How to iterate matrix multiple times?

1 view (last 30 days)
Hello everyone. I have a 3x7 matrix
La =
-5.2622 3.2610 1.0999 0 2.0590 0 0
0 0 1.0999 -2.4128 0 -0.1168 0
-5.2622 0 0 -2.4128 0 0 9.2321
I have a program below
H = [1 1 1 0 1 0 0;
0 0 1 1 0 1 0;
1 0 0 1 0 0 1];
Lb=num2cell(La);
Lb(La==0) = {[]};
Lc=reshape(Lb',1,[]);
Lf=zeros(1,21);
for R = 1:3
for L = 1+(R-1)*7:R*7;
Le=[Lc{setdiff(1+(R-1)*7:R*7, L)}];
Lf(L)=sign(prod(Le))*min(abs(Le));
end
end
Lg=reshape(Lf,7,3)';
Lh=H.*Lg;
Li=H.*[sum(Lh([2 3],:)); sum(Lh([1 3],:)); sum(Lh([1 2],:))];
Lj=La+Li
The obtained result is
Lj =
-7.6750 3.2610 1.2167 0 2.0590 0 0
0 0 -0.9591 -7.6750 0 -0.1168 0
-4.1623 0 0 -2.5296 0 0 9.2321
Lj is the new value of La after being processed. I want Lj to be reprocessed again just like La, for 5 times. So there will be different value of Lj everytime it being processed. I have tried something, such as adding
for i = 1:5
%do something
end
but seems like it's not worked because i got the same value for every i-th. I think there are mistake when i tried the looping.
How is the correct way to do that? Thank you very much.

Accepted Answer

KSSV
KSSV on 16 Jan 2022
La = [ -5.2622 3.2610 1.0999 0 2.0590 0 0
0 0 1.0999 -2.4128 0 -0.1168 0
-5.2622 0 0 -2.4128 0 0 9.2321] ;
H = [1 1 1 0 1 0 0;
0 0 1 1 0 1 0;
1 0 0 1 0 0 1];
iwant = zeros(3,7,5) ;
for i = 1:5
Lb=num2cell(La);
Lb(La==0) = {[]};
Lc=reshape(Lb',1,[]);
Lf=zeros(1,21);
for R = 1:3
for L = 1+(R-1)*7:R*7
Le=[Lc{setdiff(1+(R-1)*7:R*7, L)}];
Lf(L)=sign(prod(Le))*min(abs(Le));
end
end
Lg=reshape(Lf,7,3)';
Lh=H.*Lg;
Li=H.*[sum(Lh([2 3],:)); sum(Lh([1 3],:)); sum(Lh([1 2],:))];
Lj=La+Li ;
La = Lj;
iwant(:,:,i) = Lj ;
end
iwant
  1 Comment
Jenjen Ahmad Zaeni
Jenjen Ahmad Zaeni on 16 Jan 2022
It really works! Thank you very much, i really appreciate it.

Sign in to comment.

More Answers (1)

Simon Chan
Simon Chan on 16 Jan 2022
Edited: Simon Chan on 16 Jan 2022
Write your code as a function and save it as a m-file (or use the attached file):
function Lj = processLa(La)
H = [1 1 1 0 1 0 0;
0 0 1 1 0 1 0;
1 0 0 1 0 0 1];
Lb=num2cell(La);
Lb(La==0) = {[]};
Lc=reshape(Lb',1,[]);
Lf=zeros(1,21);
for R = 1:3
for L = 1+(R-1)*7:R*7
Le=[Lc{setdiff(1+(R-1)*7:R*7, L)}];
Lf(L)=sign(prod(Le))*min(abs(Le));
end
end
Lg=reshape(Lf,7,3)';
Lh=H.*Lg;
Li=H.*[sum(Lh([2 3],:)); sum(Lh([1 3],:)); sum(Lh([1 2],:))];
Lj=La+Li;
end
And then execute the following:
La=[-5.2622 3.2610 1.0999 0 2.0590 0 0;
0 0 1.0999 -2.4128 0 -0.1168 0;
-5.2622 0 0 -2.4128 0 0 9.2321];
%
for k = 1:5
La = processLa(La);
end
  1 Comment
Jenjen Ahmad Zaeni
Jenjen Ahmad Zaeni on 16 Jan 2022
It works too. I will doing this way in the future project. I really appreciate it. Thank you very much!

Sign in to comment.

Community Treasure Hunt

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

Start Hunting!

Translated by