On May 8, 9:31 am, Steven D'Aprano <[EMAIL PROTECTED]> wrote: > On Mon, 07 May 2007 20:45:52 -0700, Michael Tobis wrote: > > I have a reasonably elegant solution but it's a bit verbose (a couple > > dozen lines which I'll post later if there is interest). Is there some > > clever Pythonism I didn't spot? > > Peering into my crystal ball, I see that your algorithm does the wrong > thing, and contains bugs too. You certainly don't need to partion the > sequence into three sub-sequences, or do that trick with the metaclass, > and it is more efficient to use list.extend() than sum. > > Hang on... stupid crystal ball... that's somebody else's code. Maybe you > should just post your code here, so we can look at it? > > In the meantime, here's a simple generator to do permutations with > repetition: > > def permute_with_repetitions(seq): > if len(seq) <= 1: > yield list(seq) > else: > for i, item in enumerate(seq): > for tail in permute_with_repetitions(seq[:i] + seq[i+1:]): > yield [item] + tail > > It doesn't do a test for the sequence containing duplicates, and I leave > it as anexerciseto do selections of fewer items. > > -- > Steven.
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