How to find the intensity percentage in a gray scale image?
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Dear all,
I have 8bit gray scale image "I". I want to find "X" which is the 99.5% of the maximum intensity value. So if the image has at least one white pixel (255), then "X" should be 253.725.
I tried the following code but I'm getting different answer:
I = getimage;
X = (max(I(:))/100) *99.5;
Why the value for "X" is 255 not 253.725?
Any help will be appreciated.
Meshoo
1 Comment
Accepted Answer
More Answers (2)
Image Analyst
on 30 Jul 2014
Your code should be okay. Though it would be better to just use the original image if you have it rather than use getimage();
clc;
workspace;
% Read in sample image and divide by 2.
grayImage = imread('pout.tif') / 2;
subplot(1, 2, 1);
% Display at full range 0-255.
imshow(grayImage, [])
title('Original image, scaled to 0-255 for display', 'FontSize', 22);
% Enlarge figure to full screen.
set(gcf, 'Units', 'Normalized', 'OuterPosition', [0 0 1 1]);
% Get max of original image and what is returned from getimage
maxGL = max(grayImage(:))
I = getimage(gca);
maxGL2 = max(I(:))
% They should be the same. maxGL2 won't be 255 even though displayed image is 255.
X = (max(I(:))/100) *99.5
% Let's compute and display the histogram.
[pixelCount, grayLevels] = imhist(grayImage);
subplot(1, 2, 2);
bar(grayLevels, pixelCount);
grid on;
title('Histogram of original image', 'FontSize', 22);
xlim([0 grayLevels(end)]); % Scale x axis manually.
Anand
on 31 Jul 2014
If I is a uint8 image, then max(I(:)) will be a uint8 value and multiplying that by .995 will mean its still 255. You can see this as follows:
>> uint8(255)*99.5
ans =
255
To fix this, cast the maximum to a double, something like this:
X = double(max(I(:))/100) *99.5;
1 Comment
Image Analyst
on 31 Jul 2014
Yes, if it goes more than 255 (like if you multiply by 99.5 instead of .995), it clips. If it's less than 255, it rounds to nearest integer. Compare this:
uint8(255)*.995
ans = 254
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