how do I draw discriminatory lines in an image to calculate the area in each of those regions?

Hello, I am exploring matlab and I was wondering how do i draw circular lines that classify particular regions of that image? So that later on I can calculate the area of that region seperatley than the other regions?

I have several images and I am wondering which methods I can use to segment regions within that image from other regions in that same image.

Thank you

[Neo]

Or a method to calculate the area of a region within an image.

Answer 1

[Image Analyst]

See the FAQ:

https://matlab.fandom.com/wiki/FAQ#How_do_I_create_a_circle?

See my Image Processing Tutorial in my File Exchange:

Image Analyst's File Exchange

See attached drawing demos.

[Neo]

@Image Analyst

thanks for this, great tools but if i just draw equally-spaced shapes how can i tell the code that i will use to calculate the area that I only want to calculate the area within a specific shape? I guess the confusion is that I THINK I have to write a code that can identify the partitions in the image and note that each partition HAS ITS OWN area measurement and prevent the code from calculating the area of the whole image together.

[Image Analyst]

Somehow you have to segment the image into two or more regions, like foreground and background, or left third, middle third, right third, or whatever. Once you have those regions identified, you can call regionprops to get attributes for each region. How do you want to segment each region? By hand drawing, or according to some mathematical formula of circles?

[Neo]

regionprops sounds promising. i want to segment the image into 12 equal partitions. (What is inside the partition will differ, that's why.)

i just want to segment them in equal partitions across the image. So I don't mind HOW they're partitioned as long as they are partitioned evenly in the image and also when it is time to calculate the area of each partition the line can act as indicator that the area is only calculated for that partition.

Alt. the line can be hand drawn and as long as there is some indicator that the image is partitioned that's fine too.

[Image Analyst]

If you just want the mean gray level in 12 strips going across the image, try this:

% Demo by Image Analyst
% Find the mean gray level in 12 strips going across the image.
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 = 22;
markerSize = 40;
%--------------------------------------------------------------------------------------------------------
%    READ IN IMAGE
folder = [];
baseFileName = 'cameraman.tif';
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)

rows =    256

columns =    256

numberOfColorChannels =      1

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(1, 3, 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)

rows =    256

columns =    256

numberOfColorChannels =      1

% Maximize window.
g = gcf;
g.WindowState = 'maximized';
drawnow;
%--------------------------------------------------------------------------------------------------------
% Get mean across 12 vertical bands of the image
startingColumns = round(linspace(1, columns+1, 13))

startingColumns = 1×13

1 22 44 65 86 108 129 150 172 193 214 236 257

endingColumns = startingColumns(2:end) - 1

endingColumns = 1×12

21 43 64 85 107 128 149 171 192 213 235 256

for k = 1 : length(endingColumns)
    col1 = startingColumns(k);
    col2 = endingColumns(k);
    % Make a strip
    mask = false(size(grayImage));
    mask(:, col1:col2) = true;
    subplot(1, 3, 2);
    imshow(mask);
    impixelinfo;
    axis('on', 'image');
    caption = sprintf('Mask #%d', k);
    title(caption, 'FontSize', fontSize, 'Interpreter', 'None');
    % Crop the image so we get the image in the mask
    subImage = grayImage(:, col1 : col2);
    subplot(1, 3, 3);
    imshow(subImage);
    impixelinfo;
    axis('on', 'image');
    caption = sprintf('Subimage #%d', k);
    title(caption, 'FontSize', fontSize, 'Interpreter', 'None');
    drawnow;
    % Make measurements of mean graylevel.
    meanGL(k) = mean(grayImage(mask));
end

meanGL

meanGL = 1×12

146.379836309524 124.317294034091 73.9890252976191 60.8411458333333 85.0129616477273 95.8610491071429 126.819754464286 140.714133522727 144.698474702381 147.955357142857 143.329190340909 133.895833333333

[Neo]

I tried this algorithm it looks like it will work so I added in my image (attached) and inserted these lines to turn it to gray scale:

baseFileName = 'test1.jpg';

baseFileName = baseFileName(:, :, 1);

And got this error:

Error using images.internal.imageDisplayValidateParams>validateCData

If input is logical (binary), it must be two-dimensional.

