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 School of Physics, M013 Fax: +61 8 6488 1014 The University of Western Australia (CRICOS Provider No 00126G) AUSTRALIA http://physics.uwa.edu.au/~paul http://InternationalMathematicaSymposium.org/IMS2005/