replacing ascending numbers with continous numbers

1 visualización (últimos 30 días)
Daniel el 22 de En. de 2014
Comentada: Daniel el 22 de En. de 2014
Hello everybody,
I am looking for a fast and efficient way to convert a vector of non continous ascending numbers to continous numbers.
As an example:
The vector
[4 20 35 22 10 49]
should become
[1 3 5 4 2 6]
at the moment I'm using a for loop, look for the lowest number write the runtime index into the result vector and set the position in the original vector to an illegal value. I know this is not the best way but it was the only one I could think of.
Also I'm not doing a single vector at a time but doing this to the columns of a matrix.
Thank you
0 comentariosMostrar -2 comentarios más antiguosOcultar -2 comentarios más antiguos

Iniciar sesión para comentar.

Amit el 22 de En. de 2014
A = [4 20 35 22 10 49];
B = 1:numel(A);
[~,C] = sort(A);
B(C) = B; % Your vector
3 comentariosMostrar 1 comentario más antiguoOcultar 1 comentario más antiguo
Amit el 22 de En. de 2014
% Your matrix is named A
[m,n] = size(A);
B = 1:m;
[~,C] = sort(A);
D = zeros(m,n);
for i = 1:n
D(C(:,i),i) = B;
end
Daniel el 22 de En. de 2014
Mile grazie! Thank you very much. This is way more elegant than anything I managed to butchered.

Iniciar sesión para comentar.

Categorías

Más información sobre Logical 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