使用 OpenCV 和 Python 检测接触/重叠的圆/椭圆

Posted 2023-03-12

【中文标题】使用 OpenCV 和 Python 检测接触/重叠的圆/椭圆

【英文标题】：Detect touching/overlapping circles/ellipses with OpenCV and Python

【发布时间】：2015-01-12 00:22:32

【问题描述】：

我想测量圆的圆度（“圆”高度和宽度或椭圆参数的差异）。圆圈在图片中给出，如下所示：

在做了color2gray、阈值和边界检测等常规操作后，我得到如下图所示：

有了这个，我已经尝试了很多不同的东西：

List item Watershed with findContour（类似于this question）-> openCV 将圆圈之间的空间检测为闭合轮廓而不是圆圈，因为它们粘在一起不形成闭合轮廓 fitEllipse 存在同样的问题。我将椭圆拟合在黑色背景轮廓上，而不是介于两者之间。 只是尝试应用霍夫变换（如代码和第三张图片所示）也会导致奇怪的结果：

在此处查看代码：

import sys
import cv2
import numpy
from scipy.ndimage import label

# Application entry point
#img = cv2.imread("02_adj_grey.jpg")
img = cv2.imread("fuss02.jpg")

# Pre-processing.
img_gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)    
cv2.imwrite("SO_0_gray.png", img_gray)

#_, img_bin = cv2.threshold(img_gray, 0, 255, cv2.THRESH_OTSU | cv2.THRESH_BINARY)
_, img_bin = cv2.threshold(img_gray, 170, 255, cv2.THRESH_BINARY)
cv2.imwrite("SO_1_threshold.png", img_bin)

#blur = cv2.GaussianBlur(img,(5,5),0)
img_bin = cv2.morphologyEx(img_bin, cv2.MORPH_CLOSE, numpy.ones((3, 3), dtype=int))
cv2.imwrite("SO_2_img_bin_morphoEx.png", img_bin)

border = img_bin - cv2.erode(img_bin, None)
cv2.imwrite("SO_3_border.png", border)


circles = cv2.HoughCircles(border,cv2.cv.CV_HOUGH_GRADIENT,50,80, param1=80,param2=40,minRadius=10,maxRadius=150)
print circles

cimg = img
for i in circles[0,:]:
# draw the outer circle
    cv2.circle(cimg,(i[0],i[1]),i[2],(0,255,0),2)
    # draw the center of the circle
    cv2.circle(cimg,(i[0],i[1]),2,(0,0,255),3)
    cv2.putText(cimg,str(i[0])+str(',')+str(i[1]), (i[0],i[1]), cv2.FONT_HERSHEY_SIMPLEX, 0.4, 255)

cv2.imwrite("SO_8_cimg.png", cimg)

