http://imagej.273.s1.nabble.com/subtraction-of-images-tp3696059p3696064.html
> Hmm, with this example, I'd just set a threshold at the image B to
> select everything non-white, create a selection and remove every
> signal in image A within this selection (BG-Color = white)...
> But may be this is not what you want?
>
> Antje
>
>
>
> Oliver Bannach schrieb:
>> Hi Michael,
>> thank you for your suggestion, but that's not exactly what I was
>> looking for.
>> It should look like shown in this example (white=0, black=255):
>>
http://www.biophys.uni-duesseldorf.de/~bannach/ImageJ/example.jpg>> Any ideas how to do that?
>> Thanks
>> Oliver
>> Michael Schmid schrieb:
>>> Hi Oliver,
>>>
>>> The documentation of Image Calculator says:
>>>
>>> Subtract img1 = img1-img2
>>> Difference img1 = |img1-img2|
>>>
>>> So "Difference" will be zero for all pixels where two images
>>> are the same. Is this what you are looking for?
>>>
>>>
http://rsb.info.nih.gov/ij/docs/menus/process.html#calculator>>>
>>> By the way, Process>Math>Subtract subtracts a constant, not an
>>> image.
>>>
>>> Michael
>>> ________________________________________________________________
>>>
>>> On 4 Jun 2008, at 11:22, Oliver Bannach wrote:
>>>
>>>> Dear imagej experts
>>>>
>>>> What I like to do is basically a subtraction of images (imageA
>>>> minus imageB). But each pixel yielding a signal in imageA must
>>>> be set to zero, if it also appears in imageB, regardless of
>>>> signal intensities. In other words: Any colocalized pixel must
>>>> be black in image A. The conventional subtract process (Process-
>>>> >Math->Substract) does not work for me, since the pixel
>>>> intensities in ImageA are only reduced by values of ImageB
>>>> pixels. Can I use imageJ standard operations for this purpose or
>>>> do I need a plugin? Any help is greatly appreciated. Thanky you
>>>> in advance.
>>>>
>>>> Cheers
>>>> Oliver