overlap logical matrices in MATLAB

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Ban
Ban il 4 Ott 2023
Commentato: Star Strider il 17 Ott 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 Commenti
Dyuman Joshi
Dyuman Joshi il 4 Ott 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 il 4 Ott 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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Risposta accettata

Star Strider
Star Strider il 4 Ott 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 Commenti
Ban
Ban il 17 Ott 2023
Thanks.
Star Strider
Star Strider il 17 Ott 2023
As always, my pleasure!

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