提取内容并计算图像内的区域

Question

目标：

我需要提取并突出显示给定图像中的特定颜色[Say X]，最后我需要计算X的面积。

我需要帮助：

以圆圈的形式检测图像[下方]中的各个标签。

我做了什么：

下面是我的代码，我附上了RGB变量的图像[下方]。我已经提取了＆＃34;米色＆＃34;从给定的图像（即X）并标记它。

为了检测单个组件，我尝试使用经典的＆＃34;硬币示例[Identifying different coin values from an image using MATLAB＆＃34;来首先检测元素。但在这里，由于地区不统一，所以，我有点困惑。

代码：

clear all; clc;
close all;
I = imread('19.jpg');
%figure;imshow(I);

B = White_Beige(I); // The Function is used to threshold the RGB image 
                       to LAB colorspace. Here, we extract the Beige 
                       color present in the image and provide the final 
                       Black and White image with the presence of Beige 
                       color as Black and rest as white. 

%figure; imshow(B);
BW2 = imfill(~B,'holes');
figure; imshow(BW2);

Border = imclearborder(BW2);
[OBL,nbOBObjets] = bwlabel(Border,8);
RGB = label2rgb(OBL,'jet',[0 0 0]); 
figure, imshow(RGB);

非常感谢任何帮助

Answer 1

也许这会有所帮助：

请注意，在此代码中，＆＃34;区域＆＃34;由颜色占据在间隔（0,1）上表示为实数，其中0表示图像中没有像素是该颜色，1表示图像中的所有像素都是该颜色。

[A, map] = imread('L8IZU_2.png');

image(A); % display the image

if ~isempty(map)
    A = ind2rgb(X, map); % convert image to RGB
end

color_counts = zeros(256, 256, 256);
% color_counts(x, y, z) records the number of pixels in the image
% which have RGB color [x-1, y-1, z-1];

% For example color_counts(1, 1, 1) records the number of pixels in
% the image which have color [0, 0, 0];

% As another example, color_counts(65, 43, 11) records the number of
% pixels in the image which are of color [64, 42, 10];

time_of_last_progress_report = clock;

% traverse the image pixel by pixel
for m = 1:size(A, 1)
    for n = 1:size(A, 2)

        current_time = clock;
        if etime(current_time, time_of_last_progress_report) > 2
            percent_progress = 100*(((m - 1)*size(A, 2) + n)/(size(A, 1)*size(A, 2)));
            disp(strcat(num2str(percent_progress),'% of the image has been scanned'));
            time_of_last_progress_report = clock;
        end
        current_color = A(m, n, :);
        current_color = squeeze(current_color);
        % squeeze removes singleton dimensions
        % in other words, squeeze converts current_color from a 1x1x3 array into
        % a 3x1 array

        index = current_color + uint8(ones(size(current_color)));

        % Let us say that index = [1, 3, 4];
        % We want to increment color_counts(1, 3, 4);
        % However, color_counts([1, 3, 4]), or color_counts(index),
        % does not return a reference to one single element at that location
        % Instead, color_counts([1, 3, 4]) returns a 1x3 vector containing
        % the 1st, 3rd, and 4th elements of the color_counts array

        % we convert index into a cell array and then convert the cell
        % array into a comma separated list.
        index2 = num2cell(index);
        old_count = color_counts(index2{:});
        color_counts(index2{:}) = 1 + old_count;
    end
end

disp(' ');% line break
disp('Now sorting colors by how often they apeared....');

color_counts_linear = reshape(color_counts, 1, numel(color_counts));
[color_counts_sorted, indicies] = sort(color_counts_linear, 'descend');
[R,G,B] = ind2sub(size(color_counts),indicies);

R = R - ones(size(R));
G = G - ones(size(G));
B = B - ones(size(B));
areas = (1/(size(A, 1)*size(A, 2))) * color_counts_sorted;

% display the top 5 most common colors
disp(' '); % line break
disp('Top 5 most common colors:');
for k = 1:min(5, numel(areas))
    disp(strcat('[', num2str(R(k)), ',', num2str(B(k)), ',', num2str(G(k)),'] : ', num2str(100*areas(k)), '%'));
end