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 ?

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]

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)...

[Iain]

 img = double(HU);
 img = img - min(img(:));
 maximum = max(img(:));
 grayscale = (img / maximum) * 255;

Answer 1

[Image Analyst]

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));

[sara]

oh Image Analyst

always you give me a simple but best way ...thanks

[Image Analyst]

Can you please officially "Accept" some of them then? Thanks in advance.

[SaHaR]

Dear Image Analyst,

yesterday I asked the same question. Your solutions are always the best. Thank you.

[Image Analyst]

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
class(b8)
b8 = uint8(rescale(grayImage, 0, 255)) % If you want uint8
class(b8)

grayImage =

1×3 uint16 row vector

10 4000 65535

b8 =

0 15.528 255

ans =

'double'

b8 =

1×3 uint8 row vector

0 16 255

ans =

'uint8'

Answer 2

[Guillaume]

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);

[Christin Panjaitan]

So then, how to convert it back to the original size ? just make it like this :

return16bit = im2uint16(img8bit);

??

[Guillaume]

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]

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.

Answer 3

[Yawar Rehman]

 img = (img / 4000) * 255;

[Iain]

This'd be better....

 img = double(img);
 img = img - min(img(:));
 maximum = max(img(:));
 img = (img / maximum) * 255;