How to calculate area of a segment in gray scale image in pixels

Hi, I have a gray scale image .I want to select a segment/area of it using mouse cliks and want to know size of this sected area in pixels.Pls. help Richa

Answer 1

[Image Analyst]

See the demo below. You could also use roipolyold() instead of imfreehand() if you'd rather draw a polygon instead of an irregular shape.

% Demo to have the user freehand draw an irregular shape over
% a gray scale image, have it extract only that part to a new image,
% and to calculate the mean intensity value of the image within that shape.
% By ImageAnalyst
%
% IMPORTANT: The newsreader may break long lines into multiple lines.
% Be sure to join any long lines that got split into multiple single lines.
% These can be found by the red lines on the left side of your
% text editor, which indicate syntax errors, or else just run the 
% code and it will stop at the split lines with an error.

% Change the current folder to the folder of this m-file.
if(~isdeployed)
  cd(fileparts(which(mfilename)));
end
clc;  % Clear command window.
clear;  % Delete all variables.
close all;  % Close all figure windows except those created by imtool.
imtool close all;  % Close all figure windows created by imtool.
workspace;  % Make sure the workspace panel is showing.
fontSize = 16;

% Read in standard MATLAB gray scale demo image.
grayImage = imread('cameraman.tif');
subplot(2, 3, 1);
imshow(grayImage, []);
title('Original Grayscale Image', 'FontSize', fontSize);
set(gcf, 'Position', get(0,'Screensize')); % Maximize figure.
message = sprintf('Left click and hold to begin drawing.\nSimply lift the mouse button to finish');
uiwait(msgbox(message));
hFH = imfreehand();

% Create a binary image ("mask") from the ROI object.
binaryImage = hFH.createMask();
% Display the freehand mask.
subplot(2, 3, 2);
imshow(binaryImage);
title('Binary mask of the region', 'FontSize', fontSize);

% Calculate the area, in pixels, that they drew.
numberOfPixels1 = sum(binaryImage(:))
% Another way to calculate it that takes fractional pixels into account.
numberOfPixels2 = bwarea(binaryImage)

% Get coordinates of the boundary of the freehand drawn region.
structBoundaries = bwboundaries(binaryImage);
xy=structBoundaries{1}; % Get n by 2 array of x,y coordinates.
x = xy(:, 2); % Columns.
y = xy(:, 1); % Rows.
subplot(2, 3, 1); % Plot over original image.
hold on; % Don't blow away the image.
plot(x, y, 'LineWidth', 2);

% Burn line into image by setting it to 255 wherever the mask is true.
burnedImage = grayImage;
burnedImage(binaryImage) = 255;
% Display the image with the mask "burned in."
subplot(2, 3, 3);
imshow(burnedImage);
caption = sprintf('New image with\nmask burned into image');
title(caption, 'FontSize', fontSize);
% Mask the image and display it.
% Will keep only the part of the image that's inside the mask, zero outside mask.
blackMaskedImage = grayImage;
blackMaskedImage(~binaryImage) = 0;
subplot(2, 3, 4);
imshow(blackMaskedImage);
title('Masked Outside Region', 'FontSize', fontSize);
% Calculate the mean
meanGL = mean(blackMaskedImage(binaryImage));
% Report results.
message = sprintf('Mean value within drawn area = %.3f\nNumber of pixels = %d\nArea in pixels = %.2f', ...
meanGL, numberOfPixels1, numberOfPixels2);
msgbox(message);

% Now do the same but blacken inside the region.
insideMasked = grayImage;
insideMasked(binaryImage) = 0;
subplot(2, 3, 5);
imshow(insideMasked);
title('Masked Inside Region', 'FontSize', fontSize);

% Now crop the image.
topLine = min(x);
bottomLine = max(x);
leftColumn = min(y);
rightColumn = max(y);
width = bottomLine - topLine + 1;
height = rightColumn - leftColumn + 1;
croppedImage = imcrop(blackMaskedImage, [topLine, leftColumn, width, height]);
% Display cropped image.
subplot(2, 3, 6);
imshow(croppedImage);
title('Cropped Image', 'FontSize', fontSize);

[Richa Sharma]

Thanks.

