All permutations of a list

[Emile van Sebille]

def perms3(list):
  return [[[list[i]] + p for i in range(len(list))\
    for p in perms3(list[:i] + list[i+1:])],[[]]] [not list]

This one works, ugly as it is.  Now, if the items are
reversed so that the empty list is element 0, is there a way
to return either a 0 or 1 other that 'not not list'?



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Emile van Sebille
emile at fenx.com
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