Question

我的想法是绘制所有垂直线，这些垂直线是通过计算之前的Canny而创建的，从交叉点到对角线到另一个交叉点（此点也来自视线和对角线之间的交点）。作为参考，这里应该绘制红色垂直（Hough）线：

直到我用这种实现方式检测所有垂直线：

int main(int argc, char *argv[]) {
  std::vector<cv::Point> diagonalLine = DiagonalLines::diagonalLines(src);

  Mat wdst, cwdst, contRegion;
  vector<Vec4i> vericalLines;
  double maxLineGap = 200.0;
  double threshold = 100;

  std::vector<cv::Vec4i> elemLinesCur;                                              

  cv::Scalar mu, sigma;
  meanStdDev(src, mu, sigma);

  Canny(src, wdst, mu.val[0] - sigma.val[0], mu.val[0] + sigma.val[0], 3, false);
  cvtColor(wdst, cwdst, CV_GRAY2BGR);

  HoughLinesP(wdst, vericalLines, 1, CV_PI / 2, threshold, 50, 200);

  cv::Vec4i current, previous;
  cv::Point pt1, pt2, ppt1, ppt2;

  for (size_t i = 1; i < vericalLines.size(); i++) {
    current = vericalLines[i];
    pt1 = cv::Point(current[0], current[1]);
    pt2 = cv::Point(current[2], current[3]);

    previous = vericalLines[i - 1];
    ppt1 = cv::Point(previous[0], previous[1]);
    ppt2 = cv::Point(previous[2], previous[3]);

    if (diagonalLine[i - 1].y > pt2.y && diagonalLine[i].y < pt1.y) {
        std::cout << "Intersection: " << pt2.x << "\n";
    }

    double distanceBetweenPointsX = abs(pt1.x - ppt1.x)*sqrt(2);

    if (distanceBetweenPointsX >= 12) {
        elemLinesCur.push_back(current);

        double angle = atan2(ppt2.y - ppt1.y, ppt2.x - ppt1.x) * 180.0 / CV_PI;     ///draw only vertical lines (90 degree)
        if (angle) {
            line(cwdst, pt1, pt2, cv::Scalar(0, 0, 255), 2, CV_AA);
        }
//do some stuff
}

...这里有一个只检测对角线的方法（看起来与上面的相似）：

std::vector<cv::Point> diagonalLines(cv::Mat src) {
std::vector<cv::Point> hitPoint;

Scalar mu, sigma;
meanStdDev(src, mu, sigma);

Canny(src, ddst, mu.val[0] - sigma.val[0], mu.val[0] + sigma.val[0], 3, false);

cvtColor(ddst, cddst, CV_GRAY2BGR);
HoughLinesP(ddst, vertlines, 1, CV_PI / 180, 100, 50, 10);

cv::Point pt1, pt2;
for (size_t i = 1; i < vertlines.size(); i++) {
    cv::Vec4i current = vertlines[i];
    pt1 = cv::Point(current[0], current[1]);
    pt2 = cv::Point(current[2], current[3]);

    double angle = atan2(pt2.y - pt1.y, pt2.x - pt1.x) * 180.0 / CV_PI;
    if (angle != -90 && angle != 90) {
        //line(cddst, pt1, pt2, Scalar(0, 0, 255), 2, CV_AA);
        hitPoint.push_back(pt1);
        hitPoint.push_back(pt2);
    }
}
return hitPoint;
}

我所知道的： 我应该计算所有这些交叉点，是的，我也在if (diagonalLine[i - 1].y > pt2.y && diagonalLine[i].y < pt1.y)中尝试过，但我没有得到进一步的步骤。有人可以帮助我吗？提前谢谢！

Answer 1

OpenCV函数line()接受端点作为参数，因此您需要做的就是计算交叉点并将这些交叉点用于垂直线的端点。您可以直接根据HoughLinesP() using determinants的结果计算交叉点。

在Python中，计算交叉点的函数可能类似于

def find_intersection(line1, line2):
    # extract points
    x1, y1 = line1[0]
    x2, y2 = line1[1]
    x3, y3 = line2[0]
    x4, y4 = line2[1]
    # compute determinant
    Px = ((x1*y2 - y1*x2)*(x3-x4) - (x1-x2)*(x3*y4 - y3*x4)) / 
        ((x1-x2)*(y3-y4) - (y1-y2)*(x3-x4))
    Py = ((x1*y2 - y1*x2)*(y3-y4) - (y1-y2)*(x3*y4 - y3*x4)) /
        ((x1-x2)*(y3-y4) - (y1-y2)*(x3-x4))
    return (int(Px), int(Py))

让我们展示一下如何使用它。假设您的图片看起来像这样：

# draw image and lines
img = np.ones((500, 500, 3)) * 255
diag1 = [(0, 0), (499, 100)]
diag2 = [(0, 499), (499, 399)]
vert1 = [(100, 0), (100, 499)]
vert2 = [(400, 0), (400, 499)]
cv2.line(img, diag1[0], diag1[1], color=[0, 0, 255])
cv2.line(img, diag2[0], diag2[1], color=[0, 0, 255])
cv2.line(img, vert1[0], vert1[1], color=[0, 255, 0])
cv2.line(img, vert2[0], vert2[1], color=[0, 255, 0])

要在十字路口切断它们，只需使用该功能查找这些点，并仅在每条对角线的交叉点绘制垂直线。

# get intersection points
vert1_intersect = [find_intersection(diag1, vert1), find_intersection(diag2, vert1)]
vert2_intersect = [find_intersection(diag1, vert2), find_intersection(diag2, vert2)]

# draw vertical lines from intersection points
img = np.ones((500, 500, 3)) * 255
diag1 = [(0, 0), (499, 100)]
diag2 = [(0, 499), (499, 399)]
vert1 = [(100, 0), (100, 499)]
vert2 = [(400, 0), (400, 499)]
cv2.line(img, diag1[0], diag1[1], color=[0, 0, 255])
cv2.line(img, diag2[0], diag2[1], color=[0, 0, 255])
cv2.line(img, vert1_intersect[0], vert1_intersect[1], color=[0, 255, 0])
cv2.line(img, vert2_intersect[0], vert2_intersect[1], color=[0, 255, 0])

绘制垂直HoughLines直到某些交叉点

1 个答案: