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緑の矢印（pic 2）を線形ベースに整列/平行化する方法。 OpenCVの基本的なPCA解析を適用した結果、私は結果に満足していましたが、基点と一致する中心点の位置と角度を操作する方法を知りたいと思います。 提供されている写真では、緑色の線が数度ずれているのが見えますが、「ベース」上にあるか、またはそれと平行していてほしいです。オブジェクトの直線基底のx、yから角度を取得
ソース画像：
私は現在、何を得る：
#include <iostream>
#include <opencv2/opencv.hpp>
using namespace std;
using namespace cv;
// Function declarations
void drawAxis(Mat&, Point, Point, Scalar, const float);
double getOrientation(const vector<Point> &, Mat&);
void drawAxis(Mat& img, Point p, Point q, Scalar colour, const float scale = 0.2)
{
double angle;
double hypotenuse;
angle = atan2((double) p.y - q.y, (double) p.x - q.x); // angle in radians
hypotenuse = sqrt((double) (p.y - q.y) * (p.y - q.y) + (p.x - q.x) * (p.x - q.x));
double degrees = angle * 180/CV_PI; // convert radians to degrees (0-180 range)
cout << "Degrees: " << abs(degrees - 180) << endl; // angle in 0-360 degrees range
// Here we lengthen the arrow by a factor of scale
q.x = (int) (p.x - scale * hypotenuse * cos(angle));
q.y = (int) (p.y - scale * hypotenuse * sin(angle));
line(img, p, q, colour, 1, CV_AA);
// create the arrow hooks
p.x = (int) (q.x + 9 * cos(angle + CV_PI/4));
p.y = (int) (q.y + 9 * sin(angle + CV_PI/4));
line(img, p, q, colour, 1, CV_AA);
p.x = (int) (q.x + 9 * cos(angle - CV_PI/4));
p.y = (int) (q.y + 9 * sin(angle - CV_PI/4));
line(img, p, q, colour, 1, CV_AA);
}
double getOrientation(const vector<Point> &pts, Mat &img)
{
//Construct a buffer used by the pca analysis
int sz = static_cast<int>(pts.size());
Mat data_pts = Mat(sz, 2, CV_64FC1);
for (int i = 0; i < data_pts.rows; ++i)
{
data_pts.at<double>(i, 0) = pts[i].x;
data_pts.at<double>(i, 1) = pts[i].y;
}
//Perform PCA analysis
PCA pca_analysis(data_pts, Mat(), CV_PCA_DATA_AS_ROW);
//Store the center of the object
Point cntr = Point(static_cast<int>(pca_analysis.mean.at<double>(0, 0)),
static_cast<int>(pca_analysis.mean.at<double>(0, 1)));
//Store the eigenvalues and eigenvectors
vector<Point2d> eigen_vecs(2);
vector<double> eigen_val(2);
for (int i = 0; i < 2; ++i)
{
eigen_vecs[i] = Point2d(pca_analysis.eigenvectors.at<double>(i, 0),
pca_analysis.eigenvectors.at<double>(i, 1));
eigen_val[i] = pca_analysis.eigenvalues.at<double>(0, i);
}
// Draw the principal components
circle(img, cntr, 3, Scalar(255, 0, 255), 2);
Point p1 = cntr + 0.02 * Point(static_cast<int>(eigen_vecs[0].x * eigen_val[0]), static_cast<int>(eigen_vecs[0].y * eigen_val[0]));
Point p2 = cntr - 0.02 * Point(static_cast<int>(eigen_vecs[1].x * eigen_val[1]), static_cast<int>(eigen_vecs[1].y * eigen_val[1]));
drawAxis(img, cntr, p1, Scalar(0, 255, 0), 1);
drawAxis(img, cntr, p2, Scalar(255, 255, 0), 5);
double angle = atan2(eigen_vecs[0].y, eigen_vecs[0].x); // orientation in radians
return angle;
}
int main(int, char** argv)
{
// Load image
Mat src = imread("/path/image.jpg");
// Check if image is loaded successfully
if(!src.data || src.empty())
{
cout << "Problem loading image!!!" << endl;
return EXIT_FAILURE;
}
imshow("src", src);
// Convert image to grayscale
Mat gray;
cvtColor(src, gray, COLOR_BGR2GRAY);
// Convert image to binary
Mat bw;
threshold(gray, bw, 50, 255, CV_THRESH_BINARY | CV_THRESH_OTSU);
// Find all the contours in the thresholded image
vector<Vec4i> hierarchy;
vector<vector<Point> > contours;
findContours(bw, contours, hierarchy, CV_RETR_LIST, CV_CHAIN_APPROX_NONE);
for (size_t i = 0; i < contours.size(); ++i)
{
// Calculate the area of each contour
double area = contourArea(contours[i]);
// Ignore contours that are too small or too large
//if (area < 1e2 || 1e5 < area) continue;
if (area > 1e6) continue;
cout << "Area: " << area << endl;
// Draw each contour only for visualisation purposes
drawContours(src, contours, static_cast<int>(i), Scalar(0, 0, 255), 2, 8, hierarchy, 0);
// Find the orientation of each shape
getOrientation(contours[i], src);
}
imshow("output", src);
waitKey(0);
return 0;
}