In some conditions, a particle skeleton is an open curve with no branched point, according to a 4-connected neighborhood:

Getting the pixels indices according to their order on the curve may be usefull (computing direction, bending,fitting...).

Test curve |

The following

*SkelToPixel_2*script, depends on two other scripts:BranchedEndPoints and StructElem. The scripts must be stored in the same folder.The idea is simple, the script takes the curve at one end, walk on it searching a neighbor pixel up to the ...antepenultimate pixel (that works).Opencv yields the coordinates of pixels belonging to the contour of a particle,

*i.e.*a closed curve. I don't know how to, but I suppose it could be possible to do the same thing with opencv for an open curve.```
# -*- coding: utf-8 -*-
"""
Created on Tue Jun 7 10:35:17 2011
SkelToPixel_2
A code to convert an open curve stored in a numpy array
into a list of neihbor pixels
@author: jean-Patrick Pommier
http://dip4fish.blogspot.com
"""
import numpy as np
from scipy import ndimage as nd
import BranchedEndPoints as bep
im=np.array([[0,0,0,0,0,0,0],
[0,1,0,0,0,0,0],
[0,0,1,1,0,0,0],
[0,0,0,0,1,1,0],
[0,0,0,0,0,0,0]])
#total number of pixels in the curve
pixN=np.sum(im==1)
pixelsList=[]
#get end points by hit&miss operator
ep=bep.get_endpoints(im)
lab,n=nd.label(ep)
#where the first endpoint is
first_indices=np.where(lab==1)
firstEndPoint=np.uint8(lab==1)
#keep an image of the second end point
lastEndPoint=np.uint8(lab==2)
pixelsList.append(first_indices)
current_curve=np.copy(im)
current_point=firstEndPoint
print current_curve
#start to walk on the curve
while pixN<>2:
current_curve=current_curve-current_point
pixN=pixN-1
#this is not very smart, using a 3x3 neighborhood
#arround the current point would reduce the research
current_point=bep.get_endpoints(current_curve)-lastEndPoint
pixelsList.append(np.where(current_point==1))
#add the last pixel
pixelsList.append(np.where(lastEndPoint==1))
for i in range(0,len(pixelsList)):
print "i=",i,pixelsList[i]
```

Run from spyder, the script output is:[[0 0 0 0 0 0 0] [0 1 0 0 0 0 0] [0 0 1 1 0 0 0] [0 0 0 0 1 1 0] [0 0 0 0 0 0 0]] i= 0 (array([1]), array([1])) i= 1 (array([2]), array([2])) i= 2 (array([2]), array([3])) i= 3 (array([3]), array([4])) i= 4 (array([3]), array([5]))The script doesn't yield directly pixels indices.

Feel free to propose a nicer, faster code.

The result with another curve

And the output console:

[[0 0 0 0 0 0 0] [0 1 0 0 1 0 0] [0 0 1 0 0 1 0] [0 0 0 1 1 0 0] [0 0 0 0 0 0 0]] i= 0 (array([1]), array([1])) i= 1 (array([2]), array([2])) i= 2 (array([3]), array([3])) i= 3 (array([3]), array([4])) i= 4 (array([2]), array([5])) i= 5 (array([1]), array([4]))