convert an image from Cartesian to Polar

邮差的信 提交于 2019-12-05 07:12:21

You can think of your image as being a 2D matrix, where each pixel has an X and Y coordinate

[(1,1)    (1,2)    (1,3)   ....   (1,c)]
[(2,1)    (2,2)    (2,3)   ....   (2,c)]
[(3,1)    (3,2)    (3,3)   ....   (3,c)]
[....     ....     ....    ....   .... ]
[(r,1)    (r,2)    (r,3)   ....   (r,c)]

In the code that you posted, it maps each of these (X,Y) coordinates to it's equivalent polar coordinate (R, theta) using the center of the image floor(c/2) and floor(r/2) as the reference point.

% Map pixel value at (1,1) to it's polar equivalent
[r,theta] = cart2pol(1 - floor(r/2),1 - floor(c/2));

So whatever pixel value was used for (1,1) should now appear in your new polar coordinate space at (r,theta). It is important to note that to do this conversion, no information about the actual pixel values in the image matters, rather we just want to perform this transformation for each pixel within the image.

So first we figure out where the center of the image is:

[r, c] = size(img);
r = floor(r / 2);
c = floor(c / 2);

Then we figure out the (X,Y) coordinates for every point in the image (after the center has already been subtracted out

[X, Y] = meshgrid(-c:c-1,-r:r-1);

Now convert all of these cartesian points to polar coordinates

[theta, rho] = cart2pol(X, Y); 

All that warp now does, is say "display the value of img at (X,Y) at it's corresponding location in (theta, rho)"

warp(theta, rho, zeros(size(theta)), img);

Now it seems that you want a new 2D image where the dimensions are [nTheta, nRho]. To do this, you could use griddata to interpolate your scattered (theta, rho) image (which is displayed by warp above) to a regular grid.

% These is the spacing of your radius axis (columns)
rhoRange = linspace(0, max(rho(:)), 100);

% This is the spacing of your theta axis (rows)
thetaRange = linspace(-pi, pi, 100);

% Generate a grid of all (theta, rho) coordinates in your destination image
[T,R] = meshgrid(thetaRange, rhoRange);

% Now map the values in img to your new image domain
theta_rho_image = griddata(theta, rho, double(img), T, R);

Take a look at all the interpolation methods for griddata to figure out which is most appropriate for your scenario.

There were a couple other issues (like the rounding of the center) which caused the result to be slightly incorrect. A fully working example is provided below

% Create an image of circles
radii = linspace(0, 40, 10);

rows = 100;
cols = 100;
img = zeros(rows, cols);

for k = 1:numel(radii)
    t = linspace(0, 2*pi, 1000);
    xx = round((cos(t) * radii(k)) + (cols / 2));
    yy = round((sin(t) * radii(k)) + (rows / 2));

    toremove = xx > cols | xx < 1 | yy > rows | yy < 1;

    inds = sub2ind(size(img), xx(~toremove), yy(~toremove));

    img(inds) = 1;
end

[r,c] = size(img);
center_row = r / 2;
center_col = c / 2;

[X,Y] = meshgrid((1:c) - center_col, (1:r) - center_row);

[theta, rho] = cart2pol(X, Y);

rhoRange = linspace(0, max(rho(:)), 1000);
thetaRange = linspace(-pi, pi, 1000);

[T, R] = meshgrid(thetaRange, rhoRange);

theta_rho_image = griddata(theta, rho, double(img), T, R);

figure
subplot(1,2,1);
imshow(img);
title('Original Image')

subplot(1,2,2);
imshow(theta_rho_image);
title('Polar Image')

And the result

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