[Insight-users] itkJoinSeriesImageFilter question
freiman at cs.huji.ac.il
freiman at cs.huji.ac.il
Wed Sep 14 01:47:09 EDT 2005
Hi Luis,
Thank you for answering.
It is little hard to post all the typedefs i used because of my code
comlexity. however, the output dimension is input dimension+1 as should be.
the problem was as you described, and your solution, solve it.
thanks,
Moti
Original Message:
-----------------
From: Luis Ibanez luis.ibanez at kitware.com
Date: Tue, 13 Sep 2005 09:58:17 -0400
To: freiman at cs.huji.ac.il, insight-users at itk.org
Subject: Re: [Insight-users] itkJoinSeriesImageFilter question
HI Moti,
Please post to the list the typedef that you
used for instantiating the itk::JoinSeriesImageFilter.
Please include the typedefs used for the input image
and the output image.
This filter expect the output image dimension to be
1 plus the dimension of the input images.
Also... from you very short code snippet, it seems
that you are using the *SAME* 2D filter to provide
all the M inputs. If this is the case, this will
not work, because GetOutput() return a SmartPointer
to the output image of the filter. By the time you
run the Join image filter, all the inputs are pointing
to the same and only output of the single 2D filter
that you have.
If you want to reuse the 2D filter, you MUST disconnect
the output image from the pipeline at every run of
the 2D filter. You will need code such as
for(int i=0; i<M; i++)
{
filter2D->ChangeSomeParameters(..);
filter2D->Update();
ImageType::Pointer outputImage = filter2D->GetOutput();
outputImage->DisconnectPipeline();
join->SetInput(i, outputImage);
}
join->Update();
Regards,
Luis
-----------------------------
freiman at cs.huji.ac.il wrote:
> Hello!
> I'm trying to use the filter: itkJoinSeriesImageFilter from itk2.0
> i apply some 2d filter on each slice, then i want to set the inputs to the
> itkJoinSeriesImageFilter filter.
> when i'm using setinput (i, filter->getOutput());
> i got just he last slice, repeated as a volume.
> do someone know how can i use this filter?
> Thanks,
> Moti Freiman
>
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