How do I determine how do these black dots vary by location in an equal spaced magnetic field.

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Chanille 2023-2-15

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评论： Neo 2023-2-22

I circled some of the black dots I want to count in the red circle. I want to segment each area equally across the image in the same crescent shape as I have drawn but equally spaced apart. But I only want to segment the areas that are filled with black dots and not segment the areas with the magnetic field (where the magnet is). This is because since the magnet expels all of the black dots away to other regions if I segment and count the space where the magnet is it wouldn't give an accurate account of the number of cells when I compare it to other cells where the magnet is a different location.

How can I detect where to start segmenting such that I only segment the areas with the black dots and around the areas without them?

I attempted to make a for loop code that starts segmenting by columns and substract by 1 before it starts counting but sometimes the magnetic field varies in location so I can't always do that, so is there a way to detect that this area shouldnt counted and count around this area then uniformly segment the remainder of the image like in the crescent shape i've drawn? I attached a picture of a pond that gives an idea of what I am attempting to try. the rock is the area I don't want to count, the cirlce around it would be the starting counting and the subsequent waves produced would be the other counts until the last column in the image. Please advise for an old R user who now loves MATLAB. Thanks.

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Answer 1

Image Analyst 2023-2-15

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移动：Image Analyst 2023-2-16

Do you have some equation for where the lines are? Like ar they perfect circles centered at the magnet or have some other formula? If you do then you can make a mask for each zone and compute the mean gray level in that zone. Can you attach the original, un-annotated image?

In the meantime, I'm attaching a few demos.

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Image Analyst 2023-2-16

在 MATLAB Online 中打开

@Neo sorry that's not working, but thanks for trying. This is what I was thinking of.

