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在这两个处理查找二进制图像中对象之间的距离的帖子之后，我怎样才能只输出/计算特定对象与其余对象之间的最短距离（例如，{1->3}、{2->5 }、{3->1}、{4->7)？

脚本：

clc;
clear all;
I = rgb2gray(imread('E:/NCircles.png'));
imshow(I);
BW = imbinarize(I,'adaptive');
BW = imfill(BW, 'holes');
BW = bwlabel(BW); 

s = regionprops(BW,'Area', 'BoundingBox', 'Eccentricity', 'MajorAxisLength', 'MinorAxisLength', 'Orientation', 'Perimeter','Centroid');


imshow(BW)
hold on
for k = 1:numel(s)
    c = s(k).Centroid;
    text(c(1), c(2), sprintf('%d', k), 'HorizontalAlignment', 'center', 'VerticalAlignment', 'middle');
end

boundaries = bwboundaries(BW);
numberOfBoundaries = size(boundaries, 1);
for k = 1 : numberOfBoundaries
    thisBoundary = boundaries{k};
    plot(thisBoundary(:,2), thisBoundary(:,1), 'r', 'LineWidth', 3);
end
hold off;

% Define object boundaries
numberOfBoundaries = size(boundaries, 1)

for b1 = 1 : numberOfBoundaries
    for b2 = 1 : numberOfBoundaries
        if b1 == b2
            % Can't find distance between the region and itself
            continue;
        end
        boundary1 = boundaries{b1};
        boundary2 = boundaries{b2};
        boundary1x = boundary1(:, 2);
        boundary1y = boundary1(:, 1);
        x1=1;
        y1=1;
        x2=1;
        y2=1;
        overallMinDistance = inf; % Initialize.
        % For every point in boundary 2, find the distance to every point in boundary 1.
        for k = 1 : size(boundary2, 1)
            % Pick the next point on boundary 2.
            boundary2x = boundary2(k, 2);
            boundary2y = boundary2(k, 1);
            % For this point, compute distances from it to all points in boundary 1.
            allDistances = sqrt((boundary1x - boundary2x).^2 + (boundary1y - boundary2y).^2);
            % Find closest point, min distance.
            [minDistance(k), indexOfMin] = min(allDistances);
            if minDistance(k) < overallMinDistance
                x1 = boundary1x(indexOfMin);
                y1 = boundary1y(indexOfMin);
                x2 = boundary2x;
                y2 = boundary2y;
                overallMinDistance = minDistance(k);
            end
        end
        % Find the overall min distance
        minDistance = min(minDistance);
        % Report to command window.
        fprintf('The minimum distance from region %d to region %d is %.3f pixels\n', b1, b2, minDistance);

        % Draw a line between point 1 and 2
        line([x1, x2], [y1, y2], 'Color', 'y', 'LineWidth', 3);
    end
end

在此处输入图像描述

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1 回答 1

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给定BW并boundaries如上定义，以及从中计算到所有其他对象的距离的源对象：

source_object = 1;   % label of source object in BW

构造一个距离图像，使得每个像素的值是它与源对象的距离，使用bwdist：

% anonymous function to convert cell array of subsripts
% into cell array of indices
indsfun = @(a) sub2ind(size(BW), a(:,1), a(:,2));

% use function on all of the cell's boundary objects
object_inds = cellfun(indsfun, boundaries, 'UniformOutput', false);


source_image = zeros(size(BW));   % create image containing only source object
source_image(object_inds{source_object}) = 1;

% compute distance from source to all other pixels in image
dist_image = bwdist(source_image, 'euclidean');   % replace with desired metric
imagesc(dist_image);   % not necessary, but gives a cool image

现在，对于原始图像中的每个对象，找到其边界到源对象边界的最小距离：

min_dist = zeros(1,numel(boundaries));   % hold minimum distances

for target_object = 1:numel(boundaries)
   % get the distance values at the indices of the target object
   % and store the minimum.
   min_dist(target_object) = min(dist_image(object_inds{target_object}));
end

最后，min_dist将包含从源对象到所有其他对象的最小（边界）距离。在您的图像上运行的示例给出了以下欧几里德距离：

min_dist =

 Columns 1 through 7:
     0.00000    67.54258    60.00000   207.23416   154.48625   168.79869   319.01410

 Columns 8 through 13:
   236.05296   324.71063   344.05814   367.00000   469.07996   509.00000
于 2019-06-02T17:54:05.997 回答