matrix addition different dimension using for loops.

4 vues (au cours des 30 derniers jours)
Malini Bakthavatchalam
Malini Bakthavatchalam le 15 Août 2020
Commenté : Walter Roberson le 16 Août 2020
Hi,
I am trying to do matrix addition of 2*4 to fit in 4*4 matrix. I am using for loops to get me familiarize with loops so that i can use it for image matrix manupulation. But with my code my output is 2*4 instead of 4*4.. could someone explain me the concept or the logic i am lacking here ..
a = [11:14; 15:18];
b = [3 2 3 1; 2 1 1 1;1 3 3 2; 2 1 1 3];
for i = 0
for j = 0
for x = 1:2
for y = 1:4
c(i+x,j+y) =b(i+x,j+y)+a(x+i,y+j);
end
end
end
end
  2 commentaires
Walter Roberson
Walter Roberson le 16 Août 2020
If the first matrix were
1 2 3 4
5 6 7 8
and the second matrix was all zero, then what would you want the result of the addition to be ?
Malini Bakthavatchalam
Malini Bakthavatchalam le 16 Août 2020
1 2 3 4
5 6 7 8
1 2 3 4
5 6 7 8
This would be my resultant matrix

Connectez-vous pour commenter.

Réponse acceptée

Walter Roberson
Walter Roberson le 16 Août 2020
%this code will fail if a or b is empty, or if one size is not an exact multiple of the other
asz = size(a);
bsz = size(b);
maxsz = max(asz,bsz);
arep = maxsz ./ asz;
brep = maxsz ./ bsz;
ar = repmat(a, arep);
br = repmat(b, brep);
c = ar + br;
That is, create new matrices that contain the old matrices copied as many times as needed.
  3 commentaires
Walter Roberson
Walter Roberson le 16 Août 2020
Modifié(e) : Walter Roberson le 16 Août 2020
ar = repmat(a, arep);
can be replaced with
[arows, acols, apanes] = size(a);
ar = zeros(arows*arep(1), acols*arep(2), apanes, class(a));
for J = 1 : arep(1)
arows = (J-1)*arows+1 : J*arows;
for K = 1 : arep(2)
acols = (K-1)*acols+1 : K*acols;
ar(arows, acols, :) = a;
end
end
Walter Roberson
Walter Roberson le 16 Août 2020
I fixed a typing mistake in setting the class of ar

Connectez-vous pour commenter.

Plus de réponses (0)

Catégories

En savoir plus sur Creating and Concatenating Matrices dans Help Center et File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by