输入：

一张掩膜图

输出：

该掩膜的边界点，顺序表示

最大连通分量

输入一张mask 有多个连通分量
输出一个mask 只有一个最大的连通分量

def findmax(path):

    # 图像读取
    imagePath = path
    img = cv2.imread(imagePath, 0)
    ret, img = cv2.threshold(img, 0, 255, cv2.THRESH_BINARY)
    # cv2.imwrite('binary.png', img)


    # # 图像实例化
    img = measure.label(img, connectivity=1)  #4连通
    props = measure.regionprops(img)
    # cv2.imwrite('binary1.png', img)

    #
    # 最大区域获取
    max_area = 0
    max_index = 0
    # props只包含像素值不为零区域的属性，因此index要从1开始
    for index, prop in enumerate(props, start=1):
        if prop.area > max_area:
            max_area = prop.area
            # index 代表每个联通区域内的像素值；prop.area代表相应连通区域内的像素个数
            max_index = index

    img[img != max_index] = 0
    img[img == max_index] = 255
    # img = np.array(img,dtype='uint8')


    return img

边界跟踪

def edge(img):
    # edge = cv2.imread(r"D:\project\lab\badminton\chafen\final1.png", 0)
    img_list = img.tolist()
    # print(edge.shape)
    flag = 0
    for i in range(1, img.shape[0] - 1): # 将单独特殊点删除 否则影响后续标记
        for j in range(1, img.shape[1] - 1):  # w 列
            if img_list[i + 1][j]  + img_list[i + 1][j - 1] + img_list[i + 1][j + 1]\
                    + img_list[i][j + 1]+ img_list[i][j - 1]\
                    + img_list[i - 1][j]+ img_list[i - 1][j + 1]+ img_list[i - 1][j - 1] == 255:
                img_list[i][j] = 0

    for i in range(1, img.shape[0] - 1):  # 找到起始点
        for j in range(1, img.shape[1] - 1):  # w 列
            if img_list[i][j] == 255:
                flag = 1
                print(i, j)
                break
        if flag == 1:
            break

    flag_list = np.zeros(img.shape)
    list1 = []
    b = True
    dirs = [[0, 0], [0, -1], [1, -1], [1, 0], [1, 1], [0, 1], [-1, 1], [-1, 0], [-1, -1], [0, -1]]
    while 1:
        for m in range(0, 11):
            try:
                if img_list[i + dirs[m][0]][j + dirs[m][1]] == 0:
                    b = True
                if img_list[i + dirs[m][0]][j + dirs[m][1]] == 255 and flag_list[i + dirs[m][0]][
                    j + dirs[m][1]] == 0 and b == True:
                    list1.append([j + dirs[m][1], i + dirs[m][0]])
                    flag_list[i + dirs[m][0]][j + dirs[m][1]] = 255
                    i = i + dirs[m][0]
                    j = j + dirs[m][1]
                    b = False
                    break
            except:
                cv2.imwrite('shuzu.png', flag_list)
                return list1