can I equation of a curve from an image?

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Dekel Mashiach 2022-4-25

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https://ww2.mathworks.cn/matlabcentral/answers/1704545-can-i-equation-of-a-curve-from-an-image

评论： Matt J 2022-4-25

test1.png

Is it possible to find the equation of a curve from an image?

I want to find the equation of the line in green:

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Dekel Mashiach 2022-4-25

They are supposed to be parallel, and the green line is the middle (centerline). Simulate a route of a road that a vehicle should drive on the middle line.

Matt J 2022-4-25

Note that even if your lines are parallel in the 3D world, they will geneally not be parallel in the 2D camera image, depending on the camera perspective. Likewise, the centerline in the 2D image will not correspond to the centerline in 3D.

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Answer 1

Matt J 2022-4-25

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https://ww2.mathworks.cn/matlabcentral/answers/1704545-can-i-equation-of-a-curve-from-an-image#answer_950420

编辑：Matt J 2022-4-25

在 MATLAB Online 中打开

This assumes your Image is already cast to type logical. The slope and intercept are for the center-line in matrix (row,column) coordinates.

reg=regionprops(Image,'PixelList');
x1=reg(1).PixelList(:,2);
y1=reg(1).PixelList(:,1);
x2=reg(2).PixelList(:,2);
y2=reg(2).PixelList(:,1);
A=[x1,x1.^0,0*x1;
   x2,0*x2,x2.^0];
b=[y1;y2];
c=A\y;
slope=c(1);
intercept=mean(c(2:3));

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Dekel Mashiach 2022-4-25

编辑：Matt J 2022-4-25

在 MATLAB Online 中打开

I identify the lines by the camera and then create the green line in the middle of the lines. I want to find the green line equation. Here is the code for finding the green line:

labeledImage = bwlabel(mask);
line1 = ismember(labeledImage, 1);
line2 = ismember(labeledImage, 2);
% Get rows and columns of each line.
[r1, c1] = find(line1);
[r2, c2] = find(line2);
for k = 1 : length(r1)
    distances = sqrt((r1(k) - r2) .^ 2 + (c1(k) - c2) .^ 2);
    [minDistance, index] = min(distances);
    % Find the midPoint
    midX(k) = mean([c1(k), c2(index)]);
    midY(k) = mean([r1(k), r2(index)]);
    % Burn into mask
    mask(round(midY(k)), round(midX(k))) = true;
end
% Need to add a small amount of noise to x to make the values unique.
midX = midX + 0.001 * rand(size(midX));
midY = midY + 0.001 * rand(size(midY));
% Interpolate x and y to make sure there are no gaps.
kVec = 1 : length(midX);
kFit = linspace(1, kVec(end), 10000);
xFit = interp1(kVec, midX, kFit, 'linear');
yFit = interp1(kVec, midY, kFit, 'linear');
% Remove duplicate values
xy = unique(round([xFit(:), yFit(:)]), "rows");
% Extract individual x and y.
midX = xy(:, 1);
midY = xy(:, 2);

Matt J 2022-4-25

编辑：Matt J 2022-4-25

在 MATLAB Online 中打开

If you are fitting a single polynomial of known order, N, the generalization of my answer to N>1 is

reg=regionprops(Image,'PixelList');
x1=reg(1).PixelList(:,2);
y1=reg(1).PixelList(:,1);
x2=reg(2).PixelList(:,2);
y2=reg(2).PixelList(:,1);
e=N:-1:1;
A=[x1.^e, x1.^0, 0*x1;
   x2.^e, 0*x2, x2.^0];
b=[y1;y2];
c=A\y;  
coefficients=[c(1:N-2)',mean(c(N-1:1))]; %polynomial coefficients

Matt J 2022-4-25

编辑：Matt J 2022-4-25

I identify the lines by the camera and then create the green line in the middle of the lines. I want to find the green line equation. Here is the code for finding the green line:

If you already have the points on the green line, what prevents you from using fit(), lsqcurvefit(), or similar? Also, you have articulated that you want to model this as a polynomial, but of what order? Are you sure you don't need a piecewise polynomial? In any case, fit() offers all of these.

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