Why error function not work?
Mostrar comentarios más antiguos
The error function doesn't work with me. I do not get any result.
If anyone knows the problem, I will be thankful
for j1=1:3,
yi=w'*x(:,j1); % Network output
y=sign(yi);
if sum(y-d(1,:)') > 0
error=error+1
end
end
8 comentarios
madhan ravi
el 15 de Nov. de 2018
Editada: madhan ravi
el 15 de Nov. de 2018
w',x,d?? , provide the values
Amna El-tawil
el 15 de Nov. de 2018
Amna El-tawil
el 15 de Nov. de 2018
madhan ravi
el 15 de Nov. de 2018
size(y) is 4 by 1 and size(d(1,:)') is 3 by 1 how can you subtract them?
Amna El-tawil
el 15 de Nov. de 2018
Rik
el 15 de Nov. de 2018
The apostrophe transposes that output of d:
size(d(1,:)) % -> [1 3]
size(d(1,:)') % -> [3 1]
So neither will actually fit.
What are you trying to do? If you give a good description of your problem, we can try to help you with that, instead of fixing code that might not do what you mean.
It is very unlikely that the code shown does whatever it is required to do. The whole thing doesn't make sense. Of course, if the code was commented as it should be, we would have an idea of what it was supposed to do (hint!).
Note:
Madhan: size(y) is 4 by 1
Amna: size(y) is 4 by 3
size(y) is always going to be size(w, 2) by 1. It can't have 3 columns since the multiplication in w' * x(:, 1) guarantees it will have one column. It will has as many rows as w has column, which is no whose value was not given. If no is 3 the subtraction will succeed. ni has to be 4 for the multiplication to work.
Also note that naming a variable error is a bad idea. It's already the name of a matlab function.
madhan ravi
el 15 de Nov. de 2018
Perfect explanation @Guillaume
Respuestas (0)
Categorías
Más información sobre Loops and Conditional Statements en Centro de ayuda y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!