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Transformation of matrix indices in polar coordinates - quicker way

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Markus Altthaler
Markus Altthaler 2019 年 8 月 14 日
コメント済み: Anshuman Pal 2019 年 11 月 11 日
Hey :)
To problem is a quite simple one I have a matrix say n x m and every element represents a z-value. What I want to do is transform the incices n and m into polar cooordinates based on the image center.
The center of the matrix is denoted as (n0, m0) and the last pixel by (nMax, mMax). So I used two for loops one cycling n one cycling m and then use cart2pol to transform the coordinates and save them to an array.
%array for output data
polarCoordinates = [];
%cycle through all pixels
for n:nMax
for m:mMax
%transform coordinates for singel pixel
[theta, rho, z] = cart2pol(n-n0, m-m0, image(n,m));
%append coordinates for pixel to array
polarCoordinates = [polarCoordinates; theta, rho, z];
This approach works but given the two for loops it is very slow. So I am looking for an elegant solution to speed up the process.
I know cart2pol can handle arrays of x, y, z values for transformation, e.g. n,m,image(n,m). To form those arrays I'd need to cycle tha matrix with the for loops as well or maybe not? Any ideas?
Thanks for the help!

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Neuropragmatist 2019 年 8 月 14 日
You can just pass all of the indices to cart2pol as vectors:
mat = rand(120,100);
cent = size(mat)./2;
[rr,cc] = meshgrid(1:size(mat,1),1:size(mat,2));
[TH,R,Z] = cart2pol(rr(:)-cent(1),cc(:)-cent(2),mat(:));
polarCoordinates = [TH R Z];
Here I used meshgrid to get the row and column indices of every matrix element but something similar could probably be achieved using sub2ind.
I have followed your convention of giving cart2pol row indices then column indices but remember that row and column are not x,y (rows actually run along the y axis) so these might need to be reversed. You should check that TH and R are correctly oriented for what you expect.
Hope this helps,

  2 件のコメント

Markus Altthaler
Markus Altthaler 2019 年 8 月 14 日
Hey Metioche,
thank you for the quick response!
The method works very well. Regarding your concerns about indices not being aligend properly with x and y axes , this is a good remark but in my case irrelevant as I am mainly interested in a z value over rho polt afterwads ;)
Anshuman Pal
Anshuman Pal 2019 年 11 月 11 日
How exactly would one deal with the data getting transposed here? I tried changing the following, but it didn't work:
[rr,cc] = meshgrid(1:size(mat,1),1:size(mat,2));
[rr,cc] = meshgrid(1:size(mat,2),1:size(mat,1));


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