I tried but getting error undefined function at hFH = imfreehand();

I am using MATLAB version 7

[Image Analyst]

7 point what? I don't know when it was introduced. I have R2011a (version 7.12). Maybe you should upgrade to the newest version or else use roipolyold instead.

[Richa Sharma]

I got it on MATAB R2009a.Thank You.

[mohd akmal masud]

I have try it, but what is the unit of area? is it in cm^2 or mm^2?

[Image Analyst]

It's in pixels. To get a real world dimension you'd have to know how many of those dimensions (mm or cm) there are in a pixel, or how many pixels are in a real world dimension. See attached demo to calibrate your image spatially.

[mohd akmal masud]

ok i have try it. but i confuse.

first, i enter world unit(micron), enter distance in those unit=500. then i got the result like below:

The distance you drew is 112.57 pixels=500.0000 microns

The number of microns per pixels is 4.441784

The number of pixels per microns is 0.225153

I do it second time, i enter world unit(micron), enter distance in those unit=500. then i got the result like below:

The distance you drew is 121.04 pixels=500.0000 micons

The number of microns per pixels is 4.130859

The number of pixels per microns is 0.242080

My question is, should be the distance for one pixel is same. Meaning the result for first and second time must be SAME. I think so. But differ for first(4.441784) and second(4.130859)

[Image Analyst]

You drew two lines, each with a different distance -- 112.57 the first time and 121.04 pixels long the second time. However you told it that each distance was 500 microns so you got two different calibrations, one for each distance you drew. What do you expect it should do in the case where you say 500 microns can be anywhere from 112 to 121 pixels???

Blow up the image to full screen to get as much precision as you can. See attached function.

[mohd akmal masud]

i think i should be drew line from end top to end bottom or from the end right to end left. so from the distance value that i got is the pixel size for one pixel.

[mohd akmal masud]

thank you so much Image Analyst.

[Image Analyst]

It was 112 pixels one time and 121 pixels the other time. I guess one is for the width and one is for the height. It's possible, though unlikely, that you can have non-square pixels in which case you must have an x calibration and a y calibration. More likely though is that your field of view is not really 500 microns in each direction like you told it. Also, you can get the width and height simply from the size() function so you don't need to draw if you're just going to take the entire field of view. You only need to draw if your standard is taking up only part of the image.

[Stelios Fanourakis]

@Image Analyst. First of all, thank you so much for providing those calibration functions. They seem to help me a lot at my tasks but I need to ask you some things I don't quite understand.

By using your spatial_calibration_demo.m I upload my image (Dicom) among the ready images you have at the first window. Then I get the message that I need to do spatial calibration as a first step. I draw a straight line (horizontally) to the image. It asks me to Enter real world units and I put mm (milimeters) instead of microns. Then it asks me to Enter Distance in those units. I suppose it asks for pixels because the default value is an arbitrary 500. I give 600. At that point how am I supposed to know the actual distance of the line I draw. This value is supposed to automatically estimated by the Demo program based upon the line I draw. It may be short line it may be long line. Where this 500 pixels comes from? I put 600 in estimation based upon the ruler on x axis.

I get the message "The distance you drew is 590.69 pixels (it might be correct) = 600 microns (The value I put as a distance but supposedly, it is in mm not in microns). The number of microns per pixel is 1.015763 The number of pixels per micron is 0.984481

Can a pixel size be more than 1 mm???

There is also an option to Measure Area in the Demo panel. When I chose this option I draw an area and it gives me a number x in square microns. I don't get pixel values.

Second, I use the other function demo you provided as an answer to this post. I manually draw a contour and I get the values "Mean value within drawn area = 165.339. I suppose it means the mean gray pixel value in that area inside the contour?

Number of pixels = x Area in pixels = 2708. What exactly this number mean? It is almost similar to the number of pixels value. It differs only to a number of 13 pixels.

