Submatrices into a Structured Larger Matrix

Hi,
I need a code to create a lower triangular matrix consisting of several submatrices, where each submatrix has its own coordinates [between (xi_0,yi_0) and (xi_f,yi_f)] without any use of for loop.
i.e. Submatrices A, B and C are each 3x3 matrices. I want to create a D matrix as: [A zeros(3x3);B C]
Any help could be useful.
Thanks in advance!

3 Kommentare

Akan Selim
Akan Selim am 17 Okt. 2022
My first attempt is:
A=randn(3,18);
B=zeros(3,27);
layers=[1:3 10:15 19:27];
B(:,layers)=A
C=mat2cell(B,ones(1,1)*3,ones(3,1)*9)
vertcat(C{:})
David Hill
David Hill am 17 Okt. 2022
Hard to understand what you want. Please provide a simple example of A,B,C and the desired output.
Akan Selim
Akan Selim am 17 Okt. 2022
In this simple example, there are 3 matrices, mainly A, B and C.
I want to place these matrices into a lower triangular matrix form:
[A zeros(3x3);
B C]
A=[1 2 3;4 5 6;7 8 9]
B=[10 11 12;13 14 15;16 17 18]
C=[19 20 21;22 23 24;25 26 27]
Output: [1 2 3 0 0 0;
4 5 6 0 0 0;
7 8 9 0 0 0;
10 11 12 19 20 21
13 14 15 22 23 24
16 17 18 25 26 27]
In general form, I have over hundreds of matrices (A1, A2, A3, A4, ... A100) that should be placed as follows:
[A1
A2 A3
A4 A5 A6
...]
Thank you!

Melden Sie sich an, um zu kommentieren.

Antworten (1)

David Hill
David Hill am 17 Okt. 2022

0 Stimmen

Adjust as necessary. Need 105 3x3 matrices to fill described pattern above. I assume all A matrices are the same size (3x3).
for k=1:105
A{k}=randi(100,3);%fill A cell array with 3x3 matrices. Recommend you place all your matrices into a cell array or 3D array
A2(:,:,k)=randi(100,3);%3D array instead of cell array
end
B=zeros(1,42);
B2=B;
c=1;
for m=1:14
for n=1:m
B((m-1)*3+1:3*m,(n-1)*3+1:3*n)=A{c};%either storage system will work (A or A2)
B2((m-1)*3+1:3*m,(n-1)*3+1:3*n)=A2(:,:,c);
c=c+1;
end
end

3 Kommentare

Akan Selim
Akan Selim am 17 Okt. 2022
Bearbeitet: Akan Selim am 17 Okt. 2022
I was looking for a way that doesn't utilize any for loops, like in this example I gave earlier:
A=randn(3,18);
B=zeros(3,27);
layers=[1:3 10:15 19:27]; (columns 4:9, 16:18 are zeros)
B(:,layers)=A
C=mat2cell(B,ones(1,1)*3,ones(3,1)*9)
vertcat(C{:})
Thanks for the answer though.
David Hill
David Hill am 17 Okt. 2022
There is nothing wrong with loops, especially when having to index into a variable. You should never label variables like A1, A2, A3, ......, A100, but rather store then into an array that can be indexed.
Walter Roberson
Walter Roberson am 17 Okt. 2022
Why are you looking for a way that doesn't utilize any for loops? mat2cell() uses for loops internally -- you can read the source code.

Melden Sie sich an, um zu kommentieren.

Gefragt:

am 17 Okt. 2022

Kommentiert:

am 17 Okt. 2022

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by