Creating matric of multiple arrays

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Nicholas Moung
Nicholas Moung le 20 Jan 2021
Commenté : Adam Danz le 20 Jan 2021
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!
  2 commentaires
Bob Thompson
Bob Thompson le 20 Jan 2021
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
Nicholas Moung le 20 Jan 2021
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!

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Réponse acceptée

Bob Thompson
Bob Thompson le 20 Jan 2021
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;
  1 commentaire
Nicholas Moung
Nicholas Moung le 20 Jan 2021
Thank you so much! It does work!

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Plus de réponses (1)

Adam Danz
Adam Danz le 20 Jan 2021
Modifié(e) : Adam Danz le 20 Jan 2021
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
  2 commentaires
Nicholas Moung
Nicholas Moung le 20 Jan 2021
Thank you so much! Both of them do work!!!!! I do appreciate it. :)
Adam Danz
Adam Danz le 20 Jan 2021
No problem, I'm glad you found solutions!
For what it's worth, the reshape solution is most efficient and only requires 1 line.

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