我目前有一种方法可以检测图像中的卡片,大多数情况下,当灯光相当一致并且背景非常平静时,它可以正常工作。
以下是我用来执行此操作的代码:
Mat img = inImg.clone();
outImg = Mat(inImg.size(), CV_8UC1);
inImg.copyTo(outImg);
Mat img_fullRes = img.clone();
pyrDown(img, img);
Mat imgGray;
cvtColor(img, imgGray, CV_RGB2GRAY);
outImg_gray = imgGray.clone();
// Find Edges //
Mat detectedEdges = imgGray.clone();
bilateralFilter(imgGray, detectedEdges, 0, 185, 3, 0);
Canny( detectedEdges, detectedEdges, 20, 65, 3 );
dilate(detectedEdges, detectedEdges, Mat::ones(3,3,CV_8UC1));
Mat cdst = img.clone();
vector<Vec4i> lines;
HoughLinesP(detectedEdges, lines, 1, CV_PI/180, 60, 50, 3 );
for( size_t i = 0; i < lines.size(); i++ )
{
Vec4i l = lines[i];
// For debug
//line( cdst, cv::Point(l[0], l[1]), cv::Point(l[2], l[3]), Scalar(0,0,255), 1);
}
//cdst.copyTo(inImg);
// // Find points of intersection //
cv::Rect imgROI;
int ext = 10;
imgROI.x = ext;
imgROI.y = ext;
imgROI.width = img.size().width - ext;
imgROI.height = img.size().height - ext;
int N = lines.size();
// Creating N amount of points // N == lines.size()
cv::Point** poi = new cv::Point*[N];
for( int i = 0; i < N; i++ )
poi[i] = new cv::Point[N];
vector<cv::Point> poiList;
for( int i = 0; i < N; i++ )
{
poi[i][i] = cv::Point(-1,-1);
Vec4i line1 = lines[i];
for( int j = i + 1; j < N; j++ )
{
Vec4i line2 = lines[j];
cv::Point p = computeIntersect(line1, line2, imgROI);
if( p.x != -1 )
{
//line(cdst, p-cv::Point(2,0), p+cv::Point(2,0), Scalar(0,255,0));
//line(cdst, p-cv::Point(0,2), p+cv::Point(0,2), Scalar(0,255,0));
poiList.push_back(p);
}
poi[i][j] = p;
poi[j][i] = p;
}
}
cdst.copyTo(inImg);
if(poiList.size()==0)
{
outImg = inImg.clone();
//circle(outImg, cv::Point(100,100), 50, Scalar(255,0,0), -1);
return;
}
convexHull(poiList, poiList, false, true);
for( int i=0; i<poiList.size(); i++ )
{
cv::Point p = poiList[i];
//circle(cdst, p, 3, Scalar(255,0,0), 2);
}
//Evaluate all possible quadrilaterals
cv::Point cardCorners[4];
float metric_max = 0;
int Npoi = poiList.size();
for( int p1=0; p1<Npoi; p1++ )
{
cv::Point pts[4];
pts[0] = poiList[p1];
for( int p2=p1+1; p2<Npoi; p2++ )
{
pts[1] = poiList[p2];
if( isCloseBy(pts[1],pts[0]) )
continue;
for( int p3=p2+1; p3<Npoi; p3++ )
{
pts[2] = poiList[p3];
if( isCloseBy(pts[2],pts[1]) || isCloseBy(pts[2],pts[0]) )
continue;
for( int p4=p3+1; p4<Npoi; p4++ )
{
pts[3] = poiList[p4];
if( isCloseBy(pts[3],pts[0]) || isCloseBy(pts[3],pts[1])
|| isCloseBy(pts[3],pts[2]) )
continue;
// get the metrics
float area = getArea(pts);
cv::Point a = pts[0]-pts[1];
cv::Point b = pts[1]-pts[2];
cv::Point c = pts[2]-pts[3];
cv::Point d = pts[3]-pts[0];
float oppLenDiff = abs(a.dot(a)-c.dot(c)) + abs(b.dot(b)-d.dot(d));
float metric = area - 0.35*oppLenDiff;
if( metric > metric_max )
{
metric_max = metric;
cardCorners[0] = pts[0];
cardCorners[1] = pts[1];
cardCorners[2] = pts[2];
cardCorners[3] = pts[3];
}
}
}
}
}
// find the corners corresponding to the 4 corners of the physical card
sortPointsClockwise(cardCorners);
