I am displaying an intensity image of type double using imshow and casting it to uint8. In the figure property editor I changed the colormap to hsv. When the data cursor is placed on a pixel, the RGB values are revealed. How can I access these?

I have an intensity image of type double called f. The intensity range is -556.1230 to + 428.1032. When I plot it using the following code:

figure;
imshow(uint8(f));

and then go to the figure property editor and change the color map to hsv. The contrast for my purposes is fantastic. Next I use the data cursor to examine pixel values. The data cursor reveals the rgb values of the pixels. The g 'channel' of the rgb color map provides nice differences in intensity levels that I am looking for. These rgb values are scaled between 0 and 1. I don't know how to get at these values that I see when using the data cursor. There is a colormap tranlation automatically occurring (because at this point I never converted the data to rgb) but I don't know how to access it?

I have tried converting the intensity image of type double to a grayscale image and then to an indexed image and then to an rgb image and then to an hsv image using the following code:

f=my double intensity image I mentioned above with pixel values between -556.1230 to + 428.1032.
g=mat2gray(f,[min(f(:)) max(f(:))]);
[h,graymap]=gray2ind(g,256);
figure;imshow(uint8(h));
r=ind2rgb(h,graymap);
figure;imshow(r)
s=rgb2hsv(r);
figure;
imshow(r);

None of these images produces the rgb pixel values between 0 & 1 as I am able to see as described above.

Thanks for any help!

Answer 1

[Image Analyst]

I think you're reading off the colormap. There must be one applied.

Your data will be clipped to the range 0-255 when you cast it to uint8 for display. Values between -556 and 0 will be set to 0. Values greater than 255 (255 to 428) will be set to 255.

When you convert to RGB, you'll get an image with values between 0 and 255 for each color channel.

I hope that explains what's going on. But I still don't know what you want to do. Do you have any more questions?

[Image Analyst]

Lori's "Answer" moved here:

Thank you. That is why when I did the image type conversions I had mentioned I did not get the same values I was seeing in the Figure which automatically did the clipping.