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Selecting the 4-neighbours of a pixel [closed]
(1 answer)
Closed 9 years ago.
I know that there is a function called nlfilter in matlab. What I'm trying to find is the 4-neighbours of a pixel. Does that means a 2x2 window? Can we do that using nlfilter?
Thanks.
I think you might find this easier to comprehend if you think of it in terms of blocks rather than in terms of neighbors. So a 2x2 neighbor is actually just a 2x2 block.
If you are talking about a center pixel relative to a north, south, east, west pixel, then you would want to use a 3x3 block. Unfortunately, that block would also include northeast, northwest, southeast, southwest neighbors.
Here is an example of sliding neighborhood operations in Matlab using nlfilter.
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In the attached Google charts Pie chart the labels fit well inside the segments. Determining the length of a bit of text in HTML5 canvas is easy enough - but how do you determine whether the label will fit into a particular segment (using trigonometry) ? As you can see on the image, two of the segments don't have labels inside the segment.
EDIT: Here's an example of what I have at the moment: https://www.rgraph.net/tests/canvas.pie/in-pie-labels.html
As you see the labels for the small segments overlap. What I'm after is a way to calculate whether there's enough space for the labels at the point where they're going to be rendered. If not, I can just not draw the label like in the example image above.
Could chord size be useful to do this?
Here's the forumulae for the chord size that I found via Google:
"Chord length using trigonometry = 2 × r × sin(θ/2); where 'r' is the radius of the circle and 'θ' is the angle subtended at the center by the chord."
I sorted it (in about one hour) after 3 days of trying to calculate it with trig by using the built-in context.isPointInPath() function...
Draw the text (transparent color) to get the coordinates (x/y/w/h) of it. You might be able to get away with measuring it to get the width and height.
Draw the segment in a transparent color and do not stroke or fill it. Also, do not close the path.
Test each corner of the text rectangle (formed the x/y/w/h that you got above) using the context.isPointInPath() function. If the function returns true for each corner of the rectangle formed by the coordinates of the text, then the text will fit into the segment.
I have gone through a couple of YOLO tutorials but I am finding it some what hard to figure if the Anchor boxes for each cell the image is to be divided into is predetermined. In one of the guides I went through, The image was divided into 13x13 cells and it stated each cell predicts 5 anchor boxes(bigger than it, ok here's my first problem because it also says it would first detect what object is present in the small cell before the prediction of the boxes).
How can the small cell predict anchor boxes for an object bigger than it. Also it's said that each cell classifies before predicting its anchor boxes how can the small cell classify the right object in it without querying neighbouring cells if only a small part of the object falls within the cell
E.g. say one of the 13 cells contains only the white pocket part of a man wearing a T-shirt how can that cell classify correctly that a man is present without being linked to its neighbouring cells? with a normal CNN when trying to localize a single object I know the bounding box prediction relates to the whole image so at least I can say the network has an idea of what's going on everywhere on the image before deciding where the box should be.
PS: What I currently think of how the YOLO works is basically each cell is assigned predetermined anchor boxes with a classifier at each end before the boxes with the highest scores for each class is then selected but I am sure it doesn't add up somewhere.
UPDATE: Made a mistake with this question, it should have been about how regular bounding boxes were decided rather than anchor/prior boxes. So I am marking #craq's answer as correct because that's how anchor boxes are decided according to the YOLO v2 paper
I think there are two questions here. Firstly, the one in the title, asking where the anchors come from. Secondly, how anchors are assigned to objects. I'll try to answer both.
Anchors are determined by a k-means procedure, looking at all the bounding boxes in your dataset. If you're looking at vehicles, the ones you see from the side will have an aspect ratio of about 2:1 (width = 2*height). The ones viewed from in front will be roughly square, 1:1. If your dataset includes people, the aspect ratio might be 1:3. Foreground objects will be large, background objects will be small. The k-means routine will figure out a selection of anchors that represent your dataset. k=5 for yolov3, but there are different numbers of anchors for each YOLO version.
It's useful to have anchors that represent your dataset, because YOLO learns how to make small adjustments to the anchor boxes in order to create an accurate bounding box for your object. YOLO can learn small adjustments better/easier than large ones.
The assignment problem is trickier. As I understand it, part of the training process is for YOLO to learn which anchors to use for which object. So the "assignment" isn't deterministic like it might be for the Hungarian algorithm. Because of this, in general, multiple anchors will detect each object, and you need to do non-max-suppression afterwards in order to pick the "best" one (i.e. highest confidence).
There are a couple of points that I needed to understand before I came to grips with anchors:
Anchors can be any size, so they can extend beyond the boundaries of
the 13x13 grid cells. They have to be, in order to detect large
objects.
Anchors only enter in the final layers of YOLO. YOLO's neural network makes 13x13x5=845 predictions (assuming a 13x13 grid and 5 anchors). The predictions are interpreted as offsets to anchors from which to calculate a bounding box. (The predictions also include a confidence/objectness score and a class label.)
YOLO's loss function compares each object in the ground truth with one anchor. It picks the anchor (before any offsets) with highest IoU compared to the ground truth. Then the predictions are added as offsets to the anchor. All other anchors are designated as background.
