sort a matrix in a specific way

Hi,

I have two matrices that correspond to points coordinates and I want to sort their concatenation in a way illustrated below as following:

%first matrix
A = [0.1 0; 
     0.5 0; 
     1 0.3; 
     1 0.7; 
     0.6 1; 
     0.1 1; 
     0 0.9; 
     0 0.2];
%second matrix
B = [0 0;
     1 0;
     1 1;
     0 1];
%concatenate
C = [A ; B];
%sort matrix C in a way that all cooridnates are sorted in a square-like manner and looks like:
% C = [0 0;
% 0.1 0;
% 0.5 0;
% 1 0;
% 1 0.3;
% 1 0.7;
% 1 1;
% 0.6 1;
% 0.1 1;
% 1 0;
% 1 0.9;
% 0 0.2];

Any help would be appreicted.

[dpb]

I don't really see what is "sorted" about the result??? How did you arrive at that particular permutation?

[LH]

All coordinates are in a square-like order. If you imagine you have a square with vertices x=[0 1 1 0] and y[0 0 1 1], then if you have points along its edges, they must be order like matrix C.

Answer 1

[Matt J]

%first matrix
A = [0.1 0; 
     0.5 0; 
     1 0.3; 
     1 0.7; 
     0.6 1; 
     0.1 1; 
     0 0.9; 
     0 0.2];
%second matrix
B = [0 0;
     1 0;
     1 1;
     0 1];
%concatenate
C = [A ; B];

k = convhull(C);
C=C(k,:)

C = 13×2

0.1000 0 0.5000 0 1.0000 0 1.0000 0.3000 1.0000 0.7000 1.0000 1.0000 0.6000 1.0000 0.1000 1.0000 0 1.0000 0 0.9000

[LH]

Is there a reason behind some of the rows might disappear when using this method?

[Matt J]

Some of the points are not perfectly on the boundary of the square.

[LH]

Thanks, makes sense. Is there another command/method I can use to do the extact sorting but with keeping all points? If not, I believe coding this will be the only way.

[Matt J]

I have given a second answer which should keep all of the points.

Answer 2

[Matt J]

A = [0.1 0; 
     0.5 0; 
     1 0.3; 
     1 0.7; 
     0.6 1; 
     0.1 1; 
     0 0.9; 
     0 0.2];
%second matrix
B = [0 0;
     1 0;
     1 1;
     0 1];
%concatenate
C = [A ; B];
D=normalize(C,'center');

[~,k]=sort(atan2(D(:,2),D(:,1)));

C=C(k,:)

C = 12×2

0 0.2000 0 0 0.1000 0 0.5000 0 1.0000 0 1.0000 0.3000 1.0000 0.7000 1.0000 1.0000 0.6000 1.0000 0.1000 1.0000

[LH]

Many thanks for this. Although the first rows should be placed at the end of the matrix (if we consider going anticlockwise starting from the bottom edge of the square), but this method is better since it keeps all element.

[Matt J]

You can use circshift to choose the starting point for the list.

[LH]

Great. Thanks.