Python sort list with None at the end -

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i have homogeneous list of objects none, can contain 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 way without reinventing wheel list sorted usual python way, none values @ end of list, that:

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

i feel here can trick "key" parameter

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

this constructs tuple each element in list, if value none tuple (true, none), if value else (false, x) (where x value). since tuples sorted item item, means non-none elements come first (since false < true), , sorted value.


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