Representación de matriz de transformación de radón

Question

Estoy buscando una solución MATLAB para generar la representación matricial de un discreto Radon transform (DRT). Es decir, dada una versión vectorizada de una imagen MxN, X, me gustaría generar la matriz R de modo que R*X(:) sea un DRT de la imagen. En MATLAB, estoy esperando a ser algo como lo siguiente:

>> X = 2D_Image_Of_Size_MxN; 
>> R = DRT_Matrix_Of_Size_LPxMN; 
>> DRT = reshape(R * X(:), L, P); 

Sé que hay varias maneras de definir un DRT, así que sólo voy a decir que estoy en busca de un estándar normal de o o no demasiado fuera de lo común implicación.

Answer 1

function [ R rho theta ] = radonmatrix(drho, dtheta, M, N) 
% radonmatrix - Discrete Radon Trasnform matrix 
% 
% SYNOPSIS 
% [ R rho theta ] = radonmatrix(drho, dtheta, M, N) 
% 
% DESCRIPTION 
% Returns a matrix representation of a Discrete Radon 
% Transform (DRT). 
% 
% INPUT 
% drho  Radial spacing the the DRT. 
% dtheta Angular spacing of the DRT (rad). 
% M  Number of rows in the image. 
% N  Number of columns in the image. 
% 
% OUTPUT 
% R  LP x MN DRT matrix. The values of the L and 
%   P will depend on the radial and angular spacings. 
% rho  Vector of radial sample locations. 
% theta Vector of angular sample locations (rad). 
% 

% For each angle, we define a set of rays parameterized 
% by rho. We then find the pixels on the MxN grid that 
% are closest to each line. The elements in R corresponding 
% to those pixels are given the value of 1. 

% The maximum extent of the region of support. It's for 
% rho = 0 and theta = pi/4, the line that runs caddy-corner. 
W = sqrt(M^2 + N^2); 

rho = -W/2 : drho : W/2; 
theta = 0 : dtheta : 180 - dtheta; 

L = length(rho); 
P = length(theta); 

R = false(L*P, M*N); 

% Define a meshgrid w/ (0,0) in the middle that 
% we can use a standard coordinate system. 
[ mimg nimg ] = imggrid(1, 1, [ M N ]); 

% We loop over each angle and define all of the lines. 
% We then just figure out which indices each line goes 
% through and put a 1 there. 
for ii = 1 : P 

    phi = theta(ii) * pi/180; 

    % The equaiton is rho = m * sin(phi) + n * cos(phi). 
    % We either define a vector for m and solve for n 
    % or vice versa. We chose which one based on angle 
    % so that we never g4et close to dividing by zero. 
    if(phi >= pi/4 && phi <= 3*pi/4) 

    t = -W : min(1/sqrt(2), 1/abs(cot(phi))) : +W; 
    T = length(t); 

    rhom = repmat(rho(:), 1, T); 
    tn = repmat(t(:)', L, 1); 
    mline = (rhom - tn * cos(phi)) ./ sin(phi); 

    for jj = 1 : L 
     p = round(tn(jj,:) - min(nimg)) + 1; 
     q = round(mline(jj,:) - min(mimg)) + 1; 
     inds = p >= 1 & p <= N & q >= 1 & q <= M; 
     R((ii-1)*L + jj, unique(sub2ind([ M N ], q(inds), p(inds)))) = 1; 
    end 

    else 

    t = -W : min(1/sqrt(2), 1/abs(tan(phi))) : +W; 
    T = length(t); 

    rhon = repmat(rho(:)', T, 1);  
    tm = repmat(t(:), 1, L); 
    nline = (rhon - tm * sin(phi)) ./ cos(phi); 

    for jj = 1 : L 
     p = round(nline(:,jj) - min(nimg)) + 1; 
     q = round(tm(:,jj) - min(mimg)) + 1; 
     inds = p >= 1 & p <= N & q >= 1 & q <= M; 
     R((ii-1)*L + jj, unique(sub2ind([ M N ], q(inds), p(inds)))) = 1; 
    end 

    end 

end 

R = double(sparse(R)); 

return; 

Aquí está la función imggrid utilizada en lo anterior.

function [ m n ] = imggrid(dm, dn, sz) 
% imggrid -- Returns rectilinear coordinate vectors 
% 
% SYNOPSIS 
% [ m n ] = imggrid(dm, dn, sz) 
% 
% DESCRIPTION 
% Given the sample spacings and the image size, this 
% function returns the row and column coordinate vectors 
% for the image. Both vectors are centered about zero. 
% 
% INPUT 
% dm  Spacing between rows. 
% dn  Spacing between columns. 
% sz  2x1 vector of the image size: [ Nrows Ncols ]. 
% 
% OUTPUT 
% m  sz(1) x 1 row coordinate vector. 
% n  1 x sz(2) column coordinate vector. 

M = sz(1); 
N = sz(2); 

if(mod(M, 2) == 0) 
    m = dm * (ceil(-M/2) : floor(M/2) - 1)'; 
else 
    m = dm * (ceil(-M/2) : floor(M/2))'; 
end 

if(mod(N, 2) == 0) 
    n = dn * (ceil(-N/2) : floor(N/2) - 1); 
else 
    n = dn * (ceil(-N/2) : floor(N/2)); 
end 

Answer 2

image = phantom(); 
projection = radon(image); 
R = linsolve(reshape(image,1,[]), reshape(projection,1,[]));