replace part of matrix to another matrix

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mohammed hussein
mohammed hussein el 18 de Nov. de 2017
Comentada: mohammed hussein el 19 de Nov. de 2017
Hi i have many matrices in same size , i want to add all of them together in one matrix with replace the last quarter of each one to binning of next one until complete the matrix and the rest will be zeros . for example
a =zeros(5,5)
A=[1 2 ;3 4];
B=[5 6;7 8];
C=[9 10;11 12];
D=[13 14;15 16];
the answer
a=
1 2 0 0 0
3 5 6 0 0
0 7 9 10 0
0 0 11 13 14
0 0 0 15 16
thank you very much for helping

Respuesta aceptada

Stephen23
Stephen23 el 18 de Nov. de 2017
Editada: Stephen23 el 18 de Nov. de 2017
For any size matrices, even combinations of different sizes:
C = {[1,2;3,4],[5,6;7,8],[9,10;11,12],[13,14;15,16]};
num = numel(C);
szr = cellfun('size',C,1);
szc = cellfun('size',C,2);
csr = cumsum([0,szr-1]);
csc = cumsum([0,szc-1]);
M = zeros(1+csr(end),1+csc(end));
for k = 1:num
idr = csr(k)+(1:szr(k));
idc = csc(k)+(1:szc(k));
M(idr,idc) = C{k};
end
Giving:
>> M
M =
1 2 0 0 0
3 5 6 0 0
0 7 9 10 0
0 0 11 13 14
0 0 0 15 16
It works perfectly with any size matrices, e.g.:
C = {[1,2;3,4],[5,6;7,8],[1,2,3;4,5,6;7,8,9],[9,10;11,12],[13,14;15,16]};
gives
M =
1 2 0 0 0 0 0
3 5 6 0 0 0 0
0 7 1 2 3 0 0
0 0 4 5 6 0 0
0 0 7 8 9 10 0
0 0 0 0 11 13 14
0 0 0 0 0 15 16
  1 comentario
mohammed hussein
mohammed hussein el 19 de Nov. de 2017
thank you very much for your answer , this is exactly what i want

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Más respuestas (2)

Walter Roberson
Walter Roberson el 18 de Nov. de 2017
A=[1 2 ;3 4];
B=[5 6;7 8];
C=[9 10;11 12];
D=[13 14;15 16];
newvals = {A, B, C, D};
a = zeros(5,5);
for K = 1 : length(newvals)
a(K:K+1, K:K+1) = newvals{K};
end

Andrei Bobrov
Andrei Bobrov el 18 de Nov. de 2017
Editada: Andrei Bobrov el 18 de Nov. de 2017
In your case (without loop)
k = cat(3,A,B,D,C);
[m,n,q] = size(k);
a = full(gallery('tridiag',q*m/2+1,1,1,1));
ii = ((1:m*n-1) + (0:q-1)'*m*n)';
a(a>0) = k([ii(:);prod([m,n,q])]);
in general case (with for..end loop)
C = {[1,2;3,4],[5,6;7,8],[1,2,3;4,5,6],[9,10;11,12],[13,14;15,16]};
[m,n] = cellfun(@(x)size(x),C(:));
s = sum(m-1) + 1;
a = zeros(s,sum(n-1)+1);
jj = 1;
for ii = 1:numel(m)
id = jj + (0:m(ii)-1)' + s*(0:n(ii)-1);
a(id) = C{ii};
jj = id(end);
end
All for MATLAB >= R2016b

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