// Calculate Homography //
vector<Point2f> srcPts(4);
srcPts[0] = cardCorners[0]*2;
srcPts[1] = cardCorners[1]*2;
srcPts[2] = cardCorners[2]*2;
srcPts[3] = cardCorners[3]*2;
vector<Point2f> dstPts(4);
cv::Size outImgSize(1400,800);
dstPts[0] = Point2f(0,0);
dstPts[1] = Point2f(outImgSize.width-1,0);
dstPts[2] = Point2f(outImgSize.width-1,outImgSize.height-1);
dstPts[3] = Point2f(0,outImgSize.height-1);
Mat Homography = findHomography(srcPts, dstPts);
// Apply Homography
warpPerspective( img_fullRes, outImg, Homography, outImgSize, INTER_CUBIC );
outImg.copyTo(inImg);
computeIntersect
定义为:
cv::Point computeIntersect(cv::Vec4i a, cv::Vec4i b, cv::Rect ROI)
{
int x1 = a[0], y1 = a[1], x2 = a[2], y2 = a[3];
int x3 = b[0], y3 = b[1], x4 = b[2], y4 = b[3];
cv::Point p1 = cv::Point (x1,y1);
cv::Point p2 = cv::Point (x2,y2);
cv::Point p3 = cv::Point (x3,y3);
cv::Point p4 = cv::Point (x4,y4);
// Check to make sure all points are within the image boundrys, if not reject them.
if( !ROI.contains(p1) || !ROI.contains(p2)
|| !ROI.contains(p3) || !ROI.contains(p4) )
return cv::Point (-1,-1);
cv::Point vec1 = p1-p2;
cv::Point vec2 = p3-p4;
float vec1_norm2 = vec1.x*vec1.x + vec1.y*vec1.y;
float vec2_norm2 = vec2.x*vec2.x + vec2.y*vec2.y;
float cosTheta = (vec1.dot(vec2))/sqrt(vec1_norm2*vec2_norm2);
float den = ((float)(x1-x2) * (y3-y4)) - ((y1-y2) * (x3-x4));
if(den != 0)
{
cv::Point2f pt;
pt.x = ((x1*y2 - y1*x2) * (x3-x4) - (x1-x2) * (x3*y4 - y3*x4)) / den;
pt.y = ((x1*y2 - y1*x2) * (y3-y4) - (y1-y2) * (x3*y4 - y3*x4)) / den;
if( !ROI.contains(pt) )
return cv::Point (-1,-1);
// no-confidence metric
float d1 = MIN( dist2(p1,pt), dist2(p2,pt) )/vec1_norm2;
float d2 = MIN( dist2(p3,pt), dist2(p4,pt) )/vec2_norm2;
float no_confidence_metric = MAX(sqrt(d1),sqrt(d2));
// If end point ratios are greater than .5 reject
if( no_confidence_metric < 0.5 && cosTheta < 0.707 )
return cv::Point (int(pt.x+0.5), int(pt.y+0.5));
}
return cv::Point(-1, -1);
}
sortPointsClockWise
定义为:
void sortPointsClockwise(cv::Point a[])
{
cv::Point b[4];
cv::Point ctr = (a[0]+a[1]+a[2]+a[3]);
ctr.x /= 4;
ctr.y /= 4;
b[0] = a[0]-ctr;
b[1] = a[1]-ctr;
b[2] = a[2]-ctr;
b[3] = a[3]-ctr;
for( int i=0; i<4; i++ )
{
if( b[i].x < 0 )
{
if( b[i].y < 0 )
a[0] = b[i]+ctr;
else
a[3] = b[i]+ctr;
}
else
{
if( b[i].y < 0 )
a[1] = b[i]+ctr;
else
a[2] = b[i]+ctr;
}
}
}
getArea
定义为:
float getArea(cv::Point arr[])
{
cv::Point diag1 = arr[0]-arr[2];
cv::Point diag2 = arr[1]-arr[3];
return 0.5*(diag1.cross(diag2));
}
isCloseBy
定义为:
bool isCloseBy( cv::Point p1, cv::Point p2 )
{
int D = 10;
// Checking that X values are within 10, same for Y values.
return ( abs(p1.x-p2.x)<=D && abs(p1.y-p2.y)<=D );
}
最后dist2
:
float dist2( cv::Point p1, cv::Point p2 )
{
return float((p1.x-p2.x)*(p1.x-p2.x) + (p1.y-p2.y)*(p1.y-p2.y));
}
以下是几张测试图片及其结果:
对于非常冗长的帖子感到抱歉,但我希望有人可以提出一种方法,我可以使我的方法从图像中提取卡片更加健壮。可以更好地处理破坏性背景以及不一致的照明。
当卡片放置在具有良好照明的对比背景上时,我的方法几乎占90%的时间。但显然我需要一种更强大的方法。
有人有任何建议吗?
