How to calculate diameter of object

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Ahmed Ismail 2016-8-9

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评论： Ahmed Ismail 2016-8-10

采纳的回答： Image Analyst

在 MATLAB Online 中打开

I have the following code to find the centroids of the particles within this image <- Hyperlink:-

I would like to calculate the diameter of each of the particles so to generate the diameter distribution of the entire image.

X=Image;
template = fspecial('LoG',[51,51],3); % produces a laplacian of Gaussian curve as a template mask use to compare all the intensities in the image
y = conv2(double(X),-template,'same'); % convolves the image with the template to find all the points which correspond to the template mask
FoVY = y(500:700,730:940);
Peaks = imregionalmax(FoVY); % finds all the local maxima
imagesc(Peaks)
LPeaks = bwlabel(Peaks);
R = regionprops(LPeaks);
FoVY = X(500:700,730:940);
imagesc(FoVY);
hold on;
for i = 1:length(R)
    plot(R(i).Centroid(1),R(i).Centroid(2),'r.');
    centroid = cat(1, R.Centroid);
end
no_of_collagen=length(centroid);

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采纳的回答

Image Analyst 2016-8-9

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/298987-how-to-calculate-diameter-of-object#answer_231280

You can call regionprops() and then ask for EquivDiameter. You might get batter, more comprehensive answers if you put as much effort into your question here on the Mathworks MATLAB Central as you did for some other forum.

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Image Analyst 2016-8-10

在 MATLAB Online 中打开

I tried your code and got it working. All your areas are 1 pixel - not surprising.

clc;    % Clear the command window.
close all;  % Close all figures (except those of imtool.)
clear;  % Erase all existing variables.
workspace;  % Make sure the workspace panel is showing.
format long g;
format compact;
grayImage = imread('segmented_collagen.png');
[rows, columns, numberOfColorChannels] = size(grayImage);
if numberOfColorChannels > 1
  grayImage = double(rgb2gray(grayImage));
else
  grayImage = double(grayImage);
end
subplot(2,3,1);
imshow(grayImage, []);
title('X', 'FontSize', 20);
% Get rid of white surround
grayImage = imclearborder(grayImage);
template = fspecial('LoG',[51,51],3); % produces a laplacian of Gaussian curve as a template mask use to compare all the intensities in the image
y = conv2(double(grayImage),-template,'same'); % convolves the image with the template to find all the points which correspond to the template mask
% FoVY = y(500:700,730:940);
FoVY = y;
subplot(2,3,2);
imshow(FoVY, []);
title('FoVY', 'FontSize', 20);
Peaks = imregionalmax(FoVY); % finds all the local maxima
% Get rid of white surround
Peaks = imclearborder(Peaks);
subplot(2,3,3);
imagesc(Peaks);
title('Peaks', 'FontSize', 20);
[LPeaks, numberOfPeaks] = bwlabel(Peaks);
R = regionprops(LPeaks, 'Centroid', 'EquivDiameter', 'Area');
allDiameters = sort([R.EquivDiameter], 'descend')
allAreas = sort([R.Area], 'descend')
% Show a histogram of the diameters
subplot(2,3,4);
histogram(allDiameters);
grid on;
title('Histogram of Diameters', 'FontSize', 20);
% FoVY = X(500:700,730:940);
FoVY = grayImage;
subplot(2,3,5);
imagesc(FoVY);
hold on;
for i = 1:length(R)
    plot(R(i).Centroid(1),R(i).Centroid(2),'r.');
    centroid = cat(1, R.Centroid);
end
title('FoVY', 'FontSize', 20);
no_of_collagen=length(centroid);
% Set up figure properties:
% Enlarge figure to full screen.
set(gcf, 'Units', 'Normalized', 'OuterPosition', [0 0 1 1]);
% Get rid of tool bar and pulldown menus that are along top of figure.
set(gcf, 'Toolbar', 'none', 'Menu', 'none');
% Give a name to the title bar.
set(gcf, 'Name', 'Demo by ImageAnalyst', 'NumberTitle', 'Off') 
message = sprintf('Number of peaks = %d\nCheck command window for diameters.', numberOfPeaks);
uiwait(helpdlg(message));

I had trouble downloading your original image from imgur. You should have inserted it here where it would have been easier for us.

Image Analyst 2016-8-10

It would be difficult, even manually with imfreehand. I can't really tell where one starts and the other stops. Why isn't the count enough? If you know the total area, and the count, you can get the average area of the collagen.

Ahmed Ismail 2016-8-10

That is true, I guess with some images you have to be realistic with what you can extract. Thanks for the help.

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