How to create a new variable within each run in a for loop?

For an assignment of mine, I am using a for loop to obtain a new matrix "A" for each number within an "x" range.

However, every time the loop runs, the variable A remains unchanged, and so this would make it difficult to distinguish

the matrix with different numbers of "x". Is there a way where each time the loop runs, I can set a new variable equal to "A" continuously?

So then I would get A1, A2, A3, A4, A5...etc. I would appreciate any help!

clc

clear all

%define variables

x=0:0.01:2;

for k=1:length(x) %x as a range

alpha=atan(1./(2-x(k))); %alpha angle

beta=atan((2-x(k))./1); %beta angle

gamma=atan(x(k)); %gamma angle

A=zeros(10); % setting up zero matrices

b=[0;2;0;1;0;0;0;0;0;0]; % for later substitution

%Define the A matrix

A(1,1:2)=[-1, cos(alpha)];

A(2,2)=[sin(alpha)];

A(3,1:5)=[1,0,0,0,-1];

A(4,3)=[-1];

A(5,2:7)=[sin(beta),0,0,0,sin(gamma),-1];

A(6,2:5)=[-cos(beta),-1,0,-cos(gamma)];

A(7,6:8)=[1,0,-1];

A(8,7:9)=[-1,0,1];

A(9,4:10)=[1,sin(gamma),0,0,0,0,-1];

A(10,5:7)=[cos(gamma),0,1];

A

xrow=pinv(A)*b %gaussian method for finding unknown forces

end

[Scott MacKenzie]

What your are describing amounts to poor programming and poor use of MATLAB's capabilities. Visit this TUTORIAL for complete details. The variables you are describing can probably just be rows or columns in a matrix that is built up in the loop -- or something. Post some code and we'll have a look.

[dpb]

As Scott says, "there be dragons!" in that approach and you do NOT want to go down that path. Stephen C will probably come along shortly and reinforce the message even more strongly! :)

Another alternative, depending upon the content of A each pass is to use cell array -- but as above "show us code"; we can't guess what you're really trying to do so don't know best solution.

[Jason Wang]

I have added my code

I apologize for not adding so earlier

[Stephen23]

"However, every time the loop runs, the variable A remains unchanged"

Because you have not used any indexing based on the loop iteration variable k to store the results of each loop. How to use indexing in a loop is shown in the introductory tutorials, or the documentation on array preallocation:

https://www.mathworks.com/help/matlab/matlab_prog/preallocating-arrays.html

"Is there a way where each time the loop runs, I can set a new variable equal to "A" continuously?"

Of course, this is easy with basic MATLAB indexing.

"So then I would get A1, A2, A3, A4, A5...etc."

That would be about the worst approach.

Rather than forcing pseudo-indices into variable names, just use actual indices in to one array.

Answer 1

[DGM]

Arrays are indexable. Scalars and variable names aren't. If you find yourself embedding ordinal information in the variable names, you should probably stop and use an array.

https://www.mathworks.com/matlabcentral/answers/304528-tutorial-why-variables-should-not-be-named-dynamically-eval#answer_236120

Just store the variables in an array of some sort suitable to the type of things you're putting in it.

https://www.mathworks.com/matlabcentral/answers/304528-tutorial-why-variables-should-not-be-named-dynamically-eval#answer_236124

For example:

N = 4;
bigA = cell(N,1);
for n = 1:N
    A = rand(4);
    bigA{n} = A;
end
format compact % just for tighter web display
celldisp(bigA)

bigA{1} =
    0.1840    0.7896    0.8760    0.8886
    0.2737    0.9784    0.5529    0.7511
    0.0631    0.5860    0.6735    0.9561
    0.5812    0.2218    0.1747    0.3220
bigA{2} =
    0.5867    0.8644    0.8288    0.7139
    0.1717    0.1442    0.6000    0.7751
    0.2383    0.9257    0.1120    0.9512
    0.2443    0.7263    0.5976    0.5300
bigA{3} =
    0.9682    0.1485    0.0110    0.7669
    0.2349    0.3334    0.2995    0.8681
    0.8535    0.8289    0.3777    0.8126
    0.4849    0.6876    0.8785    0.3668
bigA{4} =
    0.3385    0.6856    0.0589    0.3541
    0.7796    0.4151    0.1089    0.9215
    0.0120    0.2100    0.3975    0.1568
    0.7010    0.5085    0.3946    0.0702

Using a cell has some advantages if the matrix size varies or is extremely large, but a ND array works and may be easier to use in some case.

It's also possible that the whole issue of storage could be avoided by simply calculating the appropriate matrix at the point in the code where it needs to be used. To know whether that would be an option would require further insight into your task and the code you've written so far.

EDIT:

Not really knowing what you're going to do with the results, I'm assuming you want to keep both A and xrow:

x=0:0.01:2;
results = cell(numel(x),2);
for k=1:length(x)  %x as a range
    alpha=atan(1./(2-x(k)));  %alpha angle
    beta=atan((2-x(k))./1);   %beta angle
    gamma=atan(x(k));        %gamma angle
    A=zeros(10); % setting up zero matrices
    b=[0;2;0;1;0;0;0;0;0;0]; % for later substitution
    
    %Define the A matrix
    A(1,1:2)=[-1, cos(alpha)];
    A(2,2)=[sin(alpha)];
    A(3,1:5)=[1,0,0,0,-1];
    A(4,3)=[-1];
    A(5,2:7)=[sin(beta),0,0,0,sin(gamma),-1];
    A(6,2:5)=[-cos(beta),-1,0,-cos(gamma)];
    A(7,6:8)=[1,0,-1];
    A(8,7:9)=[-1,0,1];
    A(9,4:10)=[1,sin(gamma),0,0,0,0,-1];
    A(10,5:7)=[cos(gamma),0,1];
    
    xrow=pinv(A)*b; %gaussian method for finding unknown forces
    
    results{k,1} = A;
    results{k,2} = xrow;
end

This stores A and xrow for each point in the parameter sweep. Again, the same could be done with plain numeric arrays by stacking A on dim3 and side-concatenating xrow on dim2 or so.