Creating matric of multiple arrays

x=0:12;

k=5;

a=k+x;

b=k+2x;

c=k+4x;

d=kx;

I'd like to create 2*2 matrix of four arrays (a,b,c,d).

Z=[a b; c d];

How could I realize this task by using for loop? Or do you have any other way to realize it? I ask for your advice! Thank you in advance!

[Bob Thompson]

What are you looking to loop? Your setup looks fine for a single iteration, so 'looping' should just be a matter of identifying what you want to change, and putting it, and the affected equations, inside a loop.

[Nicholas Moung]

This is just an example. The real formulae need the iteration. For instant, a,b,c and d are 1*100 arrays (when x=1:100;), and the matrix Z will be the 2*2 matrix formed by those a,b,c and d and each of the element of this matrix Z must be 1*100 array. And the formulae for a,b,c and d are just examples. Thank you for discussing!

Answer 1

[Bob Thompson]

You should be able to accomplish what you're looking for with some matrix indexing, no loop necessary.

x = 1:100;
k = 12;
a = x + k; % Makes a 100x1 size array, 13:112
b = 2*x + k; % Makes a 100x1 size array following the same formula
c = 4*x + k;
d = x * k;
% Adding a third dimension of Z allows you to basically stack each of the elements of a, b, c, and d
% into the one 2x2 format. Z(:,:,1) is a 2x2 of [a(1), b(1); c(1), d(1)], and each subsequent 'sheet'
% takes you to the next set of elements in the four matrices.
Z(1,1,:) = a;
Z(1,2,:) = b;
Z(2,1,:) = c;
Z(2,2,:) = d;

[Nicholas Moung]

Thank you so much! It does work!

Answer 2

[Adam Danz]

Perhaps this (scroll down to see vectorized version)?

x=0:12;
k=5;
a=k+x;
b=k+2*x;
c=k+4*x;
d=k*x;

Z = nan(2,2,numel(x));
for i = 1:numel(x)
    Z(:,:,i) = [a(i) b(i); c(i) d(i)];
end

disp(Z)

(:,:,1) =

     5     5
     5     0


(:,:,2) =

     6     7
     9     5


(:,:,3) =

     7     9
    13    10


(:,:,4) =

     8    11
    17    15


(:,:,5) =

     9    13
    21    20


(:,:,6) =

    10    15
    25    25


(:,:,7) =

    11    17
    29    30


(:,:,8) =

    12    19
    33    35


(:,:,9) =

    13    21
    37    40


(:,:,10) =

    14    23
    41    45


(:,:,11) =

    15    25
    45    50


(:,:,12) =

    16    27
    49    55


(:,:,13) =

    17    29
    53    60

If so, you don't need a loop.

Z = reshape([a;c;b;d],2,2,numel(x))  % Assuming a,b,c,d are row vectors

Z = 
Z(:,:,1) =

     5     5
     5     0


Z(:,:,2) =

     6     7
     9     5


Z(:,:,3) =

     7     9
    13    10


Z(:,:,4) =

     8    11
    17    15


Z(:,:,5) =

     9    13
    21    20


Z(:,:,6) =

    10    15
    25    25


Z(:,:,7) =

    11    17
    29    30


Z(:,:,8) =

    12    19
    33    35


Z(:,:,9) =

    13    21
    37    40


Z(:,:,10) =

    14    23
    41    45


Z(:,:,11) =

    15    25
    45    50


Z(:,:,12) =

    16    27
    49    55


Z(:,:,13) =

    17    29
    53    60

[Nicholas Moung]

Thank you so much! Both of them do work!!!!! I do appreciate it. :)

[Adam Danz]

No problem, I'm glad you found solutions!

For what it's worth, the reshape solution is most efficient and only requires 1 line.