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From: jabgars kohli on 9 Aug 2010 16:25
Hello Chritopher
I was trying the code that you posted with the same image, but I'm getting an error
saying
"Error using ==> rgb2gray>parse_inputs at 82
MAP must be a m x 3 array.

Error in ==> rgb2gray at 35
X = parse_inputs(varargin{:});"

I have a similar image that i needs to threshold but even I'm get the same error.
Do you have any clue about this ?


> Hi Valentin,
>
> I haven't really tried but it seems that the key phrase here is to 'try a combination of thresholds'. ImageAnalyst is correct in saying that the expected results provided are greyscale images. These can not be obtained from pure, single threshold procedures, which produce a logical mask (i.e either 1 or 0 for each pixel).
>
> Therefore, I think what you need to do is almost quantise - that is, use a number of thresholds. For instance, say:
>
> <pre class="code">
> % read image and ensure it is greyscale for thresholding
> inImage = imread('cerveau.jpg');
> greyImage = rgb2gray(inImage);
>
> % define 'white' to be any pixel greater than thresh1
> thresh1 = 200;
> whiteLayer = greyImage > thresh1;
>
> % define 'lightGrey' to be any pixel greater than thresh2 but less than or equal to
> % thresh1
> thresh2 = 125;
> lightgreyLayer = greyImage > thresh2 & greyImage <= thresh1;
>
> % etc...
> </pre>
>
> Each layer is logical and each pixel within each layer is either 1 (true) or 0 (false). Set each layer to be a given value of grey...
>
> <pre class="code">
> layer1 = whiteLayer*200;
> layer2 = lightGreyLayer*125;
>
> % etc...
> </pre>
>
> Now simply superimpose each layer by adding them. Remember each pixel should only belong to one layer...
>
> <pre class="code">
> newGreyscale = layer1 + layer2 + ...
> </pre>
>
> Still.. it would require many layers to achieve the results expected.. but this is the sort of thing that is expected I think.
>
> The other thing is that the image of the expected result is not one expected result, it is instead BOTH of the expected results. I don't believe any single common thresholding applied to the original can produce that.
> Instead I think that it is a comparative image showing the results of both operations. Remember the question asks you to use 'thresholds both above the white peak and below it'.
> This is confusing and I'm not sure what they mean by 'white peak' so.. these are some things that may aid your solution.
>
> Hope that helps,
>
> Chris
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