OpenCV C ++ / Obj-C：检测一张纸/方形检测 [英] OpenCV C++/Obj-C: Detecting a sheet of paper / Square Detection
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问题描述
我在我的测试应用程序中成功地实现了OpenCV平方检测的例子，但是现在需要过滤输出，因为它的安静混乱 - 或者我的代码错了吗？
我对纸张的四个角点感兴趣，以减少偏斜（例如
原始图片：
代码：
双角度（cv :: Point pt1，cv :: Point pt2，cv :: 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-
}
- （std :: vector< std :: vector< cv :: Point>>）findSquaresInImage：（cv :: Mat）_image
{
std :: vector< std :: vector< cv :: Point> >正方形;
cv :: Mat pyr，timg，gray0（_image.size（），CV_8U），gray;
int thresh = 50，N = 11;
cv :: pyrDown（_image，pyr，cv :: Size（_image.cols / 2，_image.rows / 2））;
cv :: pyrUp（pyr，timg，_image.size（））;
std :: vector< std :: vector< cv :: Point> >轮廓;
for（int c = 0; c <3; c ++）{
int ch [] = {c，0};
mixChannels（& timg，1，& gray0，1，ch，1）;
for（int l = 0; l
cv :: Canny（gray0，gray，0，thresh，5） ;
cv :: dilate（gray，gray，cv :: Mat（），cv :: Point（-1，-1））;
}
else {
gray = gray0> =（1 + 1）* 255 / N;
}
cv :: findContours（gray，contour，CV_RETR_LIST，CV_CHAIN_APPROX_SIMPLE）;
std :: vector< cv :: Point>约;
for（size_t i = 0; i
cv :: approxPolyDP（cv :: Mat（contours [i]），approx，arcLength （cv :: Mat（contours [i]），true）* 0.02，true）;
if（approx.size（）== 4&& fabs（contourArea（cv :: Mat（approx）））> 1000& cv :: isContourConvex（cv :: Mat ））{
double maxCosine = 0;
for（int j = 2; j <5; j ++）
{
double cosine = fabs（angle（approximately [j％4]，approximately [j-2 ]，约[j-1]））;
maxCosine = MAX（maxCosine，cosine）;
}
if（maxCosine <0.3）{
squares.push_back（approx）;
}
}
}
}
}
return squares;
}
EDIT 17/08/2012：
要在图片上绘制检测到的正方形，请使用以下代码：
cv :: Mat debugSquares（std :: vector< std :: vector< cv :: Point>> square，cv :: Mat image）
{
for = 0; i
cv :: drawContours（image，squares，i，cv :: Scalar（255,0,0），1 ，8，std :: vector< cv :: Vec4i（），0，cv :: Point（））;
//绘制边界rect
cv :: Rect rect = boundingRect（cv :: Mat（squares [i]））;
cv :: rectangle（image，rect.tl（），rect.br（），cv :: Scalar（0,255,0），2，8，0）;
//绘制旋转矩形
cv :: RotatedRect minRect = minAreaRect（cv :: Mat（squares [i]））;
cv :: Point2f rect_points [4];
minRect.points（rect_points）;
for（int j = 0; j <4; j ++）{
cv :: line（image，rect_points [j]，rect_points [（j + 1）％4]，cv :: Scalar （0,0,255），1,8）; // blue
}
}
return image;
}
解决方案
Stackoverflow，由于我无法找到相关的实现，我决定接受挑战。
我对OpenCV中的正方形演示和生成的C ++代码做了一些修改能够检测图像中的一张纸：
void find_squares（Mat& image，vector< vector< Point> ;& square）
{
// blur会增强边缘检测
Mat blur（image）;
medianBlur（image，blurred，9）;
Mat gray0（blurred.size（），CV_8U），gray;
vector< vector< Point> >轮廓;
//在图像的每个颜色平面中找到正方形
for（int c = 0; c <3; c ++）
{
int ch [] = {c，0};
mixChannels（& blurred，1，& gray0，1，ch，1）;
//尝试几个阈值级别
const int threshold_level = 2;
for（int l = 0; l< threshold_level; l ++）
{
//使用Canny而不是零阈值！
// Canny有助于捕获带渐变阴影的正方形
if（l == 0）
{
Canny（gray0，gray，10，20，3） //
// Dilate有助于消除边缘段之间的潜在空洞
dilate（gray，gray，Mat（），Point（-1，-1）
}
else
{
gray = gray0> =（1 + 1）* 255 / threshold_level;
}
//查找轮廓并将它们存储在列表中
findContours（gray，contour，CV_RETR_LIST，CV_CHAIN_APPROX_SIMPLE）;
//测试轮廓
矢量< Point>约;
for（size_t i = 0; i
//精确比例的近似轮廓
//轮廓周长
approxPolyDP（Mat（contoururs [i]），approx，arcLength（Mat（contoururs [i]），true）* 0.02，true）;
//注意：使用区域的绝对值，因为
//区域可以为正或负 - 根据
//轮廓方向
if （approx.size（）== 4&&
fabs（contourArea（Mat（approx）））> 1000&&
isContourConvex（Mat（approx）））
{
double maxCosine = 0;
for（int j = 2; j <5; j ++）
{
double cosine = fabs（angle（approximately [j％4]，approximately [j-2 ]，约[j-1]））;
maxCosine = MAX（maxCosine，cosine）;
}
if（maxCosine <0.3）
squares.push_back（approx）;
}
}
}
}
}
$ b b
在执行此过程之后，纸张将是向量<向量<点>中的最大平方。 > ：
我让你写函数找到最大的广场。 ;）
I successfully implemented the OpenCV square-detection example in my test application, but now need to filter the output, because it's quiet messy - or is my code wrong?
