我認爲你可以使用approxPolyDP。我用C++給你程序,我認爲在python中很容易翻譯。想法正在尋找具有接近實際輪廓的四個角的形狀,然後計算四個角(角角)。而輪廓點數大於4時,新輪廓與實際輪廓之間的誤差增大。 當你認爲它不是矩形時,你必須選擇一個閾值角度(90 +/- x°)。 (對不起,我英文不好)
#include "opencv2/opencv.hpp"
#include <iostream>
using namespace cv;
using namespace std;
int main(int argc, char **argv)
{
vector<String> fileName;
fileName.push_back("2nrTo.jpg");
fileName.push_back("G3I4t.jpg");
fileName.push_back("Q4ZtM.jpg");
fileName.push_back("vWgKx.jpg");
for (int i = 0; i < static_cast<int>(fileName.size()); i++)
{
Mat mThresh;
Mat m=imread(fileName[i],CV_LOAD_IMAGE_GRAYSCALE);
Mat mc;
vector<vector<Point> > contours;
vector<Vec4i> hierarchy;
threshold(m,mThresh,80,255,THRESH_BINARY);
findContours(mThresh,contours,hierarchy, cv::RETR_CCOMP, cv::CHAIN_APPROX_NONE, cv::Point(0,0));
cout << "Image " << fileName[i] << "\n";
imshow(fileName[i],m);
mc = Mat::zeros(m.size(),CV_8UC3);
drawContours(mc,contours,0,Scalar(255,0,0),1);
vector<Point> approx;
double d=0;
do
{
d=d+1;
approxPolyDP(contours[0],approx,d,true);
}
while (approx.size()>4);
cout << "#vertices =" <<approx.size() << "\t error max= " <<d<<endl;
if (approx.size() == 4)
{
cout << "Angles\n";
Point2d u(approx[1]-approx[0]),v(approx[2]-approx[1]),w(approx[3]-approx[2]),x(approx[3]-approx[0]);
cout<<acos(u.dot(v)/norm(u)/norm(v))<<"\n";
cout<<acos(v.dot(w)/norm(v)/norm(w))<<"\n";
cout<<acos(w.dot(x)/norm(w)/norm(x))<<"\n";
cout<<acos(x.dot(u)/norm(x)/norm(u))<<"\n";
}
else
cout << "looks like a triangle\n";
contours.push_back(approx);
drawContours(mc,contours,contours.size()-1,Scalar(0,0,255),1);
imshow("Ctr",mc);
waitKey();
}
return 0;
}
與以往的圖像程序放入系統給出的結果:
Image 2nrTo.jpg
#vertices =4 error max= 17
Angles
93.3283
90.2247
90
93.553
Image G3I4t.jpg
#vertices =4 error max= 15
Angles
112.503
46.3837
110.346
48.5412
Image Q4ZtM.jpg
#vertices =4 error max= 6
Angles
88.9191
90.0297
88.9488
90
Image vWgKx.jpg
#vertices =4 error max= 49
Angles
87.0753
117.999
90.3148
114.76
請界定 「正常工作」, 「穩健」 和 「精確」。分析代碼是什麼樣的?你的預期結果是什麼?你現在得到的實際結果是什麼? –
嗨,我已編輯的問題,並提供詳細信息 – user1388142