Thank you Juan, all my researches from a well-known web search engine gave me peak_local_max as a solution, in particular, the example attached to the watershed function. Sorry about this. Maybe this function should be marked as deprecated in favor of local_maxima. About peak_local_max, there is an option indices=False that results in an array with the same shape as the image in argument. yann Le 11/04/2018 à 04:44, Juan Nunez-Iglesias a écrit :
Hi Yann, and thanks for the interest!
We actually already have this algorithm implemented in skimage.morphology.local_maxima.
peak_local_max is a bit different and I must admit I don’t understand the logic in it. I *particularly* don’t understand the following result:
In [1]: def rmax(I): ...: """ ...: Own version of regional maximum ...: This avoids plateaus problems of peak_local_max ...: I: original image, int values ...: returns: binary array, with 1 for the maxima ...: """ ...: I = I.astype('float'); ...: I = I / np.max(I) * (2**31 -2); ...: I = I.astype('int32'); ...: h = 1; ...: rec = morphology.reconstruction(I, I+h); ...: maxima = I + h - rec; ...: return maxima ...: ...:
In [2]: image = np.zeros((6, 6)) In [3]: image[1, 1] = 1 In [4]: image[3:] = 2 In [6]: image[3:-1, 3:-1] = 4
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 [8]: rmax(image) Out[8]: array([[ 0., 0., 0., 0., 0., 0.], [ 0., 1., 0., 0., 0., 0.], [ 0., 0., 0., 0., 0., 0.], [ 0., 0., 0., 1., 1., 0.], [ 0., 0., 0., 1., 1., 0.], [ 0., 0., 0., 0., 0., 0.]])
In [12]: morphology.local_maxima(image) Out[12]: array([[0, 0, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 1, 1, 0], [0, 0, 0, 1, 1, 0], [0, 0, 0, 0, 0, 0]], dtype=uint8)
In [15]: feature.peak_local_max(image) Out[15]: array([[4, 4], [4, 3], [4, 1], [3, 4], [3, 3], [3, 1], [1, 1]])
In [16]: image_peak = np.zeros_like(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.]])
Anyone else on the list care to comment???
Juan.
On 10 Apr 2018, 11:49 PM +1000, Yann GAVET <gavet@emse.fr>, wrote:
First mistake, this should work, but the discretization of the 'continuous' values should be handled with care.
def rmax(I):
"""
Own version of regional maximum
This avoids plateaus problems of peak_local_max
I: original image, int values
returns: binary array, with 1 for the maxima
"""
I = I.astype('float');
I = I / np.max(I) * (2**31 -2);
I = I.astype('int32');
h = 1;
rec = morphology.reconstruction(I, I+h);
maxima = I + h - rec;
return maxima
Le 10/04/2018 à 15:35, Yann GAVET a écrit :
Dear all,
I have been playing around with the watershed segmentation by markers with the code proposed as example:
http://scikit-image.org/docs/dev/auto_examples/segmentation/plot_watershed.h...
Unfortunately, if we use for example floating values for the radii of the circles (like r1, r2 = 20.7, 24.7), the separation is not perfect, as it gives 4 labels.
If we use the chamfer distance transform instead of the Euclidean distance transform, it is even worse.
It appears that the markers detection by regional maximum (peak_local_max) fails in the presence of plateaus. Its algorithm is basically D(I)==I, where D is the morphological dilation.
A better algorithm would be to use morphological reconstruction (see SOILLE, Pierre. /Morphological image analysis: principles and applications/. Springer Science & Business Media, 2003, p202, Eq 6.13). A proposition of the code can be the following (it should deal with float values):
import numpy as np
from skimage import morphology
def rmax(I):
"""
This avoids plateaus problems of peak_local_max
I: original image, float values
returns: binary array, with True for the maxima
"""
I = I.astype('float');
I = I / np.max(I) * 2**31;
I = I.astype('int32');
rec = morphology.reconstruction(I-1, I);
maxima = I - rec;
return maxima>0
This code is relatively fast. Notice that the matlab function imregionalmax seem to work the same way (the help is not explicit, but the results on a few tests seem to be similar).
I am afraid I do not have time to integrate it on gitlab, but this should be a good start if someone wants to work on it. If you see any problem with this code, please correct it.
thank you
best regards
-- Yann
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_______________________________________________ scikit-image mailing list scikit-image@python.org https://mail.python.org/mailman/listinfo/scikit-image
-- Yann GAVET Assistant Professor - Ecole Nationale Supérieure des Mines de Saint-Etienne 158 Cours Fauriel, CS 62362, 42023 SAINT-ETIENNE cedex 2 Tel: (33) - 4 7742 0170