l, key=lambda x: (x is None, x))
[1, 2, 3, 4, 5, 7, None]

This constructs a tuple for each element in the list, if the value is None the tuple with be (True, None), if the value is anything else it will be (False, x) (where x is the value). Since tuples are sorted item by item, this means that all non-None elements will come first (since False < True), and then be sorted by value.