Extracting edges from a shape
14 ビュー (過去 30 日間)
古いコメントを表示
Hello all.
I have an irregular form shape like in the image(s) below (created using inpolygon).

I only want to extract the outer edges of this shape, so that I can only have an idea of what the shape is (rather than the complete internal points).
I used contours (imcontour) module which highlights the regions I desire to attain (as shown in the image below).

So I only want to attain these yellow regions.
Help will be much appreciated. Thanks in advance :)
1 件のコメント
採用された回答
Image Analyst
2022 年 12 月 20 日
Alright, I created the binary image in Photoshop from your red crosshairs screenshot. It's attached. Then I wrote code for it to get the perimeter image, and also a list of (x,y) coordinates in a round-the-clock sorted fashion using bwboundaries
clc; % Clear the command window.
close all; % Close all figures (except those of imtool.)
clear; % Erase all existing variables. Or clearvars if you want.
workspace; % Make sure the workspace panel is showing.
format long g;
format compact;
fontSize = 20;
%--------------------------------------------------------------------------------------------------------
% READ IN IMAGE
folder = [];
baseFileName = 'binaryImage.png';
fullFileName = fullfile(folder, baseFileName);
% Check if file exists.
if ~exist(fullFileName, 'file')
% The file doesn't exist -- didn't find it there in that folder.
% Check the entire search path (other folders) for the file by stripping off the folder.
fullFileNameOnSearchPath = baseFileName; % No path this time.
if ~exist(fullFileNameOnSearchPath, 'file')
% Still didn't find it. Alert user.
errorMessage = sprintf('Error: %s does not exist in the search path folders.', fullFileName);
uiwait(warndlg(errorMessage));
return;
end
end
grayImage = imread(fullFileName);
% Get the dimensions of the image.
% numberOfColorChannels should be = 1 for a gray scale image, and 3 for an RGB color image.
[rows, columns, numberOfColorChannels] = size(grayImage)
if numberOfColorChannels > 1
% It's not really gray scale like we expected - it's color.
fprintf('It is not really gray scale like we expected - it is color\n');
% Extract the blue channel.
grayImage = rgb2gray(grayImage);
% Update the dimensions of the image.
% numberOfColorChannels should be = 1 for a gray scale image, and 3 for an RGB color image.
[rows, columns, numberOfColorChannels] = size(grayImage)
end
%--------------------------------------------------------------------------------------------------------
% Display the image.
subplot(2, 2, 1);
imshow(grayImage);
impixelinfo;
axis('on', 'image');
title('Original Gray Scale Image', 'FontSize', fontSize, 'Interpreter', 'None');
% Maximize window.
g = gcf;
g.WindowState = 'maximized';
drawnow;
% Binarize
binaryImage = grayImage > 128;
% Call bwperim
perimImage = bwperim(binaryImage);
%--------------------------------------------------------------------------------------------------------
% Display the image.
subplot(2, 2, 2);
imshow(perimImage);
impixelinfo;
axis('on', 'image');
title('Perimeter Image', 'FontSize', fontSize, 'Interpreter', 'None');
% Find xp and yp (unsorted)
[yp, xp] = find(perimImage);
% Find boundaries (which will be sorted)
% Plot the borders of all the blobs in the overlay above the original grayscale image
% using the coordinates returned by bwboundaries().
% bwboundaries() returns a cell array, where each cell contains the row/column coordinates for an object in the image.
subplot(2, 2, 3);
imshow(grayImage); % Optional : show the original image again. Or you can leave the binary image showing if you want.
% Here is where we actually get the boundaries for each blob.
boundaries = bwboundaries(binaryImage);
% boundaries is a cell array - one cell for each blob.
% In each cell is an N-by-2 list of coordinates in a (row, column) format. Note: NOT (x,y).
% Column 1 is rows, or y. Column 2 is columns, or x.
numberOfBoundaries = size(boundaries, 1); % Count the boundaries so we can use it in our for loop
% Here is where we actually plot the boundaries of each blob in the overlay.
hold on; % Don't let boundaries blow away the displayed image.
for k = 1 : numberOfBoundaries
thisBoundary = boundaries{k}; % Get boundary for this specific blob.
x = thisBoundary(:,2); % Column 2 is the columns, which is x.
y = thisBoundary(:,1); % Column 1 is the rows, which is y.
plot(x, y, 'r-', 'LineWidth', 3); % Plot boundary in red.
end
hold off;
caption = sprintf('%d Outlines, from bwboundaries()', numberOfBoundaries);
fontSize = 15;
title(caption, 'FontSize', fontSize);
axis('on', 'image'); % Make sure image is not artificially stretched because of screen's aspect ratio.

6 件のコメント
その他の回答 (1 件)
Image Analyst
2022 年 12 月 19 日
Snippet for you to modify:
% Plot the borders of all the blobs in the overlay above the original grayscale image
% using the coordinates returned by bwboundaries().
% bwboundaries() returns a cell array, where each cell contains the row/column coordinates for an object in the image.
imshow(originalImage); % Optional : show the original image again. Or you can leave the binary image showing if you want.
% Here is where we actually get the boundaries for each blob.
boundaries = bwboundaries(mask);
% boundaries is a cell array - one cell for each blob.
% In each cell is an N-by-2 list of coordinates in a (row, column) format. Note: NOT (x,y).
% Column 1 is rows, or y. Column 2 is columns, or x.
numberOfBoundaries = size(boundaries, 1); % Count the boundaries so we can use it in our for loop
% Here is where we actually plot the boundaries of each blob in the overlay.
hold on; % Don't let boundaries blow away the displayed image.
for k = 1 : numberOfBoundaries
thisBoundary = boundaries{k}; % Get boundary for this specific blob.
x = thisBoundary(:,2); % Column 2 is the columns, which is x.
y = thisBoundary(:,1); % Column 1 is the rows, which is y.
plot(x, y, 'r-', 'LineWidth', 2); % Plot boundary in red.
end
hold off;
caption = sprintf('%d Outlines, from bwboundaries()', numberOfBoundaries);
fontSize = 15;
title(caption, 'FontSize', fontSize);
axis('on', 'image'); % Make sure image is not artificially stretched because of screen's aspect ratio.
6 件のコメント
Image Analyst
2022 年 12 月 20 日
And you forgot to attach that code or image. Alright, I guess I'll have to create an image from the screenshots you supplied and then continue on with a demo after that. I'll need several minutes to do that.
参考
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!

