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.
Related
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 a binary itk image that I am trying to fill the holes using itk::BinaryFillholeImageFilter
using FillholeFilterType=itk::BinaryFillholeImageFilter<ImageType>;
