I have a ct image that values of pixel of it is between 0-4000 how can I convert it to an image with 0-255 ?

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I have a ct image that values of pixel of it is between 0-4000 how can I convert it to an image with 0-255 ?
sara on 5 Sep 2014
ok thanks
I think I ask my question in an other way... I should ask:how can I convert HU(hounsfield units 0-4095) to grayscale(0-255)...

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Accepted Answer

Image Analyst
Image Analyst on 5 Sep 2014
There is a function that scales an array to 0-1. It's called mat2gray. Then just multiply by 255.
image8bit = uint8(255 * mat2gray(image16bit));
Image Analyst
Image Analyst on 30 Jul 2020
There is now another way, using the new rescale() function:
grayImage = uint16([10, 4000, 65535]) % Sample uint16 image.
b8 = rescale(grayImage, 0, 255) % If you want double
b8 = uint8(rescale(grayImage, 0, 255)) % If you want uint8
grayImage =
1×3 uint16 row vector
10 4000 65535
b8 =
0 15.528 255
ans =
b8 =
1×3 uint8 row vector
0 16 255
ans =

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More Answers (2)

Guillaume on 5 Sep 2014
If the image is stored as double, then Yawar's answer is correct. If your image is stored as uint16, then it won't work, you'll have to convert the image to double first.
For most image processing function to work correctly on an image in the range 0-255, it needs to be of type uint8. If the type is double, most functions assume the range 0-1.
The following is guaranteed to work:
img8bit = im2uint8(double(img) / 4000);
This is equivalent to:
img8bit = uint8(double(img)/4000 * 255);
Guillaume on 5 Sep 2014
I'm not sure what you mean by original size. The size of the image isn't changed, just the intensity range and the type.
If you want to go back from a uint8 image to an image in the range 0-4000:
img12bitish = double(img8bit) / 255 * 4000;
%or to store it as uint16
img12bitish = uint16(double(img8bit) / 255 * 4000);
Your code return16bit = im2uint16(img8bit) will convert the image to the range 0-65535.
Christin Panjaitan
Christin Panjaitan on 5 Sep 2014
I have tried your code.
[1] img12bitish = double(img8bit) / 255 * 4000; %it might return to the original range but the original value has been changing.
For the word of "size", I mean range. Sorry, I just wrong to use the word.

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Yawar Rehman
Yawar Rehman on 5 Sep 2014
img = (img / 4000) * 255;


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