C 从边缘检测器输出修剪短线段?
我正在寻找一种算法,从边缘检测器的输出中修剪短线段。如下图(和链接)所示,检测到的一些小边缘不是“长”线。理想情况下,我只希望四边形的4个边在处理后显示出来,但如果有几条虚线,那也没什么大不了的。。。有什么建议吗C 从边缘检测器输出修剪短线段?,c,algorithm,image-processing,computer-vision,C,Algorithm,Image Processing,Computer Vision,我正在寻找一种算法,从边缘检测器的输出中修剪短线段。如下图(和链接)所示,检测到的一些小边缘不是“长”线。理想情况下,我只希望四边形的4个边在处理后显示出来,但如果有几条虚线,那也没什么大不了的。。。有什么建议吗 我怀疑这是否可以通过简单的本地操作实现。查看要保留的矩形-存在多个间隙,因此执行局部操作以删除短线段可能会严重降低所需输出的质量 因此,我会尝试通过闭合间隙、拟合多边形或类似的方式将矩形检测为重要内容,然后在第二步中丢弃剩余的不重要内容。也许你能帮上忙 更新 我刚刚使用了一个内核Ho
我怀疑这是否可以通过简单的本地操作实现。查看要保留的矩形-存在多个间隙,因此执行局部操作以删除短线段可能会严重降低所需输出的质量 因此,我会尝试通过闭合间隙、拟合多边形或类似的方式将矩形检测为重要内容,然后在第二步中丢弃剩余的不重要内容。也许你能帮上忙 更新
我刚刚使用了一个内核Hough变换对你的样本图像进行了处理,得到了四条很好的直线来拟合你的矩形。在找到边缘之前,用一个打开操作或关闭操作(或两者)预处理图像,即先腐蚀后扩张,或先扩张后腐蚀。这将删除较小的对象,但保留较大的对象大致相同
我在网上找了一些例子,我能找到的最好的例子是PDF的第41页。Hough变换可能是一个非常昂贵的操作 在您的案例中,另一个可行的方法是:
可能是找到连接的组件,然后移除小于X像素的组件(根据经验确定),然后沿水平/垂直线进行扩展,以重新连接矩形内的间隙。可以遵循两种主要技术:
使用OpenCV可以很容易地探索这两种方法(前者实际上属于algos的“轮廓分析”类别)。如果有人踩到了这条线,OpenCV 2.x将提供一个名为squares.cpp的示例,它基本上完成了这项任务 我对应用程序做了一点修改,以改进四边形的检测 代码:
#include "highgui.h"
#include "cv.h"
#include <iostream>
#include <math.h>
#include <string.h>
using namespace cv;
using namespace std;
void help()
{
cout <<
"\nA program using pyramid scaling, Canny, contours, contour simpification and\n"
"memory storage (it's got it all folks) to find\n"
"squares in a list of images pic1-6.png\n"
"Returns sequence of squares detected on the image.\n"
"the sequence is stored in the specified memory storage\n"
"Call:\n"
"./squares\n"
"Using OpenCV version %s\n" << CV_VERSION << "\n" << endl;
}
int thresh = 70, N = 2;
const char* wndname = "Square Detection Demonized";
// helper function:
// finds a cosine of angle between vectors
// from pt0->pt1 and from pt0->pt2
double angle( Point pt1, Point pt2, Point pt0 )
{
double dx1 = pt1.x - pt0.x;
double dy1 = pt1.y - pt0.y;
double dx2 = pt2.x - pt0.x;
double dy2 = pt2.y - pt0.y;
return (dx1*dx2 + dy1*dy2)/sqrt((dx1*dx1 + dy1*dy1)*(dx2*dx2 + dy2*dy2) + 1e-10);
}
// returns sequence of squares detected on the image.
// the sequence is stored in the specified memory storage
void findSquares( const Mat& image, vector<vector<Point> >& squares )
{
squares.clear();
Mat pyr, timg, gray0(image.size(), CV_8U), gray;
// karlphillip: dilate the image so this technique can detect the white square,
Mat out(image);
dilate(out, out, Mat(), Point(-1,-1));
// then blur it so that the ocean/sea become one big segment to avoid detecting them as 2 big squares.
medianBlur(out, out, 3);
// down-scale and upscale the image to filter out the noise
pyrDown(out, pyr, Size(out.cols/2, out.rows/2));
pyrUp(pyr, timg, out.size());
vector<vector<Point> > contours;
// find squares only in the first color plane
for( int c = 0; c < 1; c++ ) // was: c < 3
{
int ch[] = {c, 0};
mixChannels(&timg, 1, &gray0, 1, ch, 1);
// try several threshold levels
for( int l = 0; l < N; l++ )
{
// hack: use Canny instead of zero threshold level.