% Demo by Image Analyst
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 = 18;
markerSize = 40;
%--------------------------------------------------------------------------------------------------------
%    READ IN TEST IMAGE
folder = [];
baseFileName = 'Screenshot 2023-02-15 at 6.23.03 PM.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);
end
%--------------------------------------------------------------------------------------------------------
% Display the image.
subplot(2, 2, 1);
imshow(grayImage, []);
impixelinfo;
axis('on', 'image');
title('Original Gray Scale Image', 'FontSize', fontSize, 'Interpreter', 'None');
% 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)
% Maximize window.
g = gcf;
g.WindowState = 'maximized';
drawnow;
%--------------------------------------------------------------------------------------------------------
% Display histogram
subplot(2, 2, 2);
histogram(grayImage);
grid on;
title('Histogram of Image', 'FontSize', fontSize, 'Interpreter', 'None');
%--------------------------------------------------------------------------------------------------------
% Threshold to create mask
lowThreshold = 0;
highThreshold = 40;
% Interactively and visually set a threshold on a gray scale image.
% https://www.mathworks.com/matlabcentral/fileexchange/29372-thresholding-an-image?s_tid=srchtitle
% [lowThreshold, highThreshold] = threshold(lowThreshold, highThreshold, grayImage)
mask = grayImage >= lowThreshold & grayImage <= highThreshold;
xline(highThreshold, 'Color', 'r', 'LineWidth', 2)
subplot(2, 2, 3);
% Take only the largest blob.
mask = bwareafilt(mask, 1);
imshow(mask);
impixelinfo;
axis('on', 'image');
title('Mask Image with Centroid Marked', 'FontSize', fontSize, 'Interpreter', 'None');
%--------------------------------------------------------------------------------------------------------
% Find the centroid of the blob.
props = regionprops(mask, 'Centroid');
xCenter = props.Centroid(1);
yCenter = props.Centroid(2);
hold on;
% Plot the centroid in red for this method.
plot(xCenter, yCenter, 'r+', 'LineWidth', 2, 'MarkerSize', 50);
% An alternate way if the blob is partially off the edge of the image is to
% Get the coordinates with bwboundaries and then exclude edge pixels then to 
% Fit the coordinates to a circle.
b = bwboundaries(mask);
b = b{1};
x = b(:, 2);
y = b(:, 1);
% Exclude edge pixels
edgeIndexes = (x == 1) | (x == columns) | (y == 1) | (y == rows);
x(edgeIndexes) = [];
y(edgeIndexes) = [];
plot(x, y, 'r-', 'LineWidth', 2);
% Fit those coordinates to a circle
[xCenter, yCenter, radius, a] = circlefit(x, y);
% Plot the centroid in Cyan for this method.
plot(xCenter, yCenter, 'c+', 'LineWidth', 2, 'MarkerSize', 50);
% Allocate an array for the profile
maxDistance = ceil(sqrt(columns ^ 2 + rows ^ 2))
profileSums = zeros(1, maxDistance);
profileCounts = zeros(1, maxDistance);
% Scan the original image getting distance from centroid.
% Then add those values to the profile.
for column = 1 : columns
    for row = 1 : rows
        thisDistance = round(sqrt((column - xCenter) ^2 + (row - yCenter) ^ 2));
        if thisDistance <= 0
            continue;
        end
        if thisDistance >= 838
            continue;
        end
        profileSums(thisDistance) = profileSums(thisDistance) + double(grayImage(row, column));
        profileCounts(thisDistance) = profileCounts(thisDistance) + 1;
    end
end
% Find last index so we don't consider unused ones.
lastIndex = find(profileCounts > 0, 1, 'last')
profileCounts = profileCounts(1 : lastIndex);
profileSums = profileSums(1 : lastIndex);
% Divide the sums by the counts at each distance to get the average profile
averageRadialProfile = profileSums ./ profileCounts;
subplot(2, 2, 4); 
plot(1:length(averageRadialProfile), averageRadialProfile, 'b-', 'LineWidth', 3);
grid on;
title('Average Radial Profile', 'FontSize', fontSize);
xlabel('Distance from nearest white region', 'FontSize', fontSize);
ylabel('Gray Level', 'FontSize', fontSize);
%===========================================================================================================================================
function [xCenter, yCenter, radius, a] = circlefit(x, y)
% circlefit(): Fits a circle through a set of points in the x - y plane.
% USAGE :
% [xCenter, yCenter, radius, a] = circlefit(X, Y)
% The output is the center point (xCenter, yCenter) and the radius of the fitted circle.
% "a" is an optional output vector describing the coefficients in the circle's equation:
%     x ^ 2 + y ^ 2 + a(1) * x + a(2) * y + a(3) = 0
% by Bucher Izhak 25 - Oct - 1991
numPoints = numel(x);
xx = x .* x;
yy = y .* y;
xy = x .* y;
A = [sum(x),  sum(y),  numPoints;
    sum(xy), sum(yy), sum(y);
    sum(xx), sum(xy), sum(x)];
B = [-sum(xx + yy) ;
    -sum(xx .* y + yy .* y);
    -sum(xx .* x + xy .* y)];
a = A \ B;
xCenter = -.5 * a(1);
yCenter = -.5 * a(2);
radius  =  sqrt((a(1) ^ 2 + a(2) ^ 2) / 4 - a(3));
end

Chanille 2023-2-22

@Image Analyst

I understand. Thanks for the clarification.

Sincerely,

C.

Neo 2023-2-22

donutpractic2point0.m

@Image Analyst Check this out. But I still get an error and I have done everything suggested (or maybe I need to restart my computer not sure at this point). What do you think?

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Answer 2

Neo 2023-2-15

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编辑：Neo 2023-2-15

I think I understand your question but just in case check this:

https://www.mathworks.com/matlabcentral/answers/6200-tutorial-how-to-ask-a-question-on-answers-and-get-a-fast-answer

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Answer 3

Neo 2023-2-15

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编辑：Neo 2023-2-15

This looks like something that I recently learned to do in Matlab with one of @Image Analyst analyst demo. Look at my question with his response (can’t remember which but it was recent) and you’ll see you could use that code and change the shape of the segments to get what you want.

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Neo 2023-2-15

编辑：Neo 2023-2-15

Try this:

startingColumns = round(linspace(1, columns+1, 10))

endingColumns = startingColumns(2:end) - 1

in your for loop. This would segment 9 columns equally across your image BUT IN STRIPS.

Something like this could give you the semi circle (half moon shape segmentation) that you want:

theta = linspace( pi/2, -pi/2, 100); So maybe you could add that into the starting columns. I am not sure how to detect the non black dot area though. I will have to think about that.

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