Use NaN values in an image

Hi all,

I am trying to put all the white values of an image to NaN so those values are not used on subsequent calculations over the image (mainly a co-ocurrence matrix)

The images are 16 bit per channel .tif files. First i normalize all the values to the interval 0,1 and then try to identify the white pixels:

imRGBn=mat2gray(imRGB,[0 65535]);
for i=1:size(imRGBn,1);
    for j=1:size(imRGBn,2);
        if (imRGBn(i,j,1)==1)&(imRGBn(i,j,2)==1)&(imRGBn(i,j,3)==1);
            imRGBn(i,j,1)=NaN;
            imRGBn(i,j,2)=NaN;
            imRGBn(i,j,3)=NaN;
        end
    end
end

The code is pretty un-elegant I know, but i do not know where it is going wrong.

Any help is appreciated.

I need to convert this image to other colour spaces... does anyone know any issues with NaN values and colour space transformations?

Thanks in advance.

[Geoff Hayes]

ivel78 - you don't really mention what the problem is, only that but i do not know where it is going wrong. Are you observing any errors? If so, what are they? And if no errors, then what makes you think that something is incorrect?

According to mat2gray, the returned matrix I contains values in the range 0.0 (black) to 1.0 (full intensity or white). I take that to mean that, for your example, the matrix imRGBn is now two dimensional since it is grayscale. What are the dimensions for imRGBn? What does

 size(imRGBn)

return.

If in fact it is two dimensional then you know what is incorrect in your code and so what will have to be replaced.

Note that it is good practice not to use i and j as indexing variables in your loops since MATLAB also uses i and j to represent the imaginary number.

Also, using == to compare two numbers is valid only if the numbers are integers. When they are floating point numbers, like in this case, you are not guaranteed that (for example) the number will be identical to 1. Instead, you should consider the difference between the number and one and compare this against a small number (tolerance): if less than this tolerance, then the number could be said to be one. For example, if we use eps be our tolerance, then the number x is considered to be one if

 abs(x - 1.0) < eps

[ivel78]

Hi Geoff,

Thanks for your comments, i have slightly modified the code to change i and j and also have used your suggested way of comparing the values.

Before, the problem was that it was not doing anything, the values in the matrix were staying the same, now it is working.

Thanks very much for your help.

Ignacio

Answer 1

[Geoff Hayes]

Moved comment to here

Note that it is good practice not to use i and j as indexing variables in your loops since MATLAB also uses i and j to represent the imaginary number.

Also, using == to compare two numbers is valid only if the numbers are integers. When they are floating point numbers, like in this case, you are not guaranteed that (for example) the number will be identical to 1. Instead, you should consider the difference between the number and one and compare this against a small number (tolerance): if less than this tolerance, then the number could be said to be one. For example, if we use eps be our tolerance, then the number x is considered to be one if

Answer 2

[Image Analyst]

I'm not really sure how, or even if, graycomatrix() can handle NaNs. My recommendation is to just leave the image alone, use a matrix of NxN, where N is the number of unique gray levels in the image, and then crop off the last row and column, which essentially says that there were no pixels of the max value in the image. You can resize the matrix if you want it to be a different size.