Why does times(A,B) gives me negative values when A and B don't have any ???

3 visualizaciones (últimos 30 días)
Hello, i'm working with this code, which is a non-negative factorization :
function [G] = NN_Update_G(G_0,S,H,V,nb_iter,beta)
mustBePositive(G_0);
G=G_0;
if beta<1
gamma_beta=1/(1-beta);
elseif 1<=beta && beta <=2
gamma_beta = 1;
else
gamma_beta = 1/(beta-1);
end
G_old=G;
for k=1:nb_iter
mustBePositive(G_old);
product = (((S'*(((S*G_old*H).^(beta-2)).*V)*H'))./(S'*((S*G_old*H).^(beta-1))*H')).^gamma_beta;
mustBePositive(product);mustBePositive(G_old);
G = times(G_old,product);
mustBePositive(G);
G_old = G;
end
end
I don't know how but only the last mustBepostive(G) is triggered. It's the only one, which would mean that both G and product are positive beforehand.
I really hope someone can make some sense out of this because it's really bothering me not understanding what's the problem...
  7 comentarios
Benjamin
Benjamin el 19 de Jun. de 2025
Editada: Benjamin el 19 de Jun. de 2025
Thanks for that, didn't know I could put parameters into a file like that (:
BUT, I tried but all the parameters (even alone) are too heavy to be included here /: (Apparently some files are attached so tell me if it works I guess. PS : it's just one matrice that's too heavy, so it's missing H. I'll try to find another way to send it)
for nb_iter I use 5 because it's in another loop but feel free to try anything
I use 0 for beta
Torsten
Torsten el 19 de Jun. de 2025
Editada: Torsten el 19 de Jun. de 2025
If you apply
load ("G_zero.mat")
G_0(G_0<=0)
G_0(G_0<0)
you will see that your G_0 has 0 negative elements, but 148 zero elements.

Iniciar sesión para comentar.

Respuesta aceptada

Steven Lord
Steven Lord el 19 de Jun. de 2025
Try using mustBeNonnegative instead of mustBePositive. This will detect if your calculation underflowed to 0.
x = realmin
x = 2.2251e-308
mustBePositive(x) % true so no error
y = x*x % underflow
y = 0
mustBeNonnegative(y) % true so no error
mustBePositive(y) % false
Value must be positive.

Más respuestas (1)

dpb
dpb el 19 de Jun. de 2025
Movida: dpb el 19 de Jun. de 2025
mustBePositive(0)
Value must be positive.
shows that zero is not considered positive by mustBePositive. You don't give any klews as to what the magnitudes of inputs are, but one must surmise that one or more of the elements must have underflowed owing to a large exponent in the product expression.

Categorías

Más información sobre Large Files and Big Data en Help Center y File Exchange.

Etiquetas

Productos


Versión

R2024b

Community Treasure Hunt

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

Start Hunting!

Translated by