Re: Partitioning a list into unequal partitions
- To: mathgroup at smc.vnet.net
- Subject: [mg47501] Re: Partitioning a list into unequal partitions
- From: Paul Abbott <paul at physics.uwa.edu.au>
- Date: Wed, 14 Apr 2004 07:17:08 -0400 (EDT)
- Organization: The University of Western Australia
- References: <c5gfdp$an0$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
In article <c5gfdp$an0$1 at smc.vnet.net>, "DIAMOND Mark R." <dot at dot.dot>
wrote:
> Could someone please show me a simple (non-procedural) way of partitioning a
> list into 1,2,3 ... n disjoint sublists, where the length of the list is
> guaranteed to be correct (i.e. n*(n+1)/2)
If I understand your question (a definite example would have helped),
for a list, lst, how about
Sort[ReplaceList[lst, {___, a__, ___} -> {a}]]
This gives a list of length n*(n+1)/2 if lst is of length n.
> I can't see a way, and yet I'm sure there *must* be a one-liner using Fold.
I don't see how Fold will do what you want. Distribute might work. For
example,
PowerSet[t_List]:=Sort[Distribute[{{},{#}} & /@ t,List,List,List,Join]]
gives a sorted list of all subsets of a list.
Cheers,
Paul
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