Re: subtraction of images

Posted by Antje-2 on Jun 04, 2008; 11:29am
URL: http://imagej.273.s1.nabble.com/subtraction-of-images-tp3696059p3696062.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
>