Combination of sub-matrices generated within a for loop
2 visualizaciones (últimos 30 días)
Mostrar comentarios más antiguos
Rengin
el 13 de Ag. de 2019
Comentada: Bruno Luong
el 13 de Ag. de 2019
clear;
clc;
close all;
n=[3 1 2];
rad=[0.04 0.02 0.01];
h=[10 12 15];
for kk=1:length(n)
matrix=zeros(n(kk),n(kk));
for ii=1:n(kk)
for jj=1:n(kk)
val=rad(kk)*h(kk);
matrix(ii,jj)=val;
end
end
end
% Dear users, if you put a debug at the very last "end" (row 17), you can
% see that I aim to generate 3 sub-matrices having the size of 3x3, 1x1 and
% 2x2 (see n=[3 1 2]). I want to combine each sub-matrix one each other diagonally.
% I want to get the matrix as a result below:
matrix=[0.4 0.4 0.4 0 0 0;0.4 0.4 0.4 0 0 0;0.4 0.4 0.4 0 0 0;0 0 0 0.24 0 0;0 0 0 0 0.15 0.15;0 0 0 0 0.15 0.15];
%Is there any simple way to do it?
%Thanks in advance!
3 comentarios
madhan ravi
el 13 de Ag. de 2019
Using the previous solution of your question gives the same result, please describe the situation where that condition contradicts your needs.
Bruno Luong
el 13 de Ag. de 2019
"The calculation of Val is different."
You might try to learn deeper how computer programming actually works, if you get stuck everytime the values change then you miss the whole purpose of programming.
Respuesta aceptada
infinity
el 13 de Ag. de 2019
Hello,
There are various solutions for this question, one of the closet one to your current code could be
clear;
clc;
close all;
n=[3 1 2];
rad=[0.04 0.02 0.01];
h=[10 12 15];
matrix = zeros(sum(n));
for kk=1:length(n)
% matrix=zeros(n(kk),n(kk));
for ii=1:n(kk)
for jj=1:n(kk)
val=rad(kk)*h(kk);
matrix(ii+sum(n(1:kk-1)),jj+sum(n(1:kk-1)))=val;
end
end
end
0 comentarios
Más respuestas (0)
Ver también
Categorías
Más información sobre Operating on Diagonal Matrices en Help Center y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!