for those duplicates, of course. On the other hand, I think viewing it
as a powerset is the most 'natural' in this case. (imo permutations are
about the order of objects, not about whether the objects are included
in a set or not)
cheers,
Ozz
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Dan Bishop schreef:
You can avoid the S list my making it a generator:
def subsets(L):
if L:
for s in subsets(L[1:]):
yield s
yield s + [L[0]]
else:
yield []
Nice one. Thanks!
Ozz
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al amounts correctly.
I agree with you though that the term subset may not be the best name in
this context because of those duplicates.
cheers,
Ozz
--
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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):
retur
, 4], [8], [8, 4], [8, 7], [8, 7, 4]]
>>> map(sum,subset.subsets([4,7,8,2]))
[2, 6, 9, 13, 10, 14, 17, 21, 0, 4, 7, 11, 8, 12, 15, 19]
It's not a real solution yet, and as others have pointed out the problem
is NP complete but it might help to get you going.
cheers,
Ozz
--
h
