Store arrays produced by for loop

Hello,

I am writing a script that produces an array, vq, every iteration through a loop. I am trying to save each version of vq so I can compare each vq row by row later on. I suppose storing each vq as a column in a matrix would be ideal since each vq is the same length. This is my code:

        for m=B1(1:end,:)
            for n = B2(1:end,:)
                for p=1:length(B1)
                     H=find(time_dec > m(p) & time_dec < n(p));
                     [Alt,index]=unique(Alt(H));
                     temp=Temp(H);
                     xq=30.5:17.5:2000;
                     vq=interp1(Alt,temp(index),xq)';
                     
                     
                    
                end    
            end
        
        end

Any suggestions?

Thank you very much.

[Stephen23]

What sizes are B1 and B2 ?

[Olivia Booth-Howe]

B1 and B2 chance size every iteration but are always the same size as each other. They are vectors of length 2, 3, 4, 5, or 6.

[Olivia Booth-Howe]

The code works fine to produce the vq I want, I just need to save it as a matrix and simply putting vq(p) (how I would usually do it) doesn't work here. Thanks again!

[Jan]

"Doesn't work" is not useful to describe a problem.

This looks strange:

for m=B1(1:end,:)

"They are vectors of length 2, 3, 4, 5, or 6." - please post their size(). It matters, if they are row or column vectors.

[Olivia Booth-Howe]

Solved it! Thanks anyway.

Answer 1

[KALYAN ACHARJYA]

-The code works fine to produce the vq I want, I just need to save it as a matrix and simply putting vq(p)-

You can save it as structure cell array, as I have no idea of the length of vq would be. Hence I have choosen different variable I.

vq={}
l=1; 
for
.......
vq{l}=
l=l+1
end

or

If the length is p, then you can use vq{p}=.. too. On such case remove l.

vq={}
for
.......
vq{p}=
end

[Olivia Booth-Howe]

That is exactly what I needed, thank you so much! I really appreciate it.

[KALYAN ACHARJYA]

Always welcome @Olivia

[Jan]

+1. Exactly.