Error in images.internal.imageDisplayValidateParams (line 30)

common_args.CData = validateCData(common_args.CData,image_type);

Error in images.internal.imageDisplayParseInputs (line 79)

common_args = images.internal.imageDisplayValidateParams(common_args);

Error in imshow (line 253)

images.internal.imageDisplayParseInputs({'Parent','Border','Reduce'},preparsed_varargin{:});

Error in untitled2 (line 70)

imshow(mask);

[Image Analyst]

What is this:

baseFileName = 'test1.jpg';
baseFileName = baseFileName(:, :, 1);

??? baseFileName is a string, not a three dimensional array, so you can't get slice #1 from a string.

Anyway, it looks like your image is probably color, not gray scale. If you want to process color images then you need to decide what you want to do to each color channel, and extract them as gray scale images like this:

[r, g, b] = imsplit(rgbImage);

Now r, g, and b are 2-D gray scale images, not color images, and you can use my code.

[Neo]

Error using imsplit

Expected input number 1, I, to be one of these types:

double, single, uint8, uint16, uint32, uint64, int8, int16, int32, int64, logical

Error in imsplit (line 54)

validateattributes(I,{'numeric','logical'},{'3d','nonsparse','real','nonempty'},'imsplit','I',1);

Error in binning (line 15)

[r, g, b] = imsplit(baseFileName);

Should i convert it to something else?

[Image Analyst]

Did I say to use it on the file name string? No. I said to use it on the image variable:

[r, g, b] = imsplit(rgbImage);

Not sure why you changed the variable from a 2-D uint8 matrix to a character array string, and didn't just do

baseFileName = 'test1.jpg';
rgbImage = imread(baseFileName);
[r, g, b] = imsplit(rgbImage);
% Let's assume you want to work on the red channel image, r.  So...
mask = false(size(r));

[Neo]

@Image Analyst

I think that this answer will give the answer to my question!

I ran the algorithm but it just ouputted the gray scale version of my image with the following results:

rows =

1452

columns =

5208

numberOfColorChannels =

3

It is not really gray scale like we expected - it is color

rows =

1452

columns =

5208

numberOfColorChannels =

1

startingColumns =

Columns 1 through 12

1 435 869 1303 1737 2171 2605 3039 3473 3907 4341 4775

Column 13

5209

endingColumns =

434 868 1302 1736 2170 2604 3038 3472 3906 4340 4774 5208

It did not show the mask iteration for 1-12 or the subset image. I am racking my brain to figure out what i did wrong.

[Image Analyst]

Just upload your entire m-file plus your image.

[Neo]

Attached.

The image is successfully converted to grayscale but that's it. Both are attached below. Sorry to bother you with this.

[Neo]

@Image Analyst

typo

*that's it as in it doesn't run the code like when I run the cameraman.tif.

[Image Analyst]

So are you saying you had a typo and now that you corrected it, it works now?

[Neo]

Nope still doesn’t work I ran it again 4 times since yesterday

[Image Analyst]

You really need to think about the code. You're throwing in lines of code randomly and changing from grayscale to RGB. Be consistent. You originally wanted an algorithm that only makes sense for gray scale images, which could either be a gray scale image, or an individual channel of an RGB image. My original code worked with RGB images by converting them to gray scale. They you started doing strange things to it by doing weird things with filenames, etc. Your latest image, test1.jpg, is grayscale so of course it doesn't make sense to try to get the R, G, and B color channels from it like you tried. Plus the last version of code you uploaded has most of the last part of my code missing.

[Neo]

This is the error i get using just your original code:

Error using imsplit

Too many output arguments.