有没有人想改进我的算法或完全不同的方法？我一直在尝试许多不同的方法，但到目前为止没有运气。感谢大家的帮助。

【问题讨论】：

您在从图像中提取圆圈时遇到问题吗？我不太符合你的要求。

是的，正如您在上面的边界检测图像中看到的那样，我无法分离圆圈。在过滤等过程中会丢失很多边框。

我有一些想法。给我一点儿

【参考方案1】：

我的代码有一些错误@dhanuskha。我想是因为我使用的是不同版本的 CV。如果您需要，此代码适用于 CV 3.0。

import cv2

im = cv2.imread('input.png')
hsv = cv2.cvtColor(im, cv2.COLOR_BGR2HSV)
th, bw = cv2.threshold(hsv[:, :, 2], 0, 255, cv2.THRESH_BINARY | cv2.THRESH_OTSU)
kernel = cv2.getStructuringElement(cv2.MORPH_ELLIPSE, (3, 3))
morph = cv2.morphologyEx(bw, cv2.MORPH_CLOSE, kernel)
dist = cv2.distanceTransform(morph, cv2.DIST_L2, cv2.DIST_MASK_PRECISE)
borderSize = 75
distborder = cv2.copyMakeBorder(dist, borderSize, borderSize, borderSize, borderSize, 
                                cv2.BORDER_CONSTANT | cv2.BORDER_ISOLATED, 0)
gap = 10                                
kernel2 = cv2.getStructuringElement(cv2.MORPH_ELLIPSE, (2*(borderSize-gap)+1, 2*(borderSize-gap)+1))
kernel2 = cv2.copyMakeBorder(kernel2, gap, gap, gap, gap, 
                                cv2.BORDER_CONSTANT | cv2.BORDER_ISOLATED, 0)
distTempl = cv2.distanceTransform(kernel2, cv2.DIST_L2, cv2.DIST_MASK_PRECISE)
nxcor = cv2.matchTemplate(distborder, distTempl, cv2.TM_CCOEFF_NORMED)
mn, mx, _, _ = cv2.minMaxLoc(nxcor)
th, peaks = cv2.threshold(nxcor, mx*0.5, 255, cv2.THRESH_BINARY)
peaks8u = cv2.convertScaleAbs(peaks)
_, contours, hierarchy = cv2.findContours(peaks8u, cv2.RETR_CCOMP, cv2.CHAIN_APPROX_SIMPLE)
peaks8u = cv2.convertScaleAbs(peaks)    # to use as mask
for i in range(len(contours)):
    x, y, w, h = cv2.boundingRect(contours[i])
    _, mx, _, mxloc = cv2.minMaxLoc(dist[y:y+h, x:x+w], peaks8u[y:y+h, x:x+w])
    cv2.circle(im, (int(mxloc[0]+x), int(mxloc[1]+y)), int(mx), (255, 0, 0), 2)
    cv2.rectangle(im, (x, y), (x+w, y+h), (0, 255, 255), 2)
    cv2.drawContours(im, contours, i, (0, 0, 255), 2)

cv2.imshow('circles', im)
cv2.waitKey(0)

【参考方案2】：

这是我检测圆圈的尝试。总结

执行 BGR->HSV 转换并使用 V 通道进行处理

V 频道：

阈值，应用形态闭合，然后进行距离变换（我称之为

dist

）

dist 图片：

创建一个模板。从图像中圆圈的大小来看，半径约为 75 像素的圆盘看起来是合理的。把它的距离变换作为模板（我称之为

temp

）

温度图片：

执行模板匹配：

dist

*

temp

dist * temp 图片：

找到结果图像的局部最大值。最大值的位置对应于圆心，最大值对应于它们的半径

阈值模板匹配图像：

将圆检测为局部最大值：

我是用 C++ 做的，因为我觉得它最舒服。如果你觉得这很有用，我认为你可以轻松地将其转换为 python。请注意，以上图片未按比例绘制。希望这会有所帮助。

编辑：添加了 Python 版本

C++：

    double min, max;
    Point maxLoc;

    Mat im = imread("04Bxy.jpg");
    Mat hsv;
    Mat channels[3];
    // bgr -> hsv
    cvtColor(im, hsv, CV_BGR2HSV);
    split(hsv, channels);
    // use v channel for processing
    Mat& ch = channels[2];
    // apply Otsu thresholding
    Mat bw;
    threshold(ch, bw, 0, 255, CV_THRESH_BINARY | CV_THRESH_OTSU);
    // close small gaps
    Mat kernel = getStructuringElement(MORPH_ELLIPSE, Size(3, 3));
    Mat morph;
    morphologyEx(bw, morph, CV_MOP_CLOSE, kernel);
    // take distance transform
    Mat dist;
    distanceTransform(morph, dist, CV_DIST_L2, CV_DIST_MASK_PRECISE);
    // add a black border to distance transformed image. we are going to do
    // template matching. to get a good match for circles in the margin, we are adding a border
    int borderSize = 75;
    Mat distborder(dist.rows + 2*borderSize, dist.cols + 2*borderSize, dist.depth());
    copyMakeBorder(dist, distborder, 
            borderSize, borderSize, borderSize, borderSize, 
            BORDER_CONSTANT | BORDER_ISOLATED, Scalar(0, 0, 0));
    // create a template. from the sizes of the circles in the image, 
    // a ~75 radius disk looks reasonable, so the borderSize was selected as 75
    Mat distTempl;
    Mat kernel2 = getStructuringElement(MORPH_ELLIPSE, Size(2*borderSize+1, 2*borderSize+1));
    // erode the ~75 radius disk a bit
    erode(kernel2, kernel2, kernel, Point(-1, -1), 10);
    // take its distance transform. this is the template
    distanceTransform(kernel2, distTempl, CV_DIST_L2, CV_DIST_MASK_PRECISE);
    // match template
    Mat nxcor;
    matchTemplate(distborder, distTempl, nxcor, CV_TM_CCOEFF_NORMED);
    minMaxLoc(nxcor, &min, &max);
    // threshold the resulting image. we should be able to get peak regions.
    // we'll locate the peak of each of these peak regions
    Mat peaks, peaks8u;
    threshold(nxcor, peaks, max*.5, 255, CV_THRESH_BINARY);
    convertScaleAbs(peaks, peaks8u);
    // find connected components. we'll use each component as a mask for distance transformed image,
    // then extract the peak location and its strength. strength corresponds to the radius of the circle
    vector<vector<Point>> contours;
    vector<Vec4i> hierarchy;
    findContours(peaks8u, contours, hierarchy, CV_RETR_CCOMP, CV_CHAIN_APPROX_SIMPLE, Point(0, 0));
    for(int idx = 0; idx >= 0; idx = hierarchy[idx][0])
    
