Image interpolation from random pixels

问题

I would like to ask a question regarding single channel image interpolation. Single channel is chosen just for simplicity otherwise I'm working on multiple channel images. Assume there is a single channel image with pure black background ( pixel intensity 0) on which there are some pixels with non-zero intensity values. I want to apply an interpolation algorithm to fill the entire black area of the image with interpolated values coming from the neighboring non-zero intensity pixels.

What interpolation algorithm would you recommend for a smooth interpolation applicable to this problem?

As inputs, we of course know the location of those non-black pixels and their intensity. But the location is somewhat random ( in one row may be 10 pixels, in another row only 8).

回答1:

The regular interp2 will not work here, since your points are not located at regular intervals (Not sitting on a grid). You can either try TriScatteredInterp or download inpaint_nans from the file exchange.

Here is the solution in your case with TriScatteredInterp:

function solveStackOverflowProblem()
    im = imread('http://i.stack.imgur.com/lMaYR.png');
    im = im(:,:,2);
    [i,j] = find(im);
    y = j; x = i;
    indexes = sub2ind(size(im),i,j);
    interpolator = TriScatteredInterp(x,y,double(im(indexes)));

    [Y,X] = meshgrid( 1:size(im,2),1:size(im,1));
    reconstructedImage = interpolator(X,Y);

    figure;imshow(reconstructedImage/255)
end

回答2:

Your best solution is to use gridfit. Its designed to improve on all the native Matlab functions like TriScatteredInterp and griddata.

回答3:

Here is a solution based on using Radial Basis Functions ( in this case a Gaussian ) to build interpolations for randomly spaced points with varying intensities.

Essentially this drops a Gaussian on each point, weights it by the point intensity and sums the result.

The sharpness of the interpolation function is controllable via the standard deviation of the chosen Gaussian function.

The points:

The interpolation:

Viewed in 3D:

With a smaller standard deviation:

The code:

pts = Table[{{RandomReal[{0, 200}], RandomReal[{0, 200}]}, 
   RandomReal[]}, {20}]

dists = Function[points, 
  Plus @@ ((PDF[
          MultinormalDistribution[#, 200 IdentityMatrix[2]], {x, 
           y}] & /@ points[[All, 1]] ) points[[All, 2]])/Length@points]

DensityPlot[dists[pts], {x, 0, 200}, {y, 0, 200}, PlotPoints -> 100]

回答4:

For a relatively small number of points, the ideal way to interpolate them would be to create a triangular mesh, using only the vertices of each region to determine pixels within that region, using a weighted mean to determine each pixel's color.

To find the color of a pixel inside a triangular region, the weights to use for each pixel are, for the points, A, B, and C, respectively, (bc-A)/a, (ac-B)/b, and (a*b-C)/c. This ensures that each point's influence deteriorates to zero as the point gets closer to the opposite edge, so that transitions between triangles are smooth.

Keep in mind that you can use any type of mean, including harmonic and geometric, not just arithmetic, to do the calculation (appearances will change, but other types of means might be better anyway).

回答5:

Any interpolation will look really bad with this few non-zero points, but you can try convolving the image with a 2-D filter such as a Gaussian kernel and see if you like it.

来源：https://stackoverflow.com/questions/12429283/image-interpolation-from-random-pixels