[scikit-image] 转发:回复:Re: local maxima improvements
Yann GAVET
gavet at emse.fr
Thu Apr 12 03:17:58 EDT 2018
Hello all,
actually, local_maxima should be equivalent to h-maxima, with h=1
h-maxima (and also local_maxima) give non ponctual locations. There
might be a need for a function of ultimate erosion. Does it exist
somewhere ?
I totally agree with you, the official watershed demo should be modified.
Yann GAVET
Assistant Professor - Ecole des Mines de Saint-Etienne
158 Cours Fauriel, 42023 Saint-Etienne cedex 2, FRANCE
Tel: (33) - 4 7742 0170
On 04/12/2018 09:08 AM, imagepy at sina.com wrote:
> Hi,everyone
>
> I think we should not use peak_local_max for find watershed's seeds. why
> not use h_maxima? which can give a h tolerance.
> I think if we should replace it in the official demo? It would cause a
> misunderstanding.
>
> And scikit-image's h_maxima, h_minima is very slow. here I implements
> one with numba,
> https://github.com/Image-Py/imagepy/blob/master/imagepy/ipyalg/hydrology/findmax.py.
> you can see if it is useful.
>
> yxdragon
> ----- 原始邮件 -----
> 发件人:Stefan van der Walt <stefanv at berkeley.edu>
> 收件人:"Mailing list for scikit-image (http://scikit-image.org)"
> <scikit-image at python.org>
> 主题:Re: [scikit-image] local maxima improvements
> 日期:2018年04月12日 03点08分
>
>
> On Wed, 11 Apr 2018 12:44:45 +1000, Juan Nunez-Iglesias wrote:
> > In [7]: image
> > Out[7]:
> > array([[ 0., 0., 0., 0., 0., 0.],
> > [ 0., 1., 0., 0., 0., 0.],
> > [ 0., 0., 0., 0., 0., 0.],
> > [ 2., 2., 2., 4., 4., 2.],
> > [ 2., 2., 2., 4., 4., 2.],
> > [ 2., 2., 2., 2., 2., 2.]])
> >
> > In [15]: feature.peak_local_max(image)
> > In [17]: image_peak[tuple(feature.peak_local_max(image).T)] = 1
> >
> > In [18]: image_peak
> > Out[18]:
> > array([[ 0., 0., 0., 0., 0., 0.],
> > [ 0., 1., 0., 0., 0., 0.],
> > [ 0., 0., 0., 0., 0., 0.],
> > [ 0., 1., 0., 1., 1., 0.],
> > [ 0., 1., 0., 1., 1., 0.],
> > [ 0., 0., 0., 0., 0., 0.]])
> That output in column 1 looks highly suspect! This is a great example
> for a regression test, thanks Yann.
> Stéfan
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