Error in bin (line 16)

[r, g, b] = imsplit(rgbImage);

rgbimage attached

sorry if i changed code i am bilingual sometimes i change to make more sense to me or put in translator to follow directions easily. but i have now input just your original code and still have error

[Image Analyst]

That's not an RGB image so all that stuff about RGB images does not apply. It's just a gray scale image so my original code should work:

% Demo by Image Analyst
% Find the mean gray level in 12 strips going across the image.
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 = 22;
markerSize = 40;
%--------------------------------------------------------------------------------------------------------
%    READ IN IMAGE
folder = [];
baseFileName = 'testrgbimage.tif';
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(1, 3, 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;
%--------------------------------------------------------------------------------------------------------
% Get mean across 12 vertical bands of the image
startingColumns = round(linspace(1, columns+1, 13))
endingColumns = startingColumns(2:end) - 1
for k = 1 : length(endingColumns)
    col1 = startingColumns(k);
    col2 = endingColumns(k);
    % Make a strip
    mask = false(size(grayImage));
    mask(:, col1:col2) = true;
    subplot(1, 3, 2);
    imshow(mask);
    impixelinfo;
    axis('on', 'image');
    caption = sprintf('Mask #%d', k);
    title(caption, 'FontSize', fontSize, 'Interpreter', 'None');
    % Crop the image so we get the image in the mask
    subImage = grayImage(:, col1 : col2);
    subplot(1, 3, 3);
    imshow(subImage);
    impixelinfo;
    axis('on', 'image');
    caption = sprintf('Subimage #%d', k);
    title(caption, 'FontSize', fontSize, 'Interpreter', 'None');
    drawnow;
    % Make measurements of mean graylevel.
    meanGL(k) = mean(grayImage(mask));
end
meanGL

[Neo]

If you just want the mean gray level in 12 strips going across the image:

@Image Analyst

What does it mean to have the measure the gray level? Also how to read the output ?

startingColumns =

1 44 86 129 172 214 257 300 342 385 428 470 513

endingColumns =

43 85 128 171 213 256 299 341 384 427 469 512

[Image Analyst]

The gray level is the brightness of the image pixels.

Those values are the starting and stopping columns of the vertical strip where we are going to get the mean intensity in.

Answer 2

[Matt J]

I was wondering how do i draw circular lines

You can use drawcircle.

More generally, you can segment interactively with the Segmenter App

[Neo]

Thank you. Since I want to draw circular lines (not cirlces) can you suggest a method that could give me identical ciruclar lines to segment the regions in the image?

The segmenter app is a great start but since its option is to hand drawn the lines, the lines won't be identical to each other. I am sorry to bother you about that. Thank you for your suggestion.

Regards

[Matt J]

I want to draw circular lines (not cirlces)

I don't understand your terminology, I'm afraid. How can a line be anything but straight, and how is a "circular line" different from a circle?

[Neo]

Sorry, please see the attached image.

i want to generate these lines unfiormly in the image such that i can calculate the area within each of these regions seperatley from the others.

Sorry for that.

[Neo]

Attached again.

[Neo]

So the method can also include how to calculate the enclosed region area for each curved line.

[Matt J]

The segmenter app is a great start but since its option is to hand drawn the lines, the lines won't be identical to each other.

The lines in your images are curved, but they don't look circular. What is meant to determine their shape if they are not to be hand-drawn? Are they all meant to be shifted copies of one another? If so, you can draw one line with drawfreehand and then copy them.

[Neo]

Will that method also allow me to calculate the area within that region too? Such that each curved line is spaced evenly throughout the image.

I meant curved lines sorry for the confusion.

[Matt J]

They are separate problems. How you draw/specify the boundaries will not have anything to do with how you then calculate areas between them.

[Neo]

Area is the length x width sorry to bother you sir but that means the shape will be connected to the area. I need the curved line of a region in the image to be defined in a way that it’s connected to the calculation of the area within the region. What do you suggest?

