compare pixel values to each other? Is ther afunction in Matlab that can analyze the composition of an image as a function of distance?

Question

Neo 2022-12-15

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评论： Image Analyst 2023-1-6

采纳的回答： Image Analyst

I am interested in looking at how the image changes over a distance within the image, any recommendations?

Thanks!

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Neo 2022-12-21

移动：Rena Berman 2022-12-22

It’s not a cell it’s just an example because I’m still trying to identify the image but yes if it’s a cell I want to know the size area and density of the cells in one region and how they change over the length of the image.

Image Analyst 2022-12-22

If you have any more questions, then attach your image and code to read it in with the paperclip icon after you read this:

TUTORIAL: How to ask a question (on Answers) and get a fast answer

And somehow indicate, in a separate image, the things in the image you'd like to use in computations.

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Answer 1

Image Analyst 2022-12-22

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https://ww2.mathworks.cn/matlabcentral/answers/1879232-compare-pixel-values-to-each-other-is-ther-afunction-in-matlab-that-can-analyze-the-composition-of#answer_1133937

See my Image Processing Tutorial in my File Exchange:

Image Analyst's File Exchange

That pretty much explains the basics. You can probably adapt it to your particular image.

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Neo 2022-12-28

Thank you @Image Analyst. I used the first polygon code attachment. (draw multiple..) and I tried to get the density, area and size of the region of interest.

i have googled this extensively and I am confident that my question is posed correctly. My question stated boils down to: How do I properly call regionprops to get the area, density and size of a ROI? Here's my code:

% Demo to draw multiple polygons with roipoly() and make a single binary image from them all.

%----- Initializing steps -----

% Clean up

clc;

clear vars;

close all;

workspace; % Display the workspace panel.

fontSize = 20;

hasIPT = license('test', 'image_toolbox');

if ~hasIPT

% User does not have the toolbox installed.

message = sprintf('Sorry, but you do not seem to have the Image Processing Toolbox.\nDo you want to try to continue anyway?');

reply = questdlg(message, 'Toolbox missing', 'Yes', 'No', 'Yes');

if strcmpi(reply, 'No')

% User said No, so exit.

return;

end

% Display images to prepare for the demo.

originalImage = imread('coins.png');

[rows, columns, numberOfColorChannels] = size(originalImage);

subplot(2, 2, 1);

imshow(originalImage);

title('Original Image. DRAW POLYGON HERE!!!', 'FontSize', fontSize);

subplot(2, 2, 2);

imshow(originalImage);

subplot(2, 2, 4);

imshow(originalImage);

title('Original Image with regions burned into image', 'FontSize', fontSize);

set(gcf, 'units','normalized','outerposition',[0 0 1 1]); % Maximize figure.

set(gcf,'name','Image Analysis Demo','numbertitle','off')

%----- Ask user to draw polygons ---------------------------------------------------------------------

% Create a binary image for all the regions we will draw.

cumulativeBinaryImage = false(rows, columns);

subplot(2, 2, 4);

imshow(cumulativeBinaryImage);

title('Cumulative Binary Image', 'FontSize', fontSize);

% Create region mask, h, as an ROI object over the second image in the bottom row.

axis on;

again = true;

regionCount = 0;

while again && regionCount < 20

promptMessage = sprintf('Draw region #%d in the upper left image,\nor Quit?', regionCount + 1);

titleBarCaption = 'Continue?';

button = questdlg(promptMessage, titleBarCaption, 'Draw', 'Quit', 'Draw');

if strcmpi(button, 'Quit')

break;

end

regionCount = regionCount + 1;

% Ask user to draw freehand mask.

message = sprintf('Left click vertices in the upper left image.\nRight click the last vertex to finish.\nThen double click in the middle to accept it.');

uiwait(msgbox(message));

subplot(2, 2, 1); % Switch to image axes.

% Ask user to draw a polygon.

% Use roipolyold() if you want to finish at the right click

% and not have the user need to double-click inside the shape to confirm it.

% [thisSinglePolygonImage, xi, yi] = roipolyold();

% Use roipoly() if you want to close the polygon at the right click

% but give the user the opportunity to adjust the positions of the vertices, and then

% the user needs to double-click inside the shape to confirm/accpet it.

[thisSinglePolygonImage, xi, yi] = roipoly();

% Display the last drawn polygon.

subplot(2, 2, 3);

imshow(thisSinglePolygonImage);

caption = sprintf('Binary mask you just drew');

title(caption, 'FontSize', fontSize);

% Draw the polygon over the image in the upper right.

subplot(2, 2, 2); % Switch to upper right image axes.

hold on;

plot(xi, yi, 'r-', 'LineWidth', 2);

caption = sprintf('Original Image with %d regions in overlay.', regionCount);

title(caption, 'FontSize', fontSize);

% OR it in to the "all regions" binary image mask we're building up.

cumulativeBinaryImage = cumulativeBinaryImage | thisSinglePolygonImage;

% Display the regions mask.

subplot(2, 2, 4);

imshow(cumulativeBinaryImage);

caption = sprintf('Binary mask of the %d regions', regionCount);

title(caption, 'FontSize', fontSize);

end

% cumulativeBinaryImage is your final binary image mask that contains all the individual masks you drew.

