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Re: Flattening a hypermatrix into an ordinary matrix
*To*: mathgroup at smc.vnet.net
*Subject*: [mg126734] Re: Flattening a hypermatrix into an ordinary matrix
*From*: Bob Hanlon <hanlonr357 at gmail.com>
*Date*: Sat, 2 Jun 2012 05:48:45 -0400 (EDT)
*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com
*References*: <201206010919.FAA11851@smc.vnet.net>
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, <carlos 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.
>
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