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Re: Flattening a hypermatrix into an ordinary matrix
*To*: mathgroup at smc.vnet.net
*Subject*: [mg126745] Re: Flattening a hypermatrix into an ordinary matrix
*From*: carlos at colorado.edu
*Date*: Sun, 3 Jun 2012 05:02:51 -0400 (EDT)
*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com
*References*: <201206010919.FAA11851@smc.vnet.net> <jqcnpl$19f$1@smc.vnet.net>
On Jun 2, 3:52 am, Bob Hanlon <hanlonr... at gmail.com> wrote:
> Use Join @@@ Flatten[Transpose /@ A, 1]
>
> A11 = {{B11, B12, B13}, {B21, B22, B23}, {B31, B32, B33}};
> A12 = {{C11, C12, C13}, {C21, C22, C23}, {C31, C32, C33}};
> A21 = {{D11, D12, D13}, {D21, D22, D23}, {D31, D32, D33}};
> A22 = {{E11, E12, E13}, {E21, E22, E23}, {E31, E32, E33}};
> A = {{A11, A12}, {A21, A22}};
>
> m = 3; n = 2;
>
> AA = Table[0, {m*n}, {m*n}];
> For[i = 1, i <= n, i++,
> For[j = 1, j <= n, j++,
> For[k = 1, k <= m, k++,
> For[l = 1, l <= m, l++,
> AA[[m*(i - 1) + k, m*(j - 1) + l]] = A[[i, j, k, l]]]]]]
>
> AA == Join @@@ Flatten[Transpose /@ A, 1]
>
> True
>
> Extending to m=3 and n=3
>
> A13 = A11;
> A23 = A21;
> A31 = {{F11, F12, F13}, {F21, F22, F23}, {F31, F32, F33}};
> A32 = {{G11, G12, G13}, {G21, G22, G23}, {G31, G32, G33}};
> A33 = A31;
> A = {{A11, A12, A13}, {A21, A22, A23}, {A31, A32, A33}};
>
> m = 3; n = 3;
>
> AA = Table[0, {m*n}, {m*n}];
> For[i = 1, i <= n, i++,
> For[j = 1, j <= n, j++,
> For[k = 1, k <= m, k++,
> For[l = 1, l <= m, l++,
> AA[[m*(i - 1) + k, m*(j - 1) + l]] = A[[i, j, k, l]]]]]]
>
> AA == Join @@@ Flatten[Transpose /@ A, 1]
>
> True
>
> Bob Hanlon
>
>
>
>
>
>
>
> On Fri, Jun 1, 2012 at 5:19 AM, <car... at colorado.edu> wrote:
> > I have an n x n hypermatrix, the entries of which are m x m blocks.
> > For example A below (m=3, n=2):
>
> > A11={{B11,B12,B13},{B21,B22,B23},{B31,B32,B33}};
> > A12={{C11,C12,C13},{C21,C22,C23},{C31,C32,C33}};
> > A21={{D11,D12,D13},{D21,D22,D23},{D31,D32,D33}};
> > A22={{E11,E12,E13},{E21,E22,E23},{E31,E32,E33}};
> > A={{ A11,A12},{A21,A22}};
>
> > I want to convert this to an ordinary n*m x n*m matrix.
> > For the example I want A to become
>
> > {{B11,B12,B13,C11,C12,C13},{B21,B22,B23,C21,C22,C23},
> > {B31,B32,B33,C31,C32,C33},{D11,D12,D13,E11,E12,E13},
> > {D21,D22,D23,E21,E22,E23},{D31,D32,D33,E31,E32,E33}}
>
> > This can be easily done with C style loops as
>
> > AA=Table[0,{m*n},{m*n}];
> > For [i=1,i<=n,i++, For[j=1,j<=n,j++,
> > For [k=1,k<=m,k++, For [l=1,l<=m,l++,
> > AA[[m*(i-1)+k,m*(j-1)+l]]=A[[i,j,k,l]]
> > ]]]];
>
> > but is there a more elegant way using Flatten?
> > (Flatten[A,1] doesnt do it.) It should work also for blocks
> > of varying size for future use.
Thanks! This seems to work fine, even for variable size blocks. For
example
(this comes from an actual fluid-structure interaction FEM simulation)
A={{{{B11}},{{C11}},{{D11,D12}},{{E11}},{{F11}}},
{{{G11},{G12}},{{H11},{H12}},{{I11,I12},{I21,I22}},{{J11},{J12}},
{{K11},{K12}}},
{{{L11}},{{M11}},{{N11,N12}},{{O11}},{{P11}}},
{{{Q11}},{{R11}},{{S11,S12}},{{T11}},{{U11}}},
{{{V11}},{{W11}},{{X11,X12}},{{Y11}},{{Z11}}}};
AA = Join @@@ Flatten[Transpose /@ A, 1] ; Print["AA=",AA//
MatrixForm];
I can't use ArrayFlatten since it does not exist in versions of
Mathematica
such as 4 or 5, which some of my off-campus students use (they were
licensed
by their employers long ago)
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