# Create matrix with elements representing distance from centre of matrix

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Jason on 29 Jan 2021
Commented: Image Analyst on 31 Jan 2021
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

Image Analyst on 31 Jan 2021
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
##### 2 CommentsShowHide 1 older comment
Image Analyst on 31 Jan 2021
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.

Jason on 30 Jan 2021
Edited: Jason on 30 Jan 2021
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

darova on 31 Jan 2021
m = 5;
n = (m-1)/2;
[X,Y] = meshgrid(-n:n);
A = sqrt(X.^2+Y.^2)
surf(A) R2020b

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