What may be the best way to order columns of a matrix?
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I'd like to order each column for a matrix so that the smallest number is replaced with 1, the second smallest number is replaced with 2 and so on. Say, convert mat to mat2. Any way to do this fast?
The code for before-converted matrix: mat = [20 1 4;5 3 1;-3 1 9;9 1 1];
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/152444/image.png)
1 Kommentar
John D'Errico
am 3 Dez. 2015
PLEASE DO NOT PASTE IN AN IMAGE. That makes it impossible for us to use your array without typing it in ourselves.
Antworten (1)
Guillaume
am 3 Dez. 2015
The second return value of sort tells you which row should receive 1,2,3,4... You can use sub2ind, bsxfun or a loop to distribute the values:
mat = [20 1 4;5 3 1;-3 1 9;9 1 1];
[~, order] = sort(mat);
mat2(sub2ind(size(mat), order, repmat(1:size(mat, 2), size(mat, 1), 1))) = repmat((1:size(mat, 1))', 1, size(mat, 2));
mat2 = reshape(mat2, size(mat))
5 Kommentare
Guillaume
am 3 Dez. 2015
There is/(was?) a page in the documentation that describes the exact rules for indexing. I can't find it at the moment.
The rule is rather ill-designed in my opinion as it has exceptions, but in general:
A(I)
has the same shape as I and the shape of A is lost. Exceptions come when both are vector, then it's the shape of A.
If A does not exist prior to assignment, then it would appear that A is always a row vector.
The shape of the assigned is always lost, and values are assigned in linear order, that is by column. So in your example, [1 -1; 20 2], the values are flattened in the order [1 20 -1 2].
Similarly in mat2([2 4; 1 3]), the indices are flattened into [2 1 4 3]. In the other case, the indices are already a vector.
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