        // prepare the mask
        peaks8u.setTo(Scalar(0, 0, 0));
        drawContours(peaks8u, contours, idx, Scalar(255, 255, 255), -1);
        // find the max value and its location in distance transformed image using mask
        minMaxLoc(dist, NULL, &max, NULL, &maxLoc, peaks8u);
        // draw the circles
        circle(im, maxLoc, (int)max, Scalar(0, 0, 255), 2);
    

Python：

import cv2

im = cv2.imread('04Bxy.jpg')
hsv = cv2.cvtColor(im, cv2.COLOR_BGR2HSV)
th, bw = cv2.threshold(hsv[:, :, 2], 0, 255, cv2.THRESH_BINARY | cv2.THRESH_OTSU)
kernel = cv2.getStructuringElement(cv2.MORPH_ELLIPSE, (3, 3))
morph = cv2.morphologyEx(bw, cv2.MORPH_CLOSE, kernel)
dist = cv2.distanceTransform(morph, cv2.cv.CV_DIST_L2, cv2.cv.CV_DIST_MASK_PRECISE)
borderSize = 75
distborder = cv2.copyMakeBorder(dist, borderSize, borderSize, borderSize, borderSize, 
                                cv2.BORDER_CONSTANT | cv2.BORDER_ISOLATED, 0)
gap = 10                                
kernel2 = cv2.getStructuringElement(cv2.MORPH_ELLIPSE, (2*(borderSize-gap)+1, 2*(borderSize-gap)+1))
kernel2 = cv2.copyMakeBorder(kernel2, gap, gap, gap, gap, 
                                cv2.BORDER_CONSTANT | cv2.BORDER_ISOLATED, 0)
distTempl = cv2.distanceTransform(kernel2, cv2.cv.CV_DIST_L2, cv2.cv.CV_DIST_MASK_PRECISE)
nxcor = cv2.matchTemplate(distborder, distTempl, cv2.TM_CCOEFF_NORMED)
mn, mx, _, _ = cv2.minMaxLoc(nxcor)
th, peaks = cv2.threshold(nxcor, mx*0.5, 255, cv2.THRESH_BINARY)
peaks8u = cv2.convertScaleAbs(peaks)
contours, hierarchy = cv2.findContours(peaks8u, cv2.RETR_CCOMP, cv2.CHAIN_APPROX_SIMPLE)
peaks8u = cv2.convertScaleAbs(peaks)    # to use as mask
for i in range(len(contours)):
    x, y, w, h = cv2.boundingRect(contours[i])
    _, mx, _, mxloc = cv2.minMaxLoc(dist[y:y+h, x:x+w], peaks8u[y:y+h, x:x+w])
    cv2.circle(im, (int(mxloc[0]+x), int(mxloc[1]+y)), int(mx), (255, 0, 0), 2)
    cv2.rectangle(im, (x, y), (x+w, y+h), (0, 255, 255), 2)
    cv2.drawContours(im, contours, i, (0, 0, 255), 2)

cv2.imshow('circles', im)

【讨论】：

很好的解释和说明（+1）

这是我找到的圈子的最佳解决方案，（连同这个：ceng.anadolu.edu.tr/CV/EDCircles）

可能与此处显示的相同：docs.opencv.org/3.0-rc1/d2/dbd/tutorial_distance_transform.html