replacing ascending numbers with continous numbers

1 Ansicht (letzte 30 Tage)
Daniel
Daniel am 22 Jan. 2014
Kommentiert: Daniel am 22 Jan. 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

Akzeptierte Antwort

Amit
Amit am 22 Jan. 2014
A = [4 20 35 22 10 49];
B = 1:numel(A);
[~,C] = sort(A);
B(C) = B; % Your vector
  3 Kommentare
Amit
Amit am 22 Jan. 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
Daniel am 22 Jan. 2014
Mile grazie! Thank you very much. This is way more elegant than anything I managed to butchered.

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Produkte

Community Treasure Hunt

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

Start Hunting!

Translated by