I'm interested in the four corner points of the paper for skew reduction (like that) and further processing …
Input & Output:
Original image:
Code:
double angle( cv::Point pt1, cv::Point pt2, cv::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);
}
- (std::vector<std::vector<cv::Point> >)findSquaresInImage:(cv::Mat)_image
{
std::vector<std::vector<cv::Point> > squares;
cv::Mat pyr, timg, gray0(_image.size(), CV_8U), gray;
int thresh = 50, N = 11;
cv::pyrDown(_image, pyr, cv::Size(_image.cols/2, _image.rows/2));
cv::pyrUp(pyr, timg, _image.size());
std::vector<std::vector<cv::Point> > contours;
for( int c = 0; c < 3; c++ ) {
int ch[] = {c, 0};
mixChannels(&timg, 1, &gray0, 1, ch, 1);
for( int l = 0; l < N; l++ ) {
if( l == 0 ) {
cv::Canny(gray0, gray, 0, thresh, 5);
cv::dilate(gray, gray, cv::Mat(), cv::Point(-1,-1));
}
else {
gray = gray0 >= (l+1)*255/N;
}
cv::findContours(gray, contours, CV_RETR_LIST, CV_CHAIN_APPROX_SIMPLE);
std::vector<cv::Point> approx;
for( size_t i = 0; i < contours.size(); i++ )
{
cv::approxPolyDP(cv::Mat(contours[i]), approx, arcLength(cv::Mat(contours[i]), true)*0.02, true);
if( approx.size() == 4 && fabs(contourArea(cv::Mat(approx))) > 1000 && cv::isContourConvex(cv::Mat(approx))) {
double maxCosine = 0;
for( int j = 2; j < 5; j++ )
{
double cosine = fabs(angle(approx[j%4], approx[j-2], approx[j-1]));
maxCosine = MAX(maxCosine, cosine);
}
if( maxCosine < 0.3 ) {
squares.push_back(approx);
}
}
}
}
}
return squares;
}
EDIT 17/08/2012:
To draw the detected squares on the image use this code:
cv::Mat debugSquares( std::vector<std::vector<cv::Point> > squares, cv::Mat image )
{
for ( int i = 0; i< squares.size(); i++ ) {
// draw contour
cv::drawContours(image, squares, i, cv::Scalar(255,0,0), 1, 8, std::vector<cv::Vec4i>(), 0, cv::Point());
// draw bounding rect
cv::Rect rect = boundingRect(cv::Mat(squares[i]));
cv::rectangle(image, rect.tl(), rect.br(), cv::Scalar(0,255,0), 2, 8, 0);
// draw rotated rect
cv::RotatedRect minRect = minAreaRect(cv::Mat(squares[i]));
cv::Point2f rect_points[4];
minRect.points( rect_points );
for ( int j = 0; j < 4; j++ ) {
cv::line( image, rect_points[j], rect_points[(j+1)%4], cv::Scalar(0,0,255), 1, 8 ); // blue
}
}
return image;
}
解决方案
This is a recurring subject in Stackoverflow and since I was unable to find a relevant implementation I decided to accept the challenge.
I made some modifications to the squares demo present in OpenCV and the resulting C++ code below is able to detect a sheet of paper in the image:
void find_squares(Mat& image, vector<vector<Point> >& squares)
{
// blur will enhance edge detection
Mat blurred(image);
medianBlur(image, blurred, 9);
Mat gray0(blurred.size(), CV_8U), gray;
vector<vector<Point> > contours;
// find squares in every color plane of the image
for (int c = 0; c < 3; c++)
{
int ch[] = {c, 0};
mixChannels(&blurred, 1, &gray0, 1, ch, 1);
// try several threshold levels
const int threshold_level = 2;
for (int l = 0; l < threshold_level; l++)
{
// Use Canny instead of zero threshold level!
// Canny helps to catch squares with gradient shading
if (l == 0)
{
Canny(gray0, gray, 10, 20, 3); //
// Dilate helps to remove potential holes between edge segments
dilate(gray, gray, Mat(), Point(-1,-1));
}
else
{
gray = gray0 >= (l+1) * 255 / threshold_level;
}
// Find contours and store them in a list
findContours(gray, contours, CV_RETR_LIST, CV_CHAIN_APPROX_SIMPLE);
// Test contours
vector<Point> approx;
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);
// 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++)
{
double cosine = fabs(angle(approx[j%4], approx[j-2], approx[j-1]));
maxCosine = MAX(maxCosine, cosine);
}
if (maxCosine < 0.3)
squares.push_back(approx);
}
}
}
}
}
After this procedure is executed, the sheet of paper will be the largest square in vector<vector<Point> >:
I'm letting you write the function to find the largest square. ;)
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