Question

给定一个图像矩阵，如何获得棋盘距像素 A 小于 D 的像素的位置。我需要对所有像素执行此操作。

使用MATLAB函数bwdist无法提供所需的结果。有什么解决方案？

[D,idx] = bwdist(Img,'chessboard');

Answer 1

给出图像，像素和最大距离：

% Test image
Image = zeros(20,30);

% Maximum chessboard distance from image
maxDist = 7;

% The pixel from which to measure distance
pix = [4,19];

要查找棋盘距pix的像素是小于maxDist并且在图片范围内：

选项1：使用bwdist

% Create a binary image with all pixels zero except 'pix'
bw = zeros(size(Image));
bw(pix(1), pix(2)) = 1;

% Get the chessboard distance transform
[D,idx] = bwdist(bw,'chessboard');

% Get the linear index of 'pix' 
pixInd = sub2ind(size(bw), pix(1), pix(2));

% Find linear indices of pixels who's chessboard distance from pixel are 
% less than 'maxDist'
pointsInd = find(idx == pixInd & D < maxDist);

% Remove 'pix'
pointsInd(pointsInd == pixInd) = [];

% Get the pairs of (x,y) of the pixels
[pointsX, pointsY] = ind2sub(size(bw), pointsInd);

选项2：使用meshgrid

% Get the range of x and y indices who's chessboard distance from pixel are 
% less than 'maxDist' and in the image bounds
xRange = max((pix(1)-(maxDist-1)),1):min((pix(1)+(maxDist-1)),size(Image,1));
yRange = max((pix(2)-(maxDist-1)),1):min((pix(2)+(maxDist-1)),size(Image,2));

% Create a mesgrid to get the pairs of (x,y) of the pixels
[pointsX, pointsY] = meshgrid(xRange, yRange);
pointsX = pointsX(:);
pointsY = pointsY(:);

% Remove 'pix'
pixIndToRemove = (pointsX == pix(1) & pointsY == pix(2));
pointsX(pixIndToRemove) = [];
pointsY(pixIndToRemove) = [];

显示结果：

% Get linear indices of pixels
pointsInd = sub2ind(size(Image), pointsX, pointsY);

% To display the result, create a binary image with all found pixels 
% colored white
bwPoints = zeros(size(Image));
bwPoints(pointsInd) = 1;

% Show points
imshow(bwPoints, 'InitialMagnification', 2000)

% Show pixel grid lines
hold on
[rows, cols] = size(bwPoints);
for row = 0.5 : 1 : (rows + 0.5)
    line([0.5, cols+0.5], [row, row], 'Color', 'r', 'LineWidth', 0.5);
end
for col = 0.5 : 1 : (cols + 0.5)
    line([col, col], [0.5, rows+0.5], 'Color', 'r', 'LineWidth', 0.5);
end

效率并在所有图像像素上循环运行：

选项2 比选项1 快得多。我之所以写第一个选项1 是因为问题中提到了bwdist。通过首先计算像素，而不是将它们移动到每个像素的位置，可以改善在循环中运行选项2 的作用：

% Get the range of x and y indices who's chessboard distance from pixel 
% (0,0) are less than 'maxDist'
xRange = (-(maxDist-1)):(maxDist-1);
yRange = (-(maxDist-1)):(maxDist-1);

% Create a mesgrid to get the pairs of (x,y) of the pixels
[pointsX, pointsY] = meshgrid(xRange, yRange);
pointsX = pointsX(:);
pointsY = pointsY(:);

% Remove pixel (0,0)
pixIndToRemove = (pointsX == 0 & pointsY == 0);
pointsX(pixIndToRemove) = [];
pointsY(pixIndToRemove) = [];

for x=1:size(Image, 1)
    for y=1:size(Image, 2)
        % Get a shifted copy of 'pointsX' and 'pointsY' that is centered
        % around (x, y)
        pointsX1 = pointsX + x;
        pointsY1 = pointsY + y;

        % Remove the the pixels that are out of the image bounds        
        inBounds =...
            pointsX1 >= 1 & pointsX1 <= size(Image, 1) &...
            pointsY1 >= 1 & pointsY1 <= size(Image, 2);

        pointsX1 = pointsX1(inBounds);
        pointsY1 = pointsY1(inBounds);

        % Do stuff with 'pointsX1' and 'pointsY1'
        % ...

    end
end

Answer 2

“目的是访问棋盘距像素 A 小于 D 的像素的位置。对所有像素都执行了...”

由于 D 正在创建正方形选择区域，因此只需使用简单的数学运算即可。

例如：如果 D 为3，则从像素A的[x，y]位置开始...

//# we minus D by 1 since you want less than D (not equal / higher)

Start-X = pixelA.x - (D-1); //from the left
End-X = pixelA.y + (D-1); //to the right

Start-Y = pixelA.y - (D-1); //from the top
End-Y = pixelA.y + (D-1); //to the bottom

这将为您提供一个代表所需选择区域的正方形周长。

请看下面的示例图片：
每个正方形是一个像素。如果“皇冠”图标表示像素 A 和 D 为3（其中您的“小于 D ” 表示 D 的最大长度为2个像素），您能看到上面的伪代码如何应用吗？

图像矩阵中的棋盘距离

2 个答案: