r = 5;
A_r5 = (S_Vals(1,1)*u(:,1)*(V(:,1)).')+(S_Vals(2,1)*u(:,2)*(V(:,2)).')+(S_Vals(3,1)*u(:,3)*(V(:,3)).')+(S_Vals(4,1)*u(:,4)*(V(:,4)).')+(S_Vals(5,1)*u(:,5)*(V(:,5)).');
x_5 = uint8(A_r5);
figure(2)
imshow(x_5);

Réponses (2)

Image Analyst
Image Analyst le 16 Fév 2022
Yes, that's very complicated to look at. I'd break it into more bite-sized terms, like
term1 = whatever
term2 = whatever
A_r5 = term1 * term2 + term3 or whatever.

11 commentaires

I want to sum the following in a for loop
term 1 = A(1,1)*U(:,1),V(:,1)
term 2 = A(2,1)*U(:,2),V(:,2)
term 1 = A(3,1)*U(:,3),V(:,3)
A_r = term1 + term2 + term3
thanks
Image Analyst
Image Analyst le 16 Fév 2022
I don't understand the expressions. Please check the math.
A(1,1)*U(:,1) is one term, but V(:, 1) is another term and you just put them on the same line with a comman between them. What's that supposed to do?
Im sorry the correct expression is
term 1 = A(1,1)*U(:,1)*V(:,1)
term 2 = A(2,1)*U(:,2)*V(:,2)
term 1 = A(3,1)*U(:,3)*V(:,3)
A_r = term1 + term2 + term3
for any given number of terms
Do you mean like
[rows, columns] = size(A);
A_r = zeros(rows, 1);
% Rows must equal the number of columns in both U and V or this won't work.
for row = 1 : rows
term1(row) = A(row,1)*U(:, row)*V(:, row)
end
A_r = sum(term1);
Try
A = sum(S .* U .* V);
Im trying create a simple for loop to compute the summation for
r = 100
i = 1:r
A100 = S(i,1)*U(:,i)*V(:,i);
Image Analyst
Image Analyst le 16 Fév 2022
Are those matrix products or element by element products?
S,U, and V are all matrices
Image Analyst
Image Analyst le 16 Fév 2022
I still don't know what you want so I'll give others a chance. You're multiplying a row or column of a matrix (which would be a 1-D vector) times a matrix and I don't know if you want every row multiplied by that row or what. I'm totally confused.
THANKS anyway
DGM
DGM le 16 Fév 2022
Need to transpose V(:,row).'

Connectez-vous pour commenter.

DGM
DGM le 16 Fév 2022
Modifié(e) : DGM le 16 Fév 2022
Depends what you mean by "simple". Consider the following three options. Method 2 is compact, but it's slower for large inputs. Method 3 is relatively fast for large inputs. Both method 2 and 3 have an advantage in that they can be scaled to more than 5 "pages".
N = 1000; % large inputs
u = rand(N,5);
V = rand(N,5);
S_Vals = rand(5,1);
% original method
a = timeit(@() testA(S_Vals,u,V))
a = 0.0031
% simple permutation and elementwise math
b = timeit(@() testB(S_Vals,u,V))
b = 0.0132
% looped version of original
c = timeit(@() testC(S_Vals,u,V))
c = 0.0012
% time ratio
[a/b a/c]
ans = 1×2
0.2365 2.6443
function testA(S_Vals,u,V)
A_r5 = (S_Vals(1,1)*u(:,1)*(V(:,1)).') ...
+(S_Vals(2,1)*u(:,2)*(V(:,2)).') ...
+(S_Vals(3,1)*u(:,3)*(V(:,3)).') ...
+(S_Vals(4,1)*u(:,4)*(V(:,4)).') ...
+(S_Vals(5,1)*u(:,5)*(V(:,5)).');
end
function testB(S_Vals,u,V)
A_r5 = sum(permute(S_Vals,[3 2 1]).*permute(u,[1 3 2]).*permute(V,[3 1 2]),3);
end
function testC(S_Vals,u,V)
A_r5 = zeros(size(u,1),size(V,1));
for k = 1:size(S_Vals(1))
A_r5 = A_r5 + (S_Vals(k,1)*u(:,k)*(V(:,k)).');
end
end

Catégories

En savoir plus sur Loops and Conditional Statements dans Centre d'aide et File Exchange

Tags

Community Treasure Hunt

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

Start Hunting!

Translated by