Distinct NaN from other values in image

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I want to plot image with different values and easily distinct them from NaN values. As default NaN values are imaged with the same color as the minimum value. I have tried a range of possibilites but each provide their own issue. Here is one example:
The solution from here, image all other values as one color
imagesc(CP)
sc(isnan(CP), 'bone', [1 1 1]);
  3 Comments
Image Analyst
Image Analyst on 25 Nov 2013
The answer has been accepted. I don't see a question in the last comment. And no m-file was attached.
Oliver Woodford
Oliver Woodford on 2 Jan 2014
You did not use sc correctly. Try:
sc(CP, 'bone', 'r')

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Accepted Answer

Iain
Iain on 21 Nov 2013
This can let you plot NaN & any other logically findable section of the image as white
a = rand(5);
a(3,3) = NaN;
b = imagesc(a);
set(b,'AlphaData',~isnan(a))
Alternatively, you can deliberately control the "colormap" settings to ensure that "invalid" values show up as a colour that isn't used by the colour map.
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More Answers (1)

Image Analyst
Image Analyst on 20 Nov 2013
I don't know what "plot" means when talking about an image. I don't know why you want to plot anything on an image, like markers to "cover up" the nan locations. Why not just set the nan's to the min gray level and view it like a normal image?
CP(isnan(CP)) = min(CP(:)); % Set nan values to the min value
image(CP);
colormap(bone(256));
colorbar;
  3 Comments
Agnete
Agnete on 21 Nov 2013
The problem is illustrated in the attached where it is seen that the polygon in northeast corner blends into the background
Image Analyst
Image Analyst on 22 Nov 2013
Sorry but I don't understand this. One just sort of looks like a very blurred version of the other. Both have the same pattern of triangles on them, and I don't even know what a triangle means/represents. Your image is an array. You can't have nan values have the same value as your minimum value. An element can have only one value, either a nan or some number. It can't have both at the same time. So "Nan values are imaged with the same color as my minimum value" makes no sense to me.

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