what should I do so that only gray image converts into binary , not the black backgroung while moving the threshold.

Dear Image analyst , I have a gray image ( a selected ROI) displayed in a black background screen. I am trying to use your adjustable threshold method to convert this gray image into binary.while moving the threshold cursor to convert this gray image into binary image , the black background is also changing from black to white and is including into total pixel calculations and in histogram. But I want only my ROI to convert from gray to binary not the background.

Answer 1

[Image Analyst]

Just change your thresholds. Instead of going from 0 to 150, go from 150 to 255. If that's not right, post a screenshot and describe it in more detail.

[MD]

Dear Image analyst

Thanks. I changed the threshold from 150-255 as you advised. but almost all the binary image information disappears (please see the images ( binary and grayscale image) below. my original grayscale image is an ROI taken from a breast dicom image. what I want actually is- Adjusting the threshold at a level where I decide the binary 1 as the glandular tissue and binary 0 as the adipose tissue. so the no. of pixels correspond to 1 and no. of pixels correspond to 0 are also very important. Because the black background is automatically adding in calculating pixels correspond to 0 , I am getting the wrong number of black, white, or total pixels. please help me so that this black background does not covert into binary while I move the threshold cursor. thanks again

[Image Analyst]

150 was just an example. Use the sliders to go from whatever value works best up to 255.

[MD]

Dear Image analyst

Thanks for your kind help. I found that if I start threshold from 17 , I can avoid background and only ROI converts into binary. but the main problems I face:

1. There is big peak in histogram which is obviously because of the huge no. of black background pixels. This peak misinterprets my findings. How can I delete this big peak, so that histogram will plot only the black and white pixels of ROI , not the background?

2. At the same time, I am calculating the no. of white and black pixels in the ROI. But again , because of the huge no. of black pixels in background , I am getting wrong number of black or white pixels. How can I calculate the black and white pixels inside the ROI or object only which will not include the black pixels in background. please help me sir.

[Image Analyst]

1. There might be a radio button to choose log or linear display that should shorten the big spikes. Or maybe I haven't uploaded that version yet. If you don't have that, let me know and I'll attach my latest version. There is no option currently to totally suppress any intensity bins such as those at 0 or 255. I'm not sure what you consider the ROI, but if it's defined by intensity, then that histogram is there - it's what is between the two vertical bars of course.

2. You have to threshold to find the binary image. Then you have to do a holefill with imfill().

filledBinaryImage = imfill(binaryImage, 'holes');

Then get those pixels, which will include any black pixels inside the objects, actually ALL pixels of any gray level:

pixelsInROI = grayImage(filledBinaryImage);

Then count pure black and pure white pixels in the gray image that are only inside the ROI;

numBlackPixels = sum(pixelsInROI == 0);
numWhitePixels = sum(pixelsInROI == 255);

[MD]

Dear Image Analyst

Thank you very much for your kind help. I have calculated the no. of black and white pixels correctly with the help of your codes. I have only one problem right now , that is the big spike of black background that displays at gray level 17. As you advised before , there is a radio button to display log or linear in histogram . but I dont have the update version of your threhold method.If you think the updated version can help me to suppress the big spike of black background and will display mainly the object , please send the the updated version. So kind of you sir. thanks.

[Image Analyst]

See attachments. Please accept my answer if I've helped you.