Finally, I'd like to have a similar function, that takes the area out of a automated segmented mask-contour (e.g. chanvese method) and calculates the pixels and the area. Not manually handdrawn contour but an automated one. I believe it is only a modification of the above functions

[Image Analyst]

It doesn't ask for pixels. It measures the pixels you drew and asks you for real world units and real world distance in those units. It says

userPrompt = {'Enter real world units (e.g. microns):','Enter distance in those units:'};

so you can enter whatever characters you want. You can leave it as microns, or enter cm or mm or km or miles or furlongs or whatever units name you want.

You got to know the distance you measured. If you drew across the diameter of the moon in your picture, you must know if it's 50 cm because it's a poster on your wall, or 2159 miles because it's a picture of the real moon. I can't tell you that - you gotta know.

The mean intensity value it gives you after you draw an area is the mean intensity value of the pixels in the area you drew. It seems rather straightforward to me. Where is there any possibly ambiguity in that description?

When you draw an area, it makes a binary mask and some pixels are "in" the area, and the rest are outside of the area. The area is the number of pixels considered to be in the area. Only full pixels are in the area - there are no partial pixels because the line cut them in half.

I don't know what you call " the number of pixels value". The area in pixels is the number of whole pixels in the are. I don't know how you are getting a difference of 13 pixels because I don't know what other area you're summing up to get the other area.

When you use activecontour() to do chan-vese you get a mask. You can then use that to get the area (in pixels and real world units if you calibrated) or the mean intensity within that area.

[Stelios Fanourakis]

I get the "Area in pixels" with a difference of 13 pixels from the "Number of pixels". If we assume that the second one (number of pixels) are for pixels inside the contour then what this difference of 13 pixels stands for and what the meaning of having "Area in pixels" label if it stands for the same thing as above.

I'll try to embed the chan vese active contour code to your demo so I can get the values for the contour.

I'd like the user to draw a line on the image and have the programe calculate the distance in mm or cm or whatever, based upon the line the user drew. The user might not know the distance in image. He can draw a short line (a few mm or cm) or he can draw a long line but it needs for the programe to calculate the distance on the image that the user defined by drawing a line and measuring the pixels in it so pixel spacing can be estimated.

[Stelios Fanourakis]

What's this 500 in the distance value?

[Image Analyst]

That's just a default value. Just get rid of it if you don't like it:

def = {'microns', ''};

or put in any other default value you want.

Same for the units name. If you don't like microns, change it to kilometers or miles or whatever units you like.

[Stelios Fanourakis]

How can you modify that code to calculate the distance in (mm,cm,pix) based on the line that the user draws?

[Image Analyst]

No need to modify it. It already does that. First you calibrate it by drawing a line of a known distance, then you can draw additional lines all day long and it will apply the calibration factor and tell you the length of them.

[Stelios Fanourakis]

That’s the point I cannot understand. When the user draws a line, demo asks the user to insert the distance value. The user may not know the length of the line or the distance in the image. Demo should calculate this for the user. Am I clear to my question?

[Stelios Fanourakis]

So the first time I input a value of a known distance (random one) and then I draw random lines and are automatically calibrated?

[Image Analyst]

Again, if you don't know the distance of anything in your image to use as a spatial calibration standard, then you can't calibrate. Remember my story about the moon. If all you have is a photo of the moon and you pick two pixels on it, you can't possibly know the distance between them if you don't have a calibration standard. The distance could be 50 cm if it's a poster on my wall, or it could be a thousand miles if it's the actual moon you took from a telescope. I suggest you talk this concept over with another person who may be able to explain it in away that you can understand, because apparently I'm not getting the point across.

With my demo, first you draw a line and tell it what distance it is in real world units, cm or miles or whatever. Then for every other line you draw, it will tell you the distance in calibrated real world units (e.g. cm or miles) as well as the distance in pixels. Again, the other person will help you to understand that.

[Stelios Fanourakis]

Yes, sorry for that. Thank you ImageAnalyst