Python中连通域分割Two-Pass算法的原理与实现详解

import matplotlib.pyplot as plt
from scipy.ndimage import binary_erosion
import numpy as np
path = r"coin.png"
img = plt.imread(path).astype(float)
img = np.mean(img, axis=2)
th = 0.513    # climb(img, 0.1, 0, 0.01)
b = img>0.4

def drawImg(im1, im2, c1='jet', c2='jet'):
    fig = plt.figure()
    ax = fig.add_subplot(121)
    plt.imshow(im1, cmap=c1)
    plt.axis('off')
    ax = fig.add_subplot(122)
    plt.imshow(im2, cmap=c2)
    plt.axis('off')
    plt.show()

b = img>0.4
bb = binary_erosion(b, np.ones([5,5]))

def firstPass(inImg):
    L = 0       # 标记号
    outImg = inImg.astype(float)
    h, w = inImg.shape
    dct = {}
    for i,j in product(range(h), range(w)):
        if inImg[i,j] == 0:
            continue
        neighbors = []   # 记录符合要求的邻域前景
        if i-1> 0 and inImg[i-1, j]>0:
            neighbors.append(outImg[i-1, j])
        if j-1 > 0 and inImg[i, j-1] > 0:
            neighbors.append(outImg[i, j-1])
        if len(neighbors) == 0:
            L += 1
            outImg[i,j] = L
            dct[L] = [[i],[j]]
        else:
            tmpL = min(neighbors)
            outImg[i,j] = tmpL
            dct[tmpL][0].append(i)
            dct[tmpL][1].append(j)
    return outImg, dct

c1, d = firtPass(bb)
drawImg(bb, c1)

def secondPass(outImg, dct):
    outImg = outImg * 1
    print('version')
    for i,j in zip(*np.where(outImg!=0)):
        Ls = [outImg[i,j]]
        if i-1>0 and outImg[i-1,j] != 0:
            Ls.append(outImg[i-1,j])
        if j-1 > 0 and outImg[i, j-1] != 0:
            Ls.append(outImg[i, j-1])
        Ls = np.unique(Ls)
        if len(Ls)<2:
            continue
        minL = np.min(Ls)
        for L in Ls:
            if L == minL:
                continue
            y,x = dct[L]
            outImg[y,x] = minL
            dct[minL][0].extend(dct[L][0])
            dct[minL][1].extend(dct[L][1])
            del dct[L]
    u = np.unique(outImg)
    u = np.sort(u)  # 排序
    N = len(u) - 1  # 此为图标数
    for i in range(1, N+1):
        outImg[outImg==u[i]] = i
    N = len(np.unique(outImg))
    return N, outImg

n, c2 = secondPass(c1, d)
drawImg(c1, c2)