Why I see only the last value of my row vector in the workspace?
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I have the following row vectors in my program:
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/969540/image.png)
When I run my program, I see in the workspace only the last value of these row vectors as :
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/969545/image.png)
I don't know whether my program executes using only the last values of these row vectors or all. What modification my program needs so that I can see all these row vectors in my workspace instead of a singla value? My program is:
clear all; close all; clc;
cloud = 'Homo';
T_para = zeros(5, 17, 20);
T_per = zeros(5, 17, 20);
cth=[1010:10:1200]
h = [1000:1000:4000];
a = [0.01:0.01:0.05];
r = [4:1:20];
fov = [0.2 0.5 1 2 5 10]
snr=[30 40 50 65 75];
for h = 1000:1000:4000
for fov = [0.2 0.5 1 2, 5 10]
dir_arhf = ['Tabledata_', cloud, '\', num2str(h), 'm-', num2str(fov), 'mrad'];
mkdir(dir_arhf);
for a = 0.01:0.01:0.05
for r = 4:1:20
load (['MCdatabase_', cloud, '/', num2str(a), '-', num2str(r), 'um/', num2str(h), 'm-', num2str(fov), 'mrad/I0.mat']);
load (['MCdatabase_', cloud, '/', num2str(a), '-', num2str(r), 'um/', num2str(h), 'm-', num2str(fov), 'mrad/Q0.mat']);
I_para = 1/2 * (I0 + Q0);
I_per = 1/2 * (I0 - Q0);
hh = genHeight(h).^2;
beta_para = sum(I_para, 2) .* hh';
beta_per = sum(I_per, 2) .* hh';
T_para(a * 100, r-3, :) = beta_para';
T_per(a * 100, r-3, :) = beta_per';
Norm_para(a * 100, r-3, :)=(T_para(a * 100, r-3, :)./max(T_para(a * 100, r-3, :)));
Norm_per(a * 100, r-3, :)=(T_per(a * 100, r-3, :)./max(T_per(a * 100, r-3, :)));
for snr=[30, 40, 50, 65, 75]
snr_para(a * 100, r-3, :)=awgn(T_para(a * 100, r-3, :),snr);
snr_per(a * 100, r-3, :)=awgn(T_per(a * 100, r-3, :),snr);
costf=sqrt((snr_para(a * 100, r-3, :)-Norm_para(a * 100, r-3, :)).^2+(snr_per(a * 100, r-3, :)-Norm_per(a * 100, r-3, :)).^2);
save([dir_arhf, '\TABLE.mat'], 'T_per', 'T_para');
end
end
end
end
end
1 commentaire
Stephen23
le 19 Avr 2022
Look at your FOR loops:
h = [1000:1000:4000]; % this is totally unused
..
for h = 1000:1000:4000 % becaue you redefined h here
The same for all of those other variables.
Réponse acceptée
KSSV
le 19 Avr 2022
You are using the vector as the loop index so obviously you will find only the last value. You can check it your self.
for a = 0.01:0.01:0.05
a
end
fprintf('%f\n',a)
You see, only the last value is printed, as it is the last index value. If you want it as a vector, you need to run like shown below.
a = 0.01:0.01:0.05;
for i = 1:length(a)
a(i)
end
% after loop a
a
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