Create matrix with elements representing distance from centre of matrix

Hello. If i have a 3x3, 5x5 or 7x7 matrix or generally a nxn matrix where n is odd, i want each element to represent the distance from the centre of the matrix. I.e the top left element in a 3x3 matrix to represent up one row and left 1 column. I.e (1,-1).

I then want to unravel this matrix into a 1D vector but in a serpentine fashion rather than raster scan

Answer 1

[Image Analyst]

Jason if you want a for loop way of doing it, you can do this:

rows = 3;
columns = 4;
distances = zeros(rows, columns);
midRow = mean([1, rows])
midCol = mean([1, columns])
for col = 1 : columns
	for row = 1 : rows
		distances(row, col) = sqrt((row - midRow) .^ 2 + (col - midCol) .^ 2);
	end
end
distances

Some results:

For 3,4:

distances =

1.8028 1.118 1.118 1.8028

1.5 0.5 0.5 1.5

1.8028 1.118 1.118 1.8028

For 3,3

distances =

1.4142 1 1.4142

1 0 1

1.4142 1 1.4142

For 4,4

distances =

2.1213 1.5811 1.5811 2.1213

1.5811 0.70711 0.70711 1.5811

1.5811 0.70711 0.70711 1.5811

2.1213 1.5811 1.5811 2.1213

[Jason]

Hi IA. Im moving a microscope stage so need both the x and y distances rather than r.

[Image Analyst]

Jason, it's so trivial, I'm sure you did it by now, but for what it's worth:

rows = 3;
columns = 4;
deltax = zeros(rows, columns);
deltay = zeros(rows, columns);
midRow = mean([1, rows])
midCol = mean([1, columns])
for col = 1 : columns
	for row = 1 : rows
		deltax(row, col) = col - midCol;
		deltay(row, col) = row - midRow;
	end
end
deltax
deltay

For the number of rows and columns shown, this is what you see.

midRow =

2

midCol =

2.5

deltax =

-1.5 -0.5 0.5 1.5

-1.5 -0.5 0.5 1.5

-1.5 -0.5 0.5 1.5

deltay =

-1 -1 -1 -1

0 0 0 0

1 1 1 1

But you might want to use meshgrid(). Just make sure you have the center location properly located with the right value. Like, if you have an even number of rows, the "middle" will not be exactly at a particular row, but in between existing rows.

Answer 2

[Jason]

Here's my attempt - not sure if its the best way:

            n=5;                  % n must be odd (nxn matrix)
            y=1:n^2               % create the numbers to put into a matrix
            A=reshape(y,n,n);     % create the matrix
            A=A'
            %n=length(y);
            
            for i=2:2:n                
                r=A(i,:);
                r=fliplr(r);   %Flip every even row, create serpentine pattern
                A(i,:)=r;
            end
            A
            M=[];
            for i=1:n^2
                [r,c]=find(A==i);                   
                M(i,1)=r;
                M(i,2)=c;
            end
            M
            
            %Get middle element position 
            m=ceil(n/2)
            
            %subtract from each M
            M(:,1)=M(:,1)-m
            M(:,2)=M(:,2)-m
            M               % This is now a vector of positions relative to the origin

Answer 3

[darova]

What about this?

m = 5;
n = (m-1)/2;
[X,Y] = meshgrid(-n:n);
A = sqrt(X.^2+Y.^2)
surf(A)