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MathGroup Archive 2005

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Re: opposite of partition

  • To: mathgroup at smc.vnet.net
  • Subject: [mg57581] Re: opposite of partition
  • From: Paul Abbott <paul at physics.uwa.edu.au>
  • Date: Wed, 1 Jun 2005 06:03:30 -0400 (EDT)
  • Organization: The University of Western Australia
  • References: <d7han9$3r7$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

In article <d7han9$3r7$1 at smc.vnet.net>,
 Guy Israeli <guyi1 at netvision.net.il> wrote:

> How do I do the opposite of partition quickly?
> 
> for example:
> 
> l1= {{a, b, c, d, e}, {f, g, h, i, j}, {k, l, m, n, o}, {p, q, r, s, t}, {u, 
> v, w,
> x, y}}
> 
> and then if I partition it to blocks will result in
> 
> {{{{a, b}, {f, g}}, {{c, d}, {h, i}}}, {{{k, l}, {p, q}}, {{m, n}, {r, s}}}}

You have used

 m2 = Partition[l1, {2,2}, {2,2}]

here, and you have lost matrix entries as a result.
 
> flattening it won't help, and its messy to do it by taking all first lines 
> of the blocks, then second..
> 
> How can I do it quickly?

In this case,

  <<LinearAlgebra`

  BlockMatrix[m2]

does what you want.

Cheers,
Paul

-- 
Paul Abbott                                      Phone: +61 8 6488 2734
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        http://InternationalMathematicaSymposium.org/IMS2005/


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