感谢。
尝试dhanushka的解决方案
Mat gray, bw; pyrDown(inImg, inImg);
cvtColor(inImg, gray, CV_RGB2GRAY);
int morph_size = 3;
Mat element = getStructuringElement( MORPH_ELLIPSE, cv::Size( 4*morph_size + 1, 2*morph_size+1 ), cv::Point( morph_size, morph_size ) );
morphologyEx(gray, gray, 2, element);
threshold(gray, bw, 160, 255, CV_THRESH_BINARY);
vector<vector<cv::Point> > contours;
vector<Vec4i> hierarchy;
findContours( bw, contours, hierarchy, CV_RETR_TREE, CV_CHAIN_APPROX_SIMPLE, cv::Point(0, 0) );
int largest_area=0;
int largest_contour_index=0;
cv::Rect bounding_rect;
for( int i = 0; i< contours.size(); i++ )
{
double a=contourArea( contours[i],false); // Find the area of contour
if(a>largest_area){
largest_area=a;
largest_contour_index=i; //Store the index of largest contour
bounding_rect=boundingRect(contours[i]);
}
}
//Scalar color( 255,255,255);
rectangle(inImg, bounding_rect, Scalar(0,255,0),1, 8,0);
Mat biggestRect = inImg(bounding_rect);
Mat card1 = biggestRect.clone();
答案 0 :(得分:8)
图像处理的艺术(在我10多年的经验中)就是:艺术。没有单一的答案,并且总有不止一种方法可以做到这一点。 肯定会在某些情况下失败。
根据我在医学图像中自动检测特征的经验,构建可靠的算法需要很长时间,但事后看来,使用相对简单的算法可以获得最佳结果。但是,获取这个简单算法需要花费大量时间。
为此,一般方法总是一样的:
话说回来,我认为你需要在调整算法时回答这些问题:
这些将使图像的要求和假设得以获得。您可以依赖的假设非常强大:如果您选择正确的算法,它们会使算法快速,稳健和简单。同样,让这些要求和假设成为测试数据库的一部分。
那么我会选择什么?基于您提供的三个图像,我将从以下内容开始:
然后算法看起来像:
cornerHarris
。尝试通过组合每个象限的点来构建每个象限一个角的平行四边形。创建一个健身功能,可以给予更高的分数:
此健身功能可在以后提供大量调整功能。
返回得分最高的平行四边形。
那么为什么使用角点检测而不是霍夫变换来进行线路检测呢?在我看来,霍夫变换(接下来很慢)对背景中的模式非常敏感(这是你在第一张图片中看到的 - 它在背景中检测到比卡更强的线条),它不能除非你使用更大的箱子尺寸会使检测更加恶化,否则处理好一点曲线。
祝你好运!答案 1 :(得分:2)
更一般的方法肯定会像Rutger Nijlunsing在他的回答中所说的那样。但是,在您的情况下,至少对于提供的样本图像,一个非常简单的方法,如形态开放,然后是阈值处理,轮廓处理和凸包,将产生您想要的结果。使用缩小版本的图像进行处理,这样您就不必使用大型内核进行形态学操作。以下是以这种方式处理的图像。
pyrDown(large, rgb0);
pyrDown(rgb0, rgb0);
pyrDown(rgb0, rgb0);
Mat small;
cvtColor(rgb0, small, CV_BGR2GRAY);
Mat morph;
Mat kernel = getStructuringElement(MORPH_ELLIPSE, Size(11, 11));
morphologyEx(small, morph, MORPH_OPEN, kernel);
Mat bw;
threshold(morph, bw, 0, 255.0, CV_THRESH_BINARY | CV_THRESH_OTSU);
Mat bdry;
kernel = getStructuringElement(MORPH_ELLIPSE, Size(3, 3));
erode(bw, bdry, kernel);
subtract(bw, bdry, bdry);
// do contour processing on bdry
这种方法一般不起作用,所以我强烈建议像Rutger建议的那样。