%WHERE MY CODE ATTEMPT BEGINS

BW = a < 100;

info = regionprops('table',BW, 'density', 'area', 'size')

Image Analyst 2022-12-28

在 MATLAB Online 中打开

Not sure how you're defining "density" but maybe it's MeanIntensity. So you'd change the last 3 lines of the program to look like:

% WHERE MY CODE ATTEMPT BEGINS
info = regionprops('table', cumulativeBinaryImage, originalImage, 'MeanIntensity', 'Area', 'EquivDiameter')
allAreas = info.Area;
allDiameters = info.EquivDiameter;
allIntensities = info.MeanIntensity;

I drew 2 regions and got this:

info =

2×3 table

Area EquivDiameter MeanIntensity

____ ________________ ________________

994 35.5752738213882 80.1750503018109

4549 76.1049715130375 115.819960430864

Full demo:

% Demo to draw multiple polygons with roipoly() and make a single binary image from them all.
%----- Initializing steps -----
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;
hasIPT = license('test', 'image_toolbox');
if ~hasIPT
    % User does not have the toolbox installed.
    message = sprintf('Sorry, but you do not seem to have the Image Processing Toolbox.\nDo you want to try to continue anyway?');
    reply = questdlg(message, 'Toolbox missing', 'Yes', 'No', 'Yes');
    if strcmpi(reply, 'No')
        % User said No, so exit.
        return;
    end
end
% Display images to prepare for the demo.
originalImage = imread('coins.png');
[rows, columns, numberOfColorChannels] = size(originalImage);
subplot(2, 2, 1);
imshow(originalImage);
title('Original Image.  DRAW POLYGON HERE!!!', 'FontSize', fontSize);
subplot(2, 2, 2);
imshow(originalImage);
subplot(2, 2, 4);
imshow(originalImage);
title('Original Image with regions burned into image', 'FontSize', fontSize);
set(gcf, 'units','normalized','outerposition',[0 0 1 1]); % Maximize figure.
set(gcf,'name','Image Analysis Demo','numbertitle','off')
%----- Ask user to draw polygons ---------------------------------------------------------------------
% Create a binary image for all the regions we will draw.
cumulativeBinaryImage = false(rows, columns);
subplot(2, 2, 4);
imshow(cumulativeBinaryImage);
title('Cumulative Binary Image', 'FontSize', fontSize);
% Create region mask, h, as an ROI object over the second image in the bottom row.
axis on;
again = true;
regionCount = 0;
while again && regionCount < 20
    promptMessage = sprintf('Draw region #%d in the upper left image,\nor Quit?', regionCount + 1);
    titleBarCaption = 'Continue?';
    button = questdlg(promptMessage, titleBarCaption, 'Draw', 'Quit', 'Draw');
    if strcmpi(button, 'Quit')
        break;
    end
    regionCount = regionCount + 1;
    % Ask user to draw freehand mask.
    message = sprintf('Left click vertices in the upper left image.\nRight click the last vertex to finish.\nThen double click in the middle to accept it.');
    uiwait(msgbox(message));
    subplot(2, 2, 1); % Switch to image axes.
    
    % Ask user to draw a polygon.
    % Use roipolyold() if you want to finish at the right click
    % and not have the user need to double-click inside the shape to confirm it.
    % 	[thisSinglePolygonImage, xi, yi] = roipolyold();
    % Use roipoly() if you want to close the polygon at the right click
    % but give the user the opportunity to adjust the positions of the vertices, and then
    % the user needs to double-click inside the shape to confirm/accpet it.
    [thisSinglePolygonImage, xi, yi] = roipoly();
    
    % Display the last drawn polygon.
    subplot(2, 2, 3);
    imshow(thisSinglePolygonImage);
    caption = sprintf('Binary mask you just drew');
    title(caption, 'FontSize', fontSize);
    % Draw the polygon over the image in the upper right.
    subplot(2, 2, 2); % Switch to upper right image axes.
    hold on;
    plot(xi, yi, 'r-', 'LineWidth', 2);
    caption = sprintf('Original Image with %d regions in overlay.', regionCount);
    title(caption, 'FontSize', fontSize);
    % OR it in to the "all regions" binary image mask we're building up.
    cumulativeBinaryImage = cumulativeBinaryImage | thisSinglePolygonImage;
    % Display the regions mask.
    subplot(2, 2, 4);
    imshow(cumulativeBinaryImage);
    caption = sprintf('Binary mask of the %d regions', regionCount);
    title(caption, 'FontSize', fontSize);
end
% cumulativeBinaryImage is your final binary image mask that contains all the individual masks you drew.
% WHERE MY CODE ATTEMPT BEGINS
info = regionprops('table', cumulativeBinaryImage, originalImage, 'MeanIntensity', 'Area', 'EquivDiameter')
allAreas = info.Area;
allDiameters = info.EquivDiameter;
allIntensities = info.MeanIntensity;

Neo 2023-1-6

@Image Analyst is the drawpoint program made to address very small and hard to see regions of interests?

