# How Can I establish a raking scale of variables within an array?

3 visualizaciones (últimos 30 días)
Fed el 10 de Mzo. de 2015
Comentada: Andrew Newell el 10 de Mzo. de 2015
Hi everyone!
This is my array:
a = [0 0 1 3];
I want to define which is the highest value and therefore define a rank in this way:
a_ranking = [4 4 2 1];
where the 4th value is 1 because it was the most chosen.
This is how I found the max values (first and second):
maxH = max(a);
FirstH = find(a == maxH);
secMaxH = max(a(a~= maxH));
SecondH = find(a == secMaxH);
so, I know the in the 4th position there is the highest value and in the third the second one, but how Can I associate the value 1 and 2 respectably to the the third and fourth position of the array.
But I am lost when I have to re-write the array in this way
a_ranking = [4 4 2 1];
I don't know how to reach this point!
Basically, I need to create a array with the same length of the one I am using but indicating the ranking scale of the chosen answer. Is that clear?
##### 1 comentarioMostrar -1 comentarios más antiguosOcultar -1 comentarios más antiguos
Andrew Newell el 10 de Mzo. de 2015
When displaying code, please leave a blank line above and then indent with a double space so it displays correctly (see markup).

Iniciar sesión para comentar.

Andrew Newell el 10 de Mzo. de 2015
Editada: Andrew Newell el 10 de Mzo. de 2015
Your ranking scheme is somewhat unusual because usually the ranking does not have any gaps in it. That would be easy to implement using unique, which not only finds the unique elements and sorts them but returns indices that are, in effect, ranks:
[~,~,a_ranking] = unique(max(a)-a)
a_ranking =
3
3
2
1
Are you sure this isn't what you want?
##### 2 comentariosMostrar NingunoOcultar Ninguno
Fed el 10 de Mzo. de 2015
thank you! I've never used the function unique. In this way it is much simpler. Since I have to multiple each element of ic for a different weight on the basis of their ranking, I thought it could be better to consider the last two values as 4 instead of 3, in order to give to them the less weight. do you understand?
And by the way your solution is much better than mine :)
Andrew Newell el 10 de Mzo. de 2015
You're welcome. I don't know which weighting scheme would be better - depends on the application. If this solves the problem, can you click on "Accept this answer"?

Iniciar sesión para comentar.

### Categorías

Más información sobre Operating on Diagonal Matrices en Help Center y File Exchange.

### Community Treasure Hunt

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

Start Hunting!

Translated by