If anchors which have been assigned to objects have high IoU, their loss is small. Anchors which have not been assigned to objects should predict background by setting confidence close to zero. The final loss function is a combination from all anchors. Since YOLO tries to minimise its overall loss function, the anchor closest to ground truth gets trained to recognise the object, and the other anchors get trained to ignore it.
The following pages helped my understanding of YOLO's anchors:
https://medium.com/#vivek.yadav/part-1-generating-anchor-boxes-for-yolo-like-network-for-vehicle-detection-using-kitti-dataset-b2fe033e5807
https://github.com/pjreddie/darknet/issues/568
I think that your statement about the number of predictions of the network could be misleading. Assuming a 13 x 13 grid and 5 anchor boxes the output of the network has, as I understand it, the following shape: 13 x 13 x 5 x (2+2+nbOfClasses)
13 x 13: the grid
x 5: the anchors
x (2+2+nbOfClasses): (x, y)-coordinates of the center of the bounding box (in the coordinate system of each cell), (h, w)-deviation of the bounding box (deviation to the prior anchor boxes) and a softmax activated class vector indicating a probability for each class.
If you want to have more information about the determination of the anchor priors you can take a look at the original paper in the arxiv: https://arxiv.org/pdf/1612.08242.pdf.
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Formula for controlling the movement of a tank-like vehicle?
(10 answers)
Closed 7 years ago.
I'm trying to simulate a tank-like/skid-steered vehicle, i.e. both of the wheels (one on each side) have separate velocities, and steering is done by increasing or decreasing the velocity of one of the sides.
For example, If I set the velocity of the left wheel to 5, and the right wheel to 3, it will turn right. What I'd like to know is, given the velocities of the wheels Vl and Vr, and the distance between the wheels D, by how many degrees will the direction the vehicle is pointing in change in one tick?
I've tried looking at Formula for controlling the movement of a tank-like vehicle?, and the links on that question, but haven't come up with anything. All my best guesses have failed.
First: the really easy edge cases. if V_l and V_r are zero, don't move. If they're the same, don't turn.
Second, if only one of V_l and V_r are zero, the tank pivots around the stationary tread, and the moving tread traces out an arc of length V_big with a radius of curvature D. theta = Vbig/D, plus or minus some sign conventions based on your coordinates. (the tank base also translates some distance but the calculations for that are dependent on where the center of rotation of the tank is defined to be and your coordinate system, so that detail is left as an exercise for the reader.)
Third, symmetry concerns! Obviously tank treads turning is left/right symmetric. If the left tread is twice as fast as the right the tank should turn the same amount as if the right tread is twice as fast as the left, just in a different direction. Ditto for going backwards.
Fourth: Meat and potatoes! I'm assuming neither tank tread can slip whatsoever. The faster tread traces an arc of length V_fast on a circle of radius r+D marked out by an angle theta. If you recall your trig V_fast=(r+D)*theta. The slower wheel traces out an arc of length V_slow on a circle of radius r marked out by the same angle.(V_slow = theta*r) Divide one equation by the other, receive V_fast/V_slow = (r+D)/r. Apply algebra to provide r=D/((V_fast/V_slow)-1) Note that this explodes appropriately when V_slow is zero or when V_fast=V_slow, and you appropriately receive r=D when V_fast=2*V_slow Recall that theta=V_slow*r: theta=(V_fast-V_slow)/D
IN RADIANS, mind you That's a crucial detail.
NOTE: If you define 'turning right' as positive theta and turning left as negative theta, it all works out and theta=(V_l-V_r)/D, even for negative speeds. The tank won't turn around to face the direction of travel, it'll keep facing the correct way.
I did an thinning operation on vessels, and now I'm trying to reconstruct it.
How to expand them to normal vessels in ITK when I have a skeleton line and radius values for each pixel?
DISCLAIMER: This could be slow, but since no other answer has been suggested, here you go.
Since your question does not indicate this, I'm assuming that you're talking about a 2D image, but the following approach can be extended for 3D too. This is how I'd go about it:
Create a blank image with zero filled pixel values
Create multiple instances of disk/sphere ShapedNeighborhoodIterator each having a different radius on the blank image (choose the most common radii from the vessel width histogram).
Visit each pixel in the binary skeleton image. When you come upon a white (vessel skeleton) pixel, recollect the vessel radius at that pixel.
If you already have a ShapedNeighborhoodIterator for that radius value, take the iterator to the pixel location in the blank image and fill up a disk/sphere of white pixels centered about that pixel. If you don't have a ShapedNeighborhoodIterator for that radius value, create one and do the same operation.
Once you finish iterating over the skeletonized image, you will have a reconstructed tree in the other image. Note that step 2 is optional, but will help you achieve faster computation.
I wish to use PostGIS to select all the points within a polygon, but this question is about defining the actual polygon.
I'm looking to define a polygon that is based on a great circle, specified by two points on the earths surface defined by latitude and longitude coordinates. The polygon that I'm after should be defined by a width left and right of the the center line (the center line being the line made by the great circle)
The resulting shape would be a long curved rectangular shape.
The purpose being to select all the points within x distance of the great circle line.
I think you are confused about the kind of data you are dealing with, if you use a equidistant projection you could use something as simple as this:
ST_DWithIn(ST_MakeLine(point1, point2),distanceInSRIDunits)
There is an old discussion in the postgis mail list that will be useful to you.