replace part of matrix to another matrix

9 vues (au cours des 30 derniers jours)
mohammed hussein
mohammed hussein le 18 Nov 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

Réponse acceptée

Stephen23
Stephen23 le 18 Nov 2017
Modifié(e) : Stephen23 le 18 Nov 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 commentaire
mohammed hussein
mohammed hussein le 19 Nov 2017
thank you very much for your answer , this is exactly what i want

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Plus de réponses (2)

Walter Roberson
Walter Roberson le 18 Nov 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
  1 commentaire
mohammed hussein
mohammed hussein le 19 Nov 2017
Thank you very much for your answer

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Andrei Bobrov
Andrei Bobrov le 18 Nov 2017
Modifié(e) : Andrei Bobrov le 18 Nov 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
  1 commentaire
mohammed hussein
mohammed hussein le 19 Nov 2017
Thank you very much for your answer

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