Image Analyst 2023-1-6

No, drawpoints is for situations where you need to indicate single points/pixels on an image. drawrectangle and drawpolygon are for when you need to draw 2-D regions.

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Answer 2

Walter Roberson 2022-12-21

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The answer to your question is NO, there is no MATLAB function to analyze the composition of an image as a function of distance.

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Walter Roberson 2022-12-21

First you need to figure out where the cell boundaries are. Then you need to "label" the image, producing an array the same size in which each location gives an identifier about which of the cells the pixel belongs to (0 if not part of any cell.) Then you process the grayscale image, converting the intensity into a pixel weight.Then you call regionprops passing in the label image and the weight image, using the syntax that regionprops summarizes as

stats = regionprops(___,I,properties)

Ask for the properties Area, Centroid, MeanIntensity

The MeanIntensity will be the (average) density, provided that you supplied a valid weight image.

The conversion from intensity to pixel weight might take a bit of work. Different colors might imply different material compositions. Darker portions in the centre of a cell might relate to the cell nucleus, which would not just be thicker but denser than some of the other parts of the cell. Perhaps you are using a polarized light source, so you might be looking at texture information to get clues about the local density. Perhaps the shape of the nucleus might give information about the kind of cell it is and so influence the interpretation of the intensity conversion to local density.

Walter Roberson 2022-12-29

在 MATLAB Online 中打开

Suppose you have a pixel that is (say) [240 27 93] and you have another pixel that is [200 23 78]

plot(1:10, .1*ones(1,10), 'color', [240 27 93]/255);

hold on

plot(1:10, .11*ones(1,10), 'color', [200 23 78]/255);

hold off

ylim([.05 .115])

Are they the same material or different material? If you look at them carefully, you can see that they do not look exactly the same, so you might need to assign them different materials. Rubber versus plastic for example. Different parts of a cell structure maybe.

But are they really different materials? Are are they the same material with slightly different lighting conditions? Is one of them in a little more shadow than the other? Is one of them the same material than the other, but the system is backlit and a part is slightly thicker and so cuts off more backlight? Or has the experiment been set up carefully so that the lighting can be treated as being uniform parallel rays over the entire field so that any variation in color is not due to differences in shadow or thickness ?

If you cannot reasonably model as uniform parallel rays, then you need to be considering doing correction for illumination.

The illiumination correction that @Image Analyst tends to do is to take

im2uint8(double(img) ./ double(max(img, 3)))

which divides each location by the maximum component at that location. This treats each location as if it is potentially uniform material but subject to different lighting levels -- as opposed to a model where a pixel that is (say) 3/4 of each component of a different pixel, is different because it is inherently a different material. If you need to differentiate pixels in part by brightness instead of just by hue, then illumination correction becomes more difficult.

Image Analyst 2022-12-29

编辑：Image Analyst 2022-12-29

在 MATLAB Online 中打开

originalImage must be a gray scale image. Yours is not. You can do

originalImage = originalImage(:, :, 2); % Take green channel

It looks like you have a fluorescence image so you would not do a background division like usual. For radiography (x-rays) and fluorescence you do background subtraction. For most other situations you do background division. You'd do a background subtraction. But your background looks pretty black and uniform so that is probably not even necessary.

Just a slight correction to what @Walter Roberson said about how I do background correction. I don't divide by the max of any color channel. I divide by a smoothed version of each color channel individually. This is because the lens shading and transverse chromatic aberration of the lens might be wavelength dependent. In other words, it might taper off on the sides of the image more for one color channel than another so I need to correct for each color channel independently. This could be a whole hour long discusson but I won't go into that here. I'll just attach the image of an actual white background image taken from one of our light booths (a small JPG version -- less than 5 MB so I can upload it) and script to take vertical and horizontal profiles so you can see how the shading varies by color channel.

Note the profiles are still very noisy. This is most likely not due to illumination non-uniformities because light wouldn't vary that much on a small scale. The non-uniformities are most likely due to variations in the white substrate itself, and slightly due to optical noise remaining in the image. The overall differences between the different color channels is most likely due to chromatic aberration of the lens. The lens shading is different for optical wavelengths in the red, green, and blue regions of the spectrum. This will affect both the overall intensity level as well as the amount of shading (darkening) across the image.

The spectrum of the white substrate is almost perfectly flat - I checked with a spectrophotometer. The light is "white" light LED so it's possible there could be a difference in the overall levels of the red, green, and blue light due to the particular way the spectrum of the light interacted with the spectral responsivities of the imaging sensor. For example the light is a "warm white" so there would be more red light being emitted. Note the top and right side of the image is more neutral than the bottom or left edge of the image.

If you want, I can upload my background correction, color correction, and color calibration seminar.

Neo 2022-12-29

Yes I think that would be helpful. Just to be safe how would I perform background substraction and also masking to get rid of other specs in the image that I might not be interested in for the future.

Image Analyst 2022-12-29

Background correction, color correction, and LAB calibration seminar (Generic).zip

See attached seminar on background correction, color correction, and color calibration. I've just updated some slides from what I've posted in the past years..

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