how to store new matrix on the old one

I have a array Z(4,4) such as

[1:4;
 5:8;
 9:12;
13:16].

I would like to change the order of the matrix and store in a same name as Z with new version;this for the first iteration. for the second iteration; I want to use the previous version and do the same changes on the order to get new one.keep doing this for m iteration.

for k=1:m,
    D=[];
    for j=1:4,    
          for i=1:4,
                C=A*[i;j];
                [x]=mod(C(1),4)+1
                [y]=mod(C(2),4)+1
                D=[D;Z(x,y)];
           end;
           E=[E,D];D=[];
     end;
    NewZ(k)=E;E=[];
    Z=NewZ;
 end;

always I get this error: Subscripted assignment dimension mismatch.

Please I need a help.

[the cyclist]

What is A? Can you post some self-contained code that we can just try to run, which will exhibit the error?

[sundus]

A is [1 1;1 2]

[dpb]

and m and E?

Show the error message in its entirety in the context of the run w/ the above definitions.

[sundus]

m=3;E=[];NewZ=[];

when I run the program I got this error In an assignment A(I) = B, the number of elements in B and I must be the same.

what I want to keep the values of matrix Z and use the values of NewZ instead of each time.

thank you for sharing your ideas.

[dpb]

Don't know if it will produce what you expect or not, but the error comes from

NewZ(k)=E;

where you're trying to store an array in an element. Perhaps what you're looking for is

NewZ=[NewZ;E]; E=[];

to concatenate the E onto NewZ

[sundus]

NewZ is a new version of Z ,and I need to keep it aside by itself. at the end I will have m numbers of NewZ. Yes, I'll have an array but I don't know how to save it under the same name and reuse it.

Answer 1

[Roger Stafford]

To accomplish what you appear to have in mind you would need to modify your code to:

 for k=1:m,
   E=[];
   for j=1:4,
     D=[];    
     for i=1:4,
       C=A*[i;j];
       [x]=mod(C(1),4)+1;
       [y]=mod(C(2),4)+1;
       D=[D;Z(x,y)];
     end;
     E=[E,D];
   end;
   Z=E;
 end

There are more efficient ways of carrying out this procedure. For example, you could pre-calculate all sixteen x,y pairs (which don't depend on Z) and then carry out the m transformations more quickly.

[sundus]

Thank you a good suggestion, you mean for k=1:m, for j=1:4, for i=1:4, C=A*[i;j]; [x]=mod(C(1),4)+1; [y]=mod(C(2),4)+1; Tem=[Tem [x,y]]; end; end; Tem=[]; end;

so I need m times from the Tem vector and I don't know how to write it.Could you please help me with that.

Really I like your efficient way to carry out the procedure.

[sundus]

I will have different orders for the items of Z with each iteration, I need to store each z in an array because I need them for advance process,but I don't know how to do that. Please help.

In this solution I will not have all the copies of Z, so the problem is how to store all the copies in an array.