overlap logical matrices in MATLAB

Question

Ban 2023-10-4

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链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2029194-overlap-logical-matrices-in-matlab

评论： Star Strider 2023-10-17

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

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Dyuman Joshi 2023-10-4

"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 2023-10-4

What do you mean with "overlapped logical matrixp"? is it the & operator of the 3 matrices created in the loop?

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Answer 1

Star Strider 2023-10-4

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2029194-overlap-logical-matrices-in-matlab#answer_1325794

在 MATLAB Online 中打开

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

.

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Ban 2023-10-16

The code you suggested work. But I am facing one more problem. In the below code I need to make a matrix of the quantity 'encounter_unique'. However I noticed that when I run it for individual values of 'ecc', I get different values compared to when I run for all the values of 'ecc' and 'vp'. Somewhere value of values of 'encounter_unique' are getting added up between varying values of 'ecc' and 'vp'. Please suggest.

This is merely a representative code for the sake of brevity (not the original one).

clear;

close all;

clc;

n_vp=3;

n_ecc=2;

vp_matrix=linspace(0.115*10^-1, 6*10^-3, n_vp); % Particle intrinsic velocity

ecc_matrix=linspace(-0.9,-0.1,n_ecc);

n_encounter_rate=1 ;

outdata = cell(length(vp_matrix), length(ecc_matrix));

for jecc=1:length(ecc_matrix);

for jvp=1:length(vp_matrix);

ecc=ecc_matrix(jecc) ;

vp=vp_matrix(jvp) ;

dis_threshold=0.03;

a = false(5); % Define Initial 'a'

for n=1:3

xa_time_step=ecc*rand(1,5);

za_time_step=vp*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

d1=abs(d);

d1(d1==tril(d1)) = NaN ;

a = a | (d1 < dis_threshold);

encounter_unique=sum(sum(a));

end

[outdata(n_encounter_rate)] = {encounter_unique};

n_encounter_rate=n_encounter_rate+1 ;

end

encounter_rate_matrix=cell2mat(outdata);

Star Strider 2023-10-16

在 MATLAB Online 中打开

I am having problems understanding what you are doing. The sum function will do exactly what you ask it to, and here it is summing the number of logical 1 (true) values in the ‘a’ matrix.

n_vp=3;
n_ecc=2;
vp_matrix=linspace(0.115*10^-1, 6*10^-3, n_vp);       % Particle intrinsic velocity 
ecc_matrix=linspace(-0.9,-0.1,n_ecc);
n_encounter_rate=1 ;
outdata = cell(length(vp_matrix), length(ecc_matrix));
for jecc=1:length(ecc_matrix);
for jvp=1:length(vp_matrix);
ecc=ecc_matrix(jecc) ;
vp=vp_matrix(jvp) ;
dis_threshold=0.03;
a = false(5);                                   % Define Initial 'a'
for n=1:3
xa_time_step=ecc*rand(1,5);
za_time_step=vp*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 = a | (d1 < dis_threshold)
encounter_unique=sum(sum(a))
end
[outdata(n_encounter_rate)] = {encounter_unique};
n_encounter_rate=n_encounter_rate+1 ;
end
end
a = 5×5 logical array
   0   0   0   0   0
   0   0   0   0   1
   0   0   0   0   0
   0   0   0   0   0
   0   0   0   0   0
encounter_unique = 1
a = 5×5 logical array
   0   0   0   0   0
   0   0   0   0   1
   0   0   0   0   0
   0   0   0   0   0
   0   0   0   0   0
encounter_unique = 1
a = 5×5 logical array
   0   0   0   0   0
   0   0   1   0   1
   0   0   0   0   0
   0   0   0   0   0
   0   0   0   0   0
encounter_unique = 2
a = 5×5 logical array
   0   0   0   0   1
   0   0   0   0   0
   0   0   0   0   0
   0   0   0   0   0
   0   0   0   0   0
encounter_unique = 1
a = 5×5 logical array
   0   0   0   0   1
   0   0   1   0   0
   0   0   0   0   0
   0   0   0   0   0
   0   0   0   0   0
encounter_unique = 2
a = 5×5 logical array
   0   0   0   1   1
   0   0   1   0   0
   0   0   0   0   0
   0   0   0   0   0
   0   0   0   0   0
encounter_unique = 3
a = 5×5 logical array
   0   0   0   0   0
   0   0   0   0   0
   0   0   0   0   0
   0   0   0   0   0
   0   0   0   0   0
encounter_unique = 0
a = 5×5 logical array
   0   0   0   0   0
   0   0   0   0   0
   0   0   0   0   0
   0   0   0   0   0
   0   0   0   0   0
encounter_unique = 0
a = 5×5 logical array
   0   0   0   0   0
   0   0   0   0   0
   0   0   0   0   0
   0   0   0   0   0
   0   0   0   0   0
encounter_unique = 0
a = 5×5 logical array
   0   1   1   0   0
   0   0   1   0   0
   0   0   0   0   0
   0   0   0   0   1
   0   0   0   0   0
encounter_unique = 4
a = 5×5 logical array
   0   1   1   0   1
   0   0   1   0   1
   0   0   0   1   0
   0   0   0   0   1
   0   0   0   0   0
encounter_unique = 7
a = 5×5 logical array
   0   1   1   0   1
   0   0   1   0   1
   0   0   0   1   1
   0   0   0   0   1
   0   0   0   0   0
encounter_unique = 8
a = 5×5 logical array
   0   1   0   0   0
   0   0   0   0   0
   0   0   0   0   1
   0   0   0   0   0
   0   0   0   0   0
encounter_unique = 2
a = 5×5 logical array
   0   1   0   1   1
   0   0   1   1   1
   0   0   0   1   1
   0   0   0   0   1
   0   0   0   0   0
encounter_unique = 9
a = 5×5 logical array
   0   1   1   1   1
   0   0   1   1   1
   0   0   0   1   1
   0   0   0   0   1
   0   0   0   0   0
encounter_unique = 10
a = 5×5 logical array
   0   1   0   0   1
   0   0   0   0   0
   0   0   0   1   0
   0   0   0   0   0
   0   0   0   0   0
encounter_unique = 3
a = 5×5 logical array
   0   1   1   1   1
   0   0   1   0   1
   0   0   0   1   1
   0   0   0   0   0
   0   0   0   0   0
encounter_unique = 8
a = 5×5 logical array
   0   1   1   1   1
   0   0   1   0   1
   0   0   0   1   1
   0   0   0   0   0
   0   0   0   0   0
encounter_unique = 8
encounter_rate_matrix=cell2mat(outdata)
encounter_rate_matrix = 3×2
     2     8
     3    10
     0     8

I am lost.

I do not understand what ‘ecc’ and ‘vp’ are, or what they do.

.

Ban 2023-10-17

Thanks.

Star Strider 2023-10-17

As always, my pleasure!

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