# How to find the intensity percentage in a gray scale image?

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Meshooo 2014 年 7 月 30 日

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

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### 採用された回答

Geoff Hayes 2014 年 7 月 30 日

Meshoo - what is the answer that you are getting? 255? From your question, I think it is safe to assume that the data type for your image is an 8-bit unsigned integer. So if your maximum value is 255 then
maxVal = uint8(255);
maxValBy100 = maxVal/100; % this is now 3 because the data type is uint8
maxValBy100*99.5 % this is now 255 since 3*99.5 is greater than the max
% allowed 8-bit unsigned int (255) so is capped at that
% maximum
Since you are mixing doubles with 8-bit unsigned ints, then just cast the maximum value of I to a double before dividing by 100
X = (double(max(I(:)))/100) *99.5;
and the above will return the desired 253.725. Try this and see what happens!

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### その他の回答 (3 件)

Image Analyst 2014 年 7 月 30 日
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.
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.
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Anand 2014 年 7 月 31 日
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;
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Image Analyst 2014 年 7 月 31 日
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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Meshooo 2014 年 8 月 4 日
Thank you Image Analyst, Geoff, Anand. Now I could understand the problem.

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