I would like to measure the horizontal lengths of multiple ROI. I tried Feret's diameter, but it only gives the longest distance between any two points along the selection boundary. I tried bounding rectangle, but I suppose the rectangles are tilted to obtain the minimum bounding rectangle.
Does anyone have another idea? Because clearly, the selection boundaries fit nicely to the ROI - so how could I extract that information, i.e. the xy-coordinates of the fits? Thanks in advance
PS: I did not write ROIs because 'Region of Interests' makes no sense
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I have the following image. I want to fit ellipse into each black region. Then I want to measure the major and minor axis length. I would be grateful if any one can give me any idea. I know how to fit an ellipse over the image object.
Given two rectangles, and we know the position of four corners, widths, heights, angles.
How to compute the overlapping ratio of these two rectangles?
Can you please help me out?
A convenient way is by the Sutherland-Hodgman polygon clipping algorithm. It works by clipping one of the polygons with the four supporting lines (half-planes) of the other. In the end you get the intersection polygon (at worst an octagon) and find its area by the polygon area formula.
You'll make clipping easier by counter-rotating the polygons around the origin so that one of them becomes axis parallel. This won't change the area.
Note that this approach generalizes easily to two general convex polygons, taking O(N.M) operations. G.T. Toussaint, using the Rotating Caliper principle, reduced the workload to O(N+M), and B. Chazelle & D. P. Dobkin showed that a nonempty intersection can be detected in O(Log(N+M)) operations. This shows that there is probably a little room for improvement for the S-H clipping approach, even though N=M=4 is a tiny problem.
Use rotatedRectangleIntersection function to get contour and use contourArea function to get area and find the ratios
https://docs.opencv.org/3.0-beta/modules/imgproc/doc/structural_analysis_and_shape_descriptors.html#rotatedrectangleintersection
Lets say you have rectangle A and B the you can use the operation:
intersection_area = (A & B).area();
from this area you can calculate de respective ratio towards one of the rectangles. there will be harder more dynamic ways to do this as well.
What is Distance Transform?What is the theory behind it?if I have 2 similar images but in different positions, how does distance transform help in overlapping them?The results that distance transform function produce are like divided in the middle-is it to find the center of one image so that the other is overlapped just half way?I have looked into the documentation of opencv but it's still not clear.
Look at the picture below (you may want to increase you monitor brightness to see it better). The pictures shows the distance from the red contour depicted with pixel intensities, so in the middle of the image where the distance is maximum the intensities are highest. This is a manifestation of the distance transform. Here is an immediate application - a green shape is a so-called active contour or snake that moves according to the gradient of distances from the contour (and also follows some other constraints) curls around the red outline. Thus one application of distance transform is shape processing.
Another application is text recognition - one of the powerful cues for text is a stable width of a stroke. The distance transform run on segmented text can confirm this. A corresponding method is called stroke width transform (SWT)
As for aligning two rotated shapes, I am not sure how you can use DT. You can find a center of a shape to rotate the shape but you can also rotate it about any point as well. The difference will be just in translation which is irrelevant if you run matchTemplate to match them in correct orientation.
Perhaps if you upload your images it will be more clear what to do. In general you can match them as a whole or by features (which is more robust to various deformations or perspective distortions) or even using outlines/silhouettes if they there are only a few features. Finally you can figure out the orientation of your object (if it has a dominant orientation) by running PCA or fitting an ellipse (as rotated rectangle).
cv::RotatedRect rect = cv::fitEllipse(points2D);
float angle_to_rotate = rect.angle;
The distance transform is an operation that works on a single binary image that fundamentally seeks to measure a value from every empty point (zero pixel) to the nearest boundary point (non-zero pixel).
An example is provided here and here.
The measurement can be based on various definitions, calculated discretely or precisely: e.g. Euclidean, Manhattan, or Chessboard. Indeed, the parameters in the OpenCV implementation allow some of these, and control their accuracy via the mask size.
The function can return the output measurement image (floating point) - as well as a labelled connected components image (a Voronoi diagram). There is an example of it in operation here.
I see from another question you have asked recently you are looking to register two images together. I don't think the distance transform is really what you are looking for here. If you are looking to align a set of points I would instead suggest you look at techniques like Procrustes, Iterative Closest Point, or Ransac.
I'm going to find the most look-like rectangles among shapes. The first image is the original image with shapes which possibly be rectangles but they are not. The green rectangles in the second image is what I want. So is there a way to do this with opencv? I've tried hough lines but the result's not good
The source image:
And what I want is to find out the most look-like rectangle among these shapes, like the rectangles in green.
What I want:
A very simple approach is, after you have a rectangle bounding box around your shape, count the percentage of pixels inside the box which are white.
The higher the percentage of white pixels, the closest to a rectangle it is.
To get the bounding boxes you should take a look at either findContours from opencv, or some Blob extracting algorithm, you will find plenty of questions regarding those.
Edit:
Maybe you should first get the Minimum bounding rectangles of the shapes and then do this kind of heuristic:
Shrink the rectangle dimensions until the white-pixel percentage inside the rectangle reaches some threshold defined by you (like 90% of white pixels inside the rectangle).
To get the Minimum bounding rectangle (the smallest rectangle which contains the whole shape), you might check this tutorial:
http://docs.opencv.org/doc/tutorials/imgproc/shapedescriptors/bounding_rects_circles/bounding_rects_circles.html
One thing that might also help is doing the difference of sizes from the minimum bounding rectangle and the maximum inner rectangle (the biggest rectangle you can fit inside the white shape). The less difference there is between those rectangle's properties (width, height, area, center coordinates) the closest is the shape to a rectangle.
I have a digital image, and I want to make some calculation based on distances on it. So I need to get the Milimeter/Pixel proportion. What I'm doing right now, is to mark two points wich I know the real world distance, to calculate the Euclidian distance between them, and than obtain the proportion.
The question is, Only with two points can I make the correct Milimeter/Pixel's proportion, or do I need to use 4 points, 2 for the X-Axis and 2 for Y-axis?
If your image is of a flat surface and the camera direction is perpendicular to that surface, then your scale factor should be the same in both directions.
If your image is of a flat surface, but it is tilted relative to the camera, then marking out a rectangle of known proportions on that surface would allow you to compute a perspective transform. (See for example this question)
If your image is of a 3D scene, then of course there is no way in general to convert pixels to distances.
If you know the distance between the points A and B measured on the picture(say in inch) and you also know the number of pixels between the points, you can easily calculate the pixels/inch ratio by dividing <pixels>/<inches>.
I suggest to take the points on the picture such that the line which intersects them is either horizontal either vertical such that calculations do not have errors taking into account the pixels have a rectangular form.