My own accounting python euler problem

Ozz notvalid at wathever.com
Sun Nov 8 13:20:42 CET 2009


Oops,

> For listing all different subsets of a list (This is what I came up 
> with. Can it be implemented shorter, btw?):
> 
> def subsets(L):
>         S = []
>         if (len(L) == 1):
>                 return [L, []]

better to check for the empty set too, thus;

if (len(L) == 0):
                 return [[]]

The order of the sets looks better too;

 >>> subset.subsets([1,2,3])
[[], [1], [2], [2, 1], [3], [3, 1], [3, 2], [3, 2, 1]]

cheers,






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