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我正在處理我的項目，並且遇到問題是我不知道如何僅檢測方形區域的關鍵點之前已經被檢測到。下面是我的演示和到目前爲止，我的代碼將檢測到的外部和內部的方形要點：https://www.youtube.com/watch?feature=player_embedded&v=3U8V6PhMnZ8opencv/C++我如何只檢測在之前檢測到的方形區域內的關鍵點
這是我的代碼找到方：
const int threshold_level = 2;
for (int l = 0; l < threshold_level; l++)
{
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))) > 3000 &&
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);
}
}
}
這是我的代碼繪製正方形和角點：
const Point* p = &squares[i][0];
int n = (int)squares[i].size();
Point p1 = squares[i][0];
Point p2 = squares[i][1];
Point p3 = squares[i][2];
Point p4 = squares[i][3];
cout<<"p1 is "<<p1<<" p2 is "<<p2<<" p3 is "<<p3<<" p4 is "<<p4<<endl;
circle(image, squares[i][0], 3, Scalar(0,0,255), 5, 8, 0);
circle(image, squares[i][1], 3, Scalar(0,255,255), 5, 8, 0);
circle(image, squares[i][2], 3, Scalar(255,0,255), 5, 8, 0);
circle(image, squares[i][3], 3, Scalar(255,255,0), 5, 8, 0);
polylines(image, &p, &n, 1, true, Scalar(0,255,0), 3, CV_AA);
這是我的代碼來檢測關鍵點：
Mat gray_image;
vector<KeyPoint> keyPoints;
cvtColor(image, gray_image, CV_BGR2GRAY);
FastFeatureDetector fast(60);
fast.detect(gray_image,keyPoints);
drawKeypoints(image, keyPoints,image, Scalar::all(255), DrawMatchesFlags::DRAW_OVER_OUTIMG);