python – 缺少边缘的凹面船体

我有一个二进制掩码,我想从中提取轮廓.二进制掩码的所有“外部”边缘都是具有非常高概率的“真实边缘”.保持这些边缘固定,我的目标是插入“缺失”边缘(示例图像和下面所需的结果).我尝试过使用Delaunay三角测量,没有太大的成功(见下面的代码).但是,我甚至不确定这是最好的方法,因为我会在这个过程中从“真正的边缘”中丢失一些细节.

Delaunay三角剖分是否合适？如果是这样,下面的代码出了什么问题？有没有更好的算法来解决这类问题？怎么能用Python实现呢？

当前代码(不起作用)

import numpy as np
from scipy.spatial import Delaunay
from descartes import PolygonPatch
from shapely.geometry import MultiLineString
from shapely.ops import cascaded_union, polygonize
from skimage.draw import polygon
from skimage import segmentation, morphology

def triangulate(mask, alpha=1000):

  contours = measure.find_contours(mask, 0.8)

  points = []

  for n, contour in enumerate(contours):

    for m in xrange(0, len(contour[:, 0])):
      y = contour[:, 0][m]
      x = contour[:, 1][m]
      points.append([y, x])

  points = np.asarray(points)

  tri = Delaunay(points)
  edges = set()
  edge_points = []

  def add_edge(i, j):
    if (i, j) in edges or (j, i) in edges: return
    edges.add( (i, j) )
    edge_points.append(points[ [i, j] ])

  for ia, ib, ic in tri.vertices:
    pa = points[ia]
    pb = points[ib]
    pc = points[ic]

    # Lengths of sides of triangle
    a = math.sqrt((pa[0]-pb[0])**2 + (pa[1]-pb[1])**2)
    b = math.sqrt((pb[0]-pc[0])**2 + (pb[1]-pc[1])**2)
    c = math.sqrt((pc[0]-pa[0])**2 + (pc[1]-pa[1])**2)

    # Semiperimeter of triangle
    s = (a + b + c)/2.0

    # Area of triangle by Heron's formula
    try:
      area = math.sqrt(s*(s-a)*(s-b)*(s-c))
      circum_r = a*b*c/(4.0*area)

      # Here's the radius filter.
      #if circum_r < 1.0/alpha:
      if circum_r < alpha:
          add_edge(ia, ib)
          add_edge(ib, ic)
          add_edge(ic, ia)
    except:
      print('Triangulation error')

  m = MultiLineString(edge_points)
  triangles = list(polygonize(m))

  poly = PolygonPatch(cascaded_union(triangles), alpha=0.5)

  vertices = poly.get_path().vertices

  rr, cc = polygon(vertices[:,0], vertices[:,1])

  img = np.zeros(im.shape)
  img[rr, cc] = 1

  return img

解决方法:

正如nikie指出的那样,如果不要求性能,morphological snakes将解决您的问题.

当你增加迭代次数时,蛇往往会完美地掩盖形状,但这不是你想要的,所以你需要弄清楚有效的参数组合.

我玩了一下,找到了一个很好的近似值：

对于这个形状,我使用以下代码：

def test_concave():
    # Load the image.
    imgcolor = imread("testimages/concave_hull.jpg")/255.0
    img = rgb2gray(imgcolor)

    # g(I)
    gI = morphsnakes.gborders(img, alpha=1000, sigma=7)

    # Morphological GAC. Initialization of the level-set.
    mgac = morphsnakes.MorphGAC(gI, smoothing=3, threshold=0.55, balloon=-1)
    mgac.levelset = circle_levelset(img.shape, (50, 199), 200)

    # Visual evolution.
    ppl.figure()
    morphsnakes.evolve_visual(mgac, num_iters=62, background=imgcolor)


if __name__ == '__main__':
    print """"""
    test_concave()
    ppl.show()

标签：triangulation,python,interpolation,image-processing,delaunay
来源： https://codeday.me/bug/20190725/1530231.html