replacing ascending numbers with continous numbers

1 次查看（过去 30 天）
Daniel
Daniel 2014-1-22
评论： Daniel 2014-1-22
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

请先登录，再回答此问题。

采纳的回答

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

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Logical 的更多信息

标签

产品

Community Treasure Hunt

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

Start Hunting!

Translated by