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圖像中檢測最大的矩形我問上一個問題here及繼起我建立了下面的程序，我認爲會檢測大矩形答案的建議，但沒有檢測到矩形的。它確實在這個image上工作。使用OpenCV的
我想該解決方案不僅這個形象，但這種不同的圖像上工作。下面的代碼的主要部分是從不同的答案在SO
我的完整方案：
#include <cv.h>
#include <highgui.h>
using namespace cv;
using namespace std;
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);
}
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);
}
}
}
}
}
void find_largest_square(const vector<vector <Point> >& squares, vector<Point>& biggest_square) {
if (!squares.size()) {
return;
}
int max_width = 0;
int max_height = 0;
int max_square_idx = 0;
const int n_points = 4;
for (size_t i = 0; i < squares.size(); i++) {
Rect rectangle = boundingRect(Mat(squares[i]));
if ((rectangle.width >= max_width) && (rectangle.height >= max_height)) {
max_width = rectangle.width;
max_height = rectangle.height;
max_square_idx = i;
}
}
biggest_square = squares[max_square_idx];
}
int main(int argc, char* argv[])
{
Mat img = imread(argv[1]);
if (img.empty())
{
cout << "!!! imread() failed to open target image" << endl;
return -1;
}
vector< vector< Point> > squares;
find_squares(img, squares);
vector<Point> largest_square;
find_largest_square(squares, largest_square);
for (int i = 0; i < 4; ++i) {
line(img, largest_square[i], largest_square[(i+1)%4], Scalar(0, 255, 0), 1, CV_AA);
}
imwrite("squares.png", img);
imshow("squares", img);
waitKey(0);
return 0;
}
小心，你問太多的跟進問題，用自己的代碼非常少。你會很快聽到我們問*你迄今爲止試過了什麼。 – karlphillip
對不起，我誤解了這個問題。我稍後可能會回答。晚餐時間。 – karlphillip
:)沒問題。感謝您的指導。 – birdy