Polynomial fitting of each pixel in an image without looping
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I have 3x7 images of 256x256 pixels stored in a cell array, i.e. for each pixel i have 7 x-values, 7 y-values and 7 z-values. I want to find the coefficients for z=k1*x + k2*y + k3*x^2 + k4*y^2 + k5*x*y in a least square sense for each pixel without looping over each pixel. Is there a more efficient way to do this?
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Respuesta aceptada
ChristianW
el 11 de Feb. de 2013
Editada: ChristianW
el 13 de Feb. de 2013
Referring to your 256 sec calulation time:
Got it to 1 sec and 0.8 with parfor. (on my cpu)
dim = 256;
C = mat2cell(randi(255,dim*3,dim*7), dim*ones(1,3), dim*ones(1,7));
tic
C0 = cellfun(@(x) reshape(x,1,[]),C,'UniformOutput',false);
X = cat(1,C0{1,:});
Y = cat(1,C0{2,:});
Z = cat(1,C0{3,:});
K = cell(size(C{1}));
for i = 1:size(X,2) % 1:NumberOfPixelsPerImage
K{i} = [X(:,i), Y(:,i), X(:,i).^2, Y(:,i).^2, X(:,i).*Y(:,i)]\Z(:,i);
end
toc
Más respuestas (4)
Image Analyst
el 10 de Feb. de 2013
I don't understand your data layout. So you have a cell array with 3 rows and 7 columns. What is inside each cell? Is each cell a 256 by 256 array of either x, y, or z values, like
{256x256 x1 values}, {256x256 x2 values},....{256x256 x7 values};
{256x256 y1 values}, {256x256 y2 values},....{256x256 y7 values};
{256x256 z1 values}, {256x256 z2 values},....{256x256 z7 values};
ChristianW
el 11 de Feb. de 2013
Is the overall result just 5 scalar k values?
X = cat(1,C{1,:});
Y = cat(1,C{2,:});
Z = cat(1,C{3,:});
M = [X(:), Y(:), X(:).^2, Y(:).^2, X(:).*Y(:)];
K = M\Z(:); % Z = M*K
7 comentarios
ChristianW
el 11 de Feb. de 2013
I'll give it a second shot. I need some help with the math.
Lets talk about a single pixel only. With 7 xValues in X(:,1), each row one of the 7 pictures (analogously for Y and Z), like this:
X = [pixel1_image1
pixel1_image2
...
pixel1_image7];
With these inputs, does this function solve the equations for that pixel?
function K = fcn(X,Y,Z)
M = [X, Y, X.^2, Y.^2, X.*Y];
K = M\Z; % Z = M*K
I need a check for that function or an example input with correct output to varify.
Teja Muppirala
el 12 de Feb. de 2013
Here's an approach using sparse matrices to do it. this works in about 0.3 seconds for me, and gives the coefficients in a 5x65536 matrix K.
It should be noted that using a simple for-loop is much simpler to implement, and still works in about 0.6 seconds if you preallocate properly.
% Making random data
dim = 256;
C = mat2cell(randi(255,dim*3,dim*7), dim*ones(1,3), dim*ones(1,7));
tic
C0 = cellfun(@(x) reshape(x,1,[]),C,'UniformOutput',false);
X = cat(1,C0{1,:});
Y = cat(1,C0{2,:});
Z = cat(1,C0{3,:});
M = permute( cat(3,X,Y,X.^2,Y.^2,X.*Y), [1 3 2]);
% Generate the locations of the block-diagonal sparse entries
jloc = repmat(1:(dim^2*5),7,1);
iloc = bsxfun(@plus, repmat((1:7)',1,5) ,reshape( 7*(0:dim^2-1) , 1, 1, []));
SM = sparse(iloc(:),jloc(:),M(:));
% Do the pseudoinversion
K = (SM'*SM) \ (SM'*Z(:));
K = reshape(K,5,[]);
toc
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