overlap logical matrices in MATLAB

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Ban
Ban le 4 Oct 2023
Commenté : Star Strider le 17 Oct 2023
I have code which runs multiple times (say, 3 times). The code produces logical matrix denoted by 'a' in each loop (denoted by n). I want to get an overlapped logical matrix at the end of the loop. Please suggest.
clear;
close all;
clc;
dis_threshold=0.4;
for n=1:3
xa_time_step=rand(1,5);
za_time_step=rand(1,5);
for k1=1:length(xa_time_step)
for j1=1:length(xa_time_step)
d(j1,k1)=sqrt((xa_time_step(j1)-xa_time_step(k1)).^2+(za_time_step(j1)-za_time_step(k1)).^2);
end
end
d1=abs(d);
d1(d1==tril(d1)) = NaN ;
a=d1 < dis_threshold ;
disp(a)
end
  2 commentaires
Dyuman Joshi
Dyuman Joshi le 4 Oct 2023
"I want to get an overlapped logical matrix at the end of the loop."
What does overlap mean in this context? Take Logical OR of the outputs? Logical AND? Concatenate the arrays? Horizontally or vertically? or something else?
Fabio Freschi
Fabio Freschi le 4 Oct 2023
What do you mean with "overlapped logical matrixp"? is it the & operator of the 3 matrices created in the loop?

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Réponse acceptée

Star Strider
Star Strider le 4 Oct 2023
Use the logical or (|) function to accumulate the matrices —
dis_threshold=0.4;
a = false(5); % Define Initial 'a'
for n=1:3
xa_time_step=rand(1,5);
za_time_step=rand(1,5);
for k1=1:length(xa_time_step)
for j1=1:length(xa_time_step)
d(j1,k1)=sqrt((xa_time_step(j1)-xa_time_step(k1)).^2+(za_time_step(j1)-za_time_step(k1)).^2);
end
end
d1=abs(d);
d1(d1==tril(d1)) = NaN ;
a_temp = d1 < dis_threshold % Delete Later
a = a | (d1 < dis_threshold)
% disp(a)
end
a_temp = 5×5 logical array
0 1 1 1 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0
a = 5×5 logical array
0 1 1 1 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0
a_temp = 5×5 logical array
0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0
a = 5×5 logical array
0 1 1 1 1 0 0 0 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 0
a_temp = 5×5 logical array
0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0
a = 5×5 logical array
0 1 1 1 1 0 0 0 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 0
.
  10 commentaires
Ban
Ban le 17 Oct 2023
Thanks.
Star Strider
Star Strider le 17 Oct 2023
As always, my pleasure!

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