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我想测量圆的圆度(“圆”的高度和宽度或椭圆参数的差异)。该圆圈中的图片给出如下所示:用OpenCV和Python检测触摸/重叠圆/椭圆

做平常的东西像color2gray,阈值和边界检测之后,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:

  • 列表项与findContour的流域(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
  • 与fitEllipse相同的问题。我在椭圆形背景上安装椭圆而不是在两者之间。
  • 只是想申请霍夫transforamtion(如代码,并显示第三画面),以及导致奇怪的结果:

在这里看到的代码:

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) 

有没有人有想法来改进我的算法或完全不同的方法?到目前为止,我一直在尝试许多不同的方法,但没有成功。谢谢大家的帮助。

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你是否对从图像中提取出界的一个问题?我不太喜欢你想要的。 – rayryeng 2014-11-14 17:25:47

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是的,我无法像上面的边框检测图像中看到的那样分开圆圈。在过滤等过程中,很多边框都会丢失。 – Merlin 2014-11-15 16:37:38

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我有一些想法。给我一点点 – rayryeng 2014-11-15 16:47:23

回答

32

这是我在检测圆圈的尝试。总之

  • 执行BGR-> HSV的转换,并使用V通道进行处理

V通道:

enter image description here

  • 阈值时,应用形态学闭,则取距离变换(我会称之为dist

DIST图像:

enter image description here

  • 创建模板。从图像中圆圈的大小看,一个~75像素半径的磁盘看起来是合理的。以它的距离变换,并用它作为模板(我称之为临时

温度图像:

enter image description here

  • 执行模板匹配:DIST * temp

DIST *临时图像:

enter image description here

  • 找到最终图像的局部最大值。最大值的位置对应于圆心和最大值对应其半径

阈值模板匹配的图像:

enter image description here

检测界称为局部最大值:

enter image description here

我用C++做了这个,因为我对它很满意。我认为你可以很容易地将它转换为Python,如果你觉得这很有用。请注意,上面的图片不是按比例绘制的。希望这可以帮助。

编辑:增加了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); 
    } 

的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) 
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+1非常好的解释 – 2014-11-17 14:04:58

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我同意!非常好! – rayryeng 2014-11-18 07:41:31

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WOW。非常好。 – Merlin 2014-11-18 09:17:55