Question

我有矩形图像或带圆角的变形矩形，如下所示：

或者这个：

有没有办法让角落与matlab平方？

然后我怎样才能得到那些新角落的坐标？

谢谢

Answer 1

<强>解释

此问题与following question类似。我的答案将与我的答案类似，并进行相关修改。

我们希望找到最符合给定形状的平行四边形角。可以通过优化找到解决方案，如下所示：

找到形状的4个角的初始猜测。这可以通过找到曲率最高的边界点来完成，并使用kmean聚类将它们聚类成4组。
通过在每对相应角之间画一条线，给出这4个角创建一个平行四边形。
找到优化边界图像的Jaccard系数和生成的平行四边形图的角落。

优化将在每个角落进行，以便节省时间。

<强>结果

初始角落猜测（角落标记为蓝色）

最终结果：

<强>代码

主脚本

%reads image and binarize it
I = rgb2gray(imread('eA4ci.jpg')) > 50;

%finds boundry of largerst connected component
boundries = bwboundaries(I,8);
numPixels = cellfun(@length,boundries);
[~,idx] = max(numPixels);
B = boundries{idx};

%finds best 4 corners
[ corners ] = optimizeCorners(B);

%generate line mask given these corners, fills the result
linesMask = drawLines(size(I),corners,corners([2:4,1],:));
rectMask = imfill(linesMask,'holes');

%remove biggest CC from image, adds linesMask instead
CC = bwconncomp(I,8);
numPixels = cellfun(@numel,CC.PixelIdxList);
[~,idx] = max(numPixels);
res = I;
res(CC.PixelIdxList{idx}) = 0;
res = res | rectMask;

优化角落功能：

function [ corners] = optimizeCorners(xy)
%finds the corners which fits the most for this set of points

Y = xy(:,1);
X = xy(:,2);

%initial corners guess
corners = getInitialCornersGuess(xy);
boundriesIm = zeros(max(Y)+20,max(X)+20);
boundriesIm(sub2ind(size(boundriesIm),xy(:,1),xy(:,2))) = 1;

%R represents the search radius
R = 7;

%continue optimizing as long as there is no change in the final result
unchangedIterations = 0;
while unchangedIterations<4

    for ii=1:4
        %optimize corner ii
        currentCorner = corners(ii,:);
        bestCorner = currentCorner;
        bestRes = calcEnergy(boundriesIm,corners);
        cornersToEvaluate = corners;

        for yy=currentCorner(1)-R:currentCorner(1)+R
            for xx=currentCorner(2)-R:currentCorner(2)+R

                cornersToEvaluate(ii,:) = [yy,xx];
                res = calcEnergy(boundriesIm,cornersToEvaluate);
                if res > bestRes
                    bestRes = res;
                    bestCorner = [yy,xx];
                end
            end
        end
        if isequal(bestCorner,currentCorner)
            unchangedIterations = unchangedIterations + 1;
        else
            unchangedIterations = 0;
            corners(ii,:) = bestCorner;

        end
    end
end

end

function res = calcEnergy(boundriesIm,corners)
%calculates the score of the corners list, given the boundries image.
%the result is acutally the jaccard index of the boundries map and the
%lines map
linesMask =  drawLines(size(boundriesIm),corners,corners([2:4,1],:));
res = sum(sum(linesMask&boundriesIm)) / sum(sum(linesMask|boundriesIm));

end

获得初始角落功能：

function corners = getInitialCornersGuess(boundryPnts)
%calculates an initial guess for the 4 corners

%finds corners by performing kmeans on largest curvature pixels
[curvatureArr] = calcCurvature(boundryPnts, 5);
highCurv = boundryPnts(curvatureArr>0.3,:);
[~,C] = kmeans([highCurv(:,1),highCurv(:,2)],4);

%sorts the corners from top to bottom - preprocessing stage
C = int16(C);
corners = zeros(size(C));

%top left corners
topLeftInd = find(sum(C,2)==min(sum(C,2)));
corners(1,:) = C(topLeftInd,:);
%bottom right corners
bottomRightInd  = find(sum(C,2)==max(sum(C,2))); 
corners(3,:) = C(bottomRightInd,:);
%top right and bottom left corners
C([topLeftInd,bottomRightInd],:) = [];
topRightInd = find(C(:,2)==max(C(:,2)));
corners(4,:) = C(topRightInd,:);
bottomLeftInd = find(C(:,2)==min(C(:,2)));
corners(2,:) = C(bottomLeftInd,:);



end



function [curvatureArr] = calcCurvature(xy, halfWinSize)
%calculate the curvature of a list of points (xy) given a window size

%curvature calculation
curvatureArr = zeros(size(xy,1),1);
for t=1:halfWinSize
    y = xy(t:halfWinSize:end,1);
    x = xy(t:halfWinSize:end,2);
    dx  = gradient(x);
    ddx = gradient(dx);
    dy  = gradient(y);
    ddy = gradient(dy);
    num   = abs(dx .* ddy - ddx .* dy)  + 0.000001;
    denom = dx .* dx + dy .* dy + 0.000001;
    denom = sqrt(denom);
    denom = denom .* denom .* denom;
    curvature = num ./ denom;

    %normalizing
    if(max(curvature) > 0)
        curvature = curvature / max(curvature);
    end

    curvatureArr(t:halfWinSize:end) = curvature;

end

end

绘制线条功能：

function mask = drawLines(imgSize, P1, P2)
%generates a mask with lines, determine by P1 and P2 points

mask = zeros(imgSize);

P1 = double(P1);
P2 = double(P2);

for ii=1:size(P1,1)
    x1 = P1(ii,2); y1 = P1(ii,1);
    x2 = P2(ii,2); y2 = P2(ii,1);

    % Distance (in pixels) between the two endpoints
    nPoints = ceil(sqrt((x2 - x1).^2 + (y2 - y1).^2));

    % Determine x and y locations along the line
    xvalues = round(linspace(x1, x2, nPoints));
    yvalues = round(linspace(y1, y2, nPoints));

    % Replace the relevant values within the mask
    mask(sub2ind(size(mask), yvalues, xvalues)) = 1;
end

如何对＃34;矩形的角落进行平方＆＃34;用matlab在bw图像中

1 个答案: