Question

正如您在下面的图片中看到的，在我的一些分割结果中（通过分水岭变换方法进行分割），剩下一些残羹剩饭。我想以某种方式裁剪图像，以便只保留矩形。此操作仅基于矩形，与强度级别无关。

Answer 1

解决方案说明

我建议采用以下方法：

根据几何属性生成形状的4个角的初始猜测（详见下面的代码）。
通过在每对相应角之间画一条线，在给定这4个角的情况下创建一个四边形。
找到优化边界图像的Jaccard系数和生成的四边形地图的角落。

为了节省时间，优化阶段将在本地完成。我们将尝试用某个社区中可实现的最佳角落替换每个角落。如果4个角落中的每个角落都没有改善，我们就会停止优化阶段。

<强>码

%reads image
gray = rgb2gray(imread('Bqx51.png'));
I = gray>0;

%extract boundries
B = bwboundaries(I,8);
B = B{1};
boundriesImage = zeros(size(I));
boundriesImage(sub2ind(size(I),B(:,1),B(:,2))) = 1;

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

%generate line mask
linesMask =  drawLines(size(I),corners,corners([2:4,1],:));

%fill holes
rectMask = imfill(linesMask,'holes');

%noise reduction
rectMask = I & rectMask;
rectMask = imopen(rectMask,strel('disk',2));

%calculate result 
result = gray;
result(~rectMask) = 0;

%display results
figure,imshow([gray, 255*ones(size(I,1),1),result]);

角落优化功能

function [ corners] = optimizeCorners(pnts)
%OPTIMIZE4PTS Summary of this function goes here
%   Detailed explanation goes here

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

corners = getInitialGuess(X,Y); 
boundriesIm = zeros(max(Y),max(X));
boundriesIm(sub2ind(size(boundriesIm),pnts(:,1),pnts(:,2))) = 1;

%R represents the search radius
R = 3;

%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;
        candidateInds = sum(((repmat(currentCorner,size(X,1),1)-[Y,X]).^2),2)<(R^2);
        candidateCorners = [Y(candidateInds),X(candidateInds)];
        for jj=length(candidateCorners)
            xx = candidateCorners(jj,2);
            yy = candidateCorners(jj,1);
            cornersToEvaluate(ii,:) = [yy,xx];
            res = calcEnergy(boundriesIm,cornersToEvaluate);
            if res > bestRes
                bestRes = res;
                bestCorner = [yy,xx];
            end
        end
        if isequal(bestCorner,currentCorner)
            unchangedIterations = unchangedIterations + 1;
        else
            unchangedIterations = 0;
            corners(ii,:) = bestCorner;

        end
    end
end

end

计算能量函数

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

找到角落功能的初始猜测

function corners = getInitialGuess(X,Y)
%calculates an initial guess for the 4 corners

corners = zeros(4,2);

%preprocessing stage
minYCoords = find(Y==min(Y));
maxYCoords = find(Y==max(Y));
minXCoords = find(X==min(X));
maxXCoords = find(X==max(X));
%top corners
topRightInd = find(X(minYCoords)==max(X(minYCoords)),1,'last');
topLeftInd = find(Y(minXCoords)==min(Y(minXCoords)),1,'last');
corners(1,:) = [Y(minYCoords(topRightInd)) X((minYCoords(topRightInd)))];
corners(2,:) = [Y(minXCoords(topLeftInd)) X((minXCoords(topLeftInd)))];
%bottom corners
bottomRightInd = find(Y(maxXCoords)==max(Y(maxXCoords)),1,'last');
bottomLeftInd = find(X(minYCoords)==min(X(minYCoords)),1,'last');
corners(4,:) = [Y(maxXCoords(bottomRightInd)) X((maxXCoords(bottomRightInd)))];
corners(3,:) = [Y(maxYCoords(bottomLeftInd)) X((maxYCoords(bottomLeftInd)))];


end

drawLine函数（取自@Suever的以下answer）

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

<强>结果

第一张图片（之前和之后）：

第二张图片（之前和之后）：

<强>运行

Elapsed time is 0.033998 seconds.

可能的改进/建议

能量函数还可能包括鼓励平行线具有相似斜率的约束（在您的示例中，它们不具有相同的斜率）。
能量函数可能包括鼓励每个角度接近90度的约束。
降噪阶段（如imclose）可以在执行此方法之前完成，以消除小的伪影。
可以使用多个初始猜测运行算法并选择最佳算法。
请注意，此解决方案并未估算出最佳矩形 - 它估计最佳四边形。原因是输入图像不是矩形（线条不平行）。

修剪图像片段的剩余部分

1 个答案: