Elementwise division in matrix notation
Mostrar comentarios más antiguos
Dear all,
I am trying to write the following code in matrix notation:
A = B. / C;
A, B and C are column vectors with 1435 rows.
I have already managed to do this with multiplication, i.e.:
D = E.*F;
This is equivalent to
D = diag(F)*E;
Also in this case D, E and F are also column vectors with 1435 rows.
I want to do this because it is more foolproof and does not give results if the dimensions do not match.
Best,
Hylke
6 comentarios
Hylke Dijkstra
el 28 de En. de 2022
Torsten
el 28 de En. de 2022
If B and C are column vectors of the same length,
A = diag(C)\B
should work fine.
Hylke Dijkstra
el 28 de En. de 2022
Hylke Dijkstra
el 28 de En. de 2022
Respuesta aceptada
Más respuestas (1)
Image Analyst
el 28 de En. de 2022
Editada: Image Analyst
el 28 de En. de 2022
Try getting rid of the space between the dot and the slash:
A = B ./ C;
If the length of B and C don't match, you might want to consider if you even want to divide them. Like, WHY don't they match? If one is shorter to you want to just assign the "left over" elements to something specific, like 0 or 1 or B or C or something?
4 comentarios
Hylke Dijkstra
el 28 de En. de 2022
Image Analyst
el 28 de En. de 2022
Yes, sometimes mistakes are made, and wouldn't you want an error to be thrown in those situations? Or would you rather not know about them? Anyway, it can't decide how to proceed upon an error unless you tell it. Like
try
A = B ./ C;
catch ME
% Some code to handle different length vectors, whatever that might be.
end
Usually it doesn't just try to "fix" the problem automatically, which is a good thing because its fix may not be what you were hoping it would do.
Hylke Dijkstra
el 28 de En. de 2022
Not using diag() will also give an error. Look:
B = rand(10, 1);
C = rand(2, 1);
A = B ./ C;
Categorías
Más información sobre Creating and Concatenating Matrices 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!
