More point-finding related questions...
If I have an image with three squares of points in it, each with four corner points and other, interior points...
What is the best way to identify each as a separate square so I can process it individually as a ROI?
They may be tilted, so the sides in 2d might not look equal, but each will have the same number of points, and each can be contained in a 4-sided polygon.
Thanks!
you can do the following
Pass the centers to Kmeans Kmean
clustring or this
link
Kmean will find the clusters group of points
For every cluster You can use one of the following function todetect the rectangle(boundingRect, approxPolyDP,rotatedRect)
Thanks! I have the contours, the moments and the centroids of each point.
Stuck on kmeans... I just pass it a vector containing (x,y) of each point?
This is what I have:
findContours(threshold_output, contours, hierarchy, CV_RETR_TREE, CV_CHAIN_APPROX_SIMPLE, Point(0, 0));
/// Get the moments
vector<Moments> mu(contours.size());
for (int i = 0; i < contours.size(); i++)
{
mu[i] = moments(contours[i], false);
}
/// Get the mass centers:
vector<Point2f> centroids(contours.size());
for (int i = 0; i < contours.size(); i++)
{
centroids[i] = Point2f(mu[i].m10 / mu[i].m00, mu[i].m01 / mu[i].m00);
}
So the points are all in the centroids vector.
you can use boundingRect function the first parameter of this function is the list of points in the squar
Example:
vector<Point> contour;
contour.push_back(Point2f(0, 0));
contour.push_back(Point2f(10, 0));
contour.push_back(Point2f(10, 10));
contour.push_back(Point2f(5, 4));
Rect ResultedboundingRect = boundingRect(contour);
That gives me a bounding box for each contour though, right? What i need is to loop through a list of points, get the nearby points, and find out if a four-sided polygon can be made from them.
Picture the '5' side of a dice. Say three images of that in various places in an image. I need to identify each as a separate ROI.
thanks!
Asked: 2015-04-09 03:18:33 -0600
Seen: 1,491 times
Last updated: Apr 09 '15