// Canny helps to catch squares with gradient shading
if( l == 0 )
{
// apply Canny. Take the upper threshold from slider
// and set the lower to 0 (which forces edges merging)
Canny(gray0, gray, 0, thresh, 5);
// dilate canny output to remove potential
// holes between edge segments
dilate(gray, gray, Mat(), Point(-1,-1));
}
else
{
// apply threshold if l!=0:
// tgray(x,y) = gray(x,y) < (l+1)*255/N ? 255 : 0
gray = gray0 >= (l+1)*255/N;
}
// find contours and store them all as a list
findContours(gray, contours, CV_RETR_LIST, CV_CHAIN_APPROX_SIMPLE);
vector<Point> approx;
// test each contour
for( size_t i = 0; i < contours.size(); i++ )
{
// approximate contour with accuracy proportional
// to the contour perimeter
approxPolyDP(Mat(contours[i]), approx, arcLength(Mat(contours[i]), true)*0.02, true);
// square contours should have 4 vertices after approximation
// relatively large area (to filter out noisy contours)
// and be convex.
// Note: absolute value of an area is used because
// area may be positive or negative - in accordance with the
// contour orientation
if( approx.size() == 4 &&
fabs(contourArea(Mat(approx))) > 1000 &&
isContourConvex(Mat(approx)) )
{
double maxCosine = 0;
for( int j = 2; j < 5; j++ )
{
// find the maximum cosine of the angle between joint edges
double cosine = fabs(angle(approx[j%4], approx[j-2], approx[j-1]));
maxCosine = MAX(maxCosine, cosine);
}
// if cosines of all angles are small
// (all angles are ~90 degree) then write quandrange
// vertices to resultant sequence
if( maxCosine < 0.3 )
squares.push_back(approx);
}
}
}
}
}
// the function draws all the squares in the image
void drawSquares( Mat& image, const vector<vector<Point> >& squares )
{
for( size_t i = 1; i < squares.size(); i++ )
{
const Point* p = &squares[i][0];
int n = (int)squares[i].size();
polylines(image, &p, &n, 1, true, Scalar(0,255,0), 3, CV_AA);
}
imshow(wndname, image);
}
int main(int argc, char** argv)
{
if (argc < 2)
{
cout << "Usage: ./program <file>" << endl;
return -1;
}
static const char* names[] = { argv[1], 0 };
help();
namedWindow( wndname, 1 );
vector<vector<Point> > squares;
for( int i = 0; names[i] != 0; i++ )
{
Mat image = imread(names[i], 1);
if( image.empty() )
{
cout << "Couldn't load " << names[i] << endl;
continue;
}
findSquares(image, squares);
drawSquares(image, squares);
imwrite("out.jpg", image);
int c = waitKey();
if( (char)c == 27 )
break;
}
return 0;
}
#包括“highgui.h”
#包括“cv.h”
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使用名称空间cv;
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无效帮助()
{
库特1000&&
isContourConvex(材料(近似)))
{
双最大余弦=0;
对于(int j=2;j<5;j++)
{
//求关节边之间角度的最大余弦
双余弦=fabs(角度(约[j%4],约[j-2],约[j-1]);
最大余弦=最大值(最大余弦,余弦);
}
//如果所有角度的余弦都很小
//(所有角度约为90度)然后写入量子线
//结果序列的顶点
如果(最大余弦<0.3)
正方形。推回(大约);
}
}
}
}
}
//该函数用于绘制图像中的所有正方形
空心绘图方块(材质和图像、常量向量和方块)
{
对于(size_t i=1;i cout下面是一个简单的形态过滤解决方案,如下@Tom10行:
matlab中的解决方案:
se1 = strel('line',5,180); % linear horizontal structuring element
se2 = strel('line',5,90); % linear vertical structuring element
I = rgb2gray(imread('test.jpg'))>80; % threshold (since i had a grayscale version of the image)
Idil = imdilate(imdilate(I,se1),se2); % dilate contours so that they connect
Idil_area = bwareaopen(Idil,1200); % area filter them to remove the small components
其基本思想是将水平轮廓连接起来,形成一个大的组件,然后通过区域开放过滤器进行过滤,以获得矩形
结果:
+1用于建议Hough变换。只需在变换空间中找到四个最强的峰值,这就是你的四边形。你只寻找矩形吗?看看示例图片。矩形的边缘轮廓只有1像素薄!如果你先腐蚀,你将完全失去矩形和小边。如果你“首先放大,你可以缩小大矩形中的一些间隙,但这是另一个问题,并不能真正帮助你去除小边缘。@Levy-不,正如我在回答中明确指出的,在找到边缘之前,应该先关闭图像。当然,这不应该应用于边缘。”(但要感谢计算边缘的对象)。@tom10-谢谢你的建议,我换了