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
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Paul Abbott Phone: +61 8 6488 2734
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