在二进制边缘图像中找到闭合循环的数量 [英] Find number of close loops in Binary Edge Image

问题描述

我有一个带有闭环和自由曲线数量的二进制图像，如图所示.

I have a binary image with the number of close loops and free curves like shown in image..

到目前为止，我的方法-(python代码):

My approach so far- (python code):

• 骨骼化图像以获得1个像素的边缘
• 使用标签/bwlabel获取单个曲线/边缘
• 应用DFS获取未闭合的边缘
• 应用DFS以获得接近的边缘..但无法正常工作..

预期输出-

闭合循环数= 6，沿循环的像素总和或闭合曲线边缘上的所有(x，y)点，如 RED

Number of close loops =6, the sum of the pixels along the loop or all (x,y) points along the edge of the close curve as highlighted in RED

预期输出-

推荐答案

这是在Python/OpenCV中实现此目的的一种方法.

Here is one way to do that in Python/OpenCV.

• 将输入读取为灰度
• 二进制阈值
• 获取轮廓和层次
• 遍历轮廓和相应的层次结构，并丢弃没有父级的轮廓(外部的)，并保存和计数没有子级的轮廓(最内部的)
• 在输入上绘制轮廓
• 保存结果

输入:

import cv2
import numpy as np

# read image as grayscale
img = cv2.imread('loops.png', cv2.IMREAD_GRAYSCALE)
hh, ww = img.shape

# blacken right columns that are white
img[0:hh, ww-3:ww] = 0

# threshold
thresh = cv2.threshold(img, 0, 255, cv2.THRESH_BINARY)[1]

# get contours
contours = cv2.findContours(thresh, cv2.RETR_TREE, cv2.CHAIN_APPROX_SIMPLE)
hierarchy = contours[1] if len(contours) == 2 else contours[2]
contours = contours[0] if len(contours) == 2 else contours[1]

# get the actual inner list of hierarchy descriptions
hierarchy = hierarchy[0]

# count inner contours
count = 0
result = img.copy()
result = cv2.merge([result,result,result])
for component in zip(contours, hierarchy):
    cntr = component[0]
    hier = component[1]
    # discard outermost no parent contours and keep innermost no child contours
    # hier = indices for next, previous, child, parent
    # no parent or no child indicated by negative values
    if (hier[3] > -1) & (hier[2] < 0):
        count = count + 1
        cv2.drawContours(result, [cntr], 0, (0,0,255), 2)

# print count
print("count:",count)

# save result
cv2.imwrite("loops_result.png",result)

# show result   
cv2.imshow("result", result)
cv2.waitKey(0)
cv2.destroyAllWindows()

产生的轮廓:

Resulting contours:

计数:

count: 6

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