[Matt J]

Connected in what way? You still haven't told us what the criteria is for choosing the curved boundaries. Are you saying you want the boundaries placed such that the partitioned image regions all have the same area? But that still isn't specific enough to uniquely define the boundaries. One way to partition them into equal-area regions (among many others) is to just draw equally-spaced vertical lines.

[Neo]

@Matt J

no criteria just want them to be equally partitioned throughout the image such that all the areas are identical (what is inside each area will be different).

if i just draw equally-spaced lines how can i tell the code that i will use to calculate the area that I only want to calculate the area within each line? I guess the confusion is that i THINK i have to write a code that can identify partitions in the image and note that each partition has its own area measurement and prevent the code from calculaating area of the whole image together.

[Matt J]

and note that each partition has its own area measurement

??? but previously you said that all areas are identical: "just want them to be equally partitioned throughout the image such that all the areas are identical"

[Matt J]

In any case, you can get label maps for each of the regions using cumsum, (see below), whereupon you can then use regionprops to analyze a particular region.

BW=zeros(256); BW(:,64:64:end)=1; %vertically partiioned black/white image.

RegionLabels=cumsum(BW,2);

immontage({imdilate(BW,[1,1,1]), RegionLabels},'Bord',15,'Back','w')

[Neo]

and note that each partition has its own area measurement

By this I mean that each partition has its own contents but identical areas.

[Neo]

@Matt J

Thanks for this. I ran the attached and got a small error:

Not enough input arguments.

Error in immontage (line 18)

if iscell(Images)

i did not see regionprops when i ran the code. Sorry to bother you

[Matt J]

I ran the attached and got a small error:

The main code is what I appears in my post, not the attachment. immontage is a utitlity function for image display. You do not have to use immontage if you don't want to. You can use imshow or imagesc or whatever you normally use. It was just convenient for me to present it that way.

i did not see regionprops when i ran the code.

No, i did not demonstrate regionprops, but here is its documentation: regionprops.

[Neo]

@Image Analyst

i am not sure if this should be a seperate question but before I take the mean gray level in the strips, I would like to substract the background, correct for contrast, implement gaussian filtering and thresholding/binarization so i have added this code that i found from other matlab answers:

% Binarize the image.

mask = grayImage > lowThreshold & grayImage < highThreshold;

mask = imfill(mask, 'holes');

% Find out the areas

props = regionprops(mask, 'Area');

allAreas = sort([props.Area], 'descend')

% Take only blobs 20 pixels or larger

mask = bwareaopen(mask, 20);

% Update measurements.

props = regionprops('table', mask, grayImage, 'Area', 'MeanIntensity', 'MaxIntensity', 'MinIntensity');

allAreas = [props.Area]

meanIntensities = [props.MeanIntensity]

maxIntensities = [props.MaxIntensity]

minIntensities = [props.MinIntensity]

% Display the mask.

subplot(2, 2, 3);

imshow(mask, []);

title('Mask', 'FontSize', fontSize, 'Interpreter', 'None');

axis('on', 'image');

maskedImage = grayImage; % Initialize.

maskedImage(~mask) = 0; % Or whatever gray level you want, such as mean(grayImage(~mask));

% Display the image.

subplot(2, 2, 4);

imshow(maskedImage, []);

title('Masked Image', 'FontSize', fontSize, 'Interpreter', 'None');

axis('on', 'image');

what do you think?

[Neo]

@Image Analyst

[Image Analyst]

What? It's doing what you told it to do.

[Neo]

I have asked it in another question but I cant tell that it has been performed background substraction, gaussian and contrast correction before mean grey level analysis. How can you?

[Image Analyst]

I don't see anything being subtracted from the gray scale image in the code above. You need to read in a background image and then subtract it. Decide where you want to discuss this. I suggest your new thread just about background subtraction. (Note spelling of subtraction -- there are not two s's).