[ITK-users] fastest way of calculating highest value on watershed each line branches

[asertyuio]

Ok, I'll go for that !

Many thanks Richard, for your quick answer !

Yann



Le 16/03/2017 à 02:09, Richard Beare a écrit :
> This is often a little haphazard, but the usual approach I use is
> something along these lines:
>
> 1) Use the watershed version that marks watershed lines, and binarize
> the output so that the line is 1 and everything else is zero.
>
> 2) Filter this to produce a count of neighbours at each watershed
> voxel. You could write a custom neighbourhood filter to do this, or
> simply use a box mean filter, radius 1. The neighbour count is then
> filtervalue*filtersize - 1. You can, of course, base the following
> steps on the mean, as the transform is constant.
>
> 3) In 2D the non branch points will have 2 neighbours, branch points
> have more. Thus you can select branch points, dilate them, mask them
> out of the watershed line and label the result. Then you can use the
> label statistics filter to compute the maximum in each segment. In 3D
> the watershed boundaries are "sheets", so the neighbour count is more
> complex, but analagous.
>
> On Thu, Mar 16, 2017 at 6:41 AM, asertyuio via Insight-users
> <insight-users at itk.org <mailto:insight-users at itk.org>> wrote:
>
>     Hi all,
>
>     I'm calculating the watershed transform of an image based on some
>     markers. Now, I want to calculate the maximum value of the image
>     on each
>     branch of the lines separating watershed labels.
>
>     I haven't found anything in ITK to separate the different branches.
>
>     I'm thinking of a good way of implementing this. My idea would be
>     to use
>     a neighborhood iterator marching on the lines to list pixels belonging
>     to the different branches of the watershed lines, and then use
>     some sort
>     of point set to calculate the max intensity in the original image.
>
>     Do you think it is a good way to go, or is there some better
>     alternatives ?
>
>     Many thanks,
>
>     Yann
>
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