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Create matrix with elements representing distance from centre of matrix

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Jason
Jason 2021-1-29
评论： Image Analyst 2021-1-31
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

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Image Analyst
Image Analyst 2021-1-31
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 个评论
Jason
Jason 2021-1-31
Hi IA. Im moving a microscope stage so need both the x and y distances rather than r.
Image Analyst
Image Analyst 2021-1-31
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.

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更多回答（2 个）

Jason
Jason 2021-1-30
编辑：Jason 2021-1-30
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
darova 2021-1-31
What about this?
m = 5;
n = (m-1)/2;
[X,Y] = meshgrid(-n:n);
A = sqrt(X.^2+Y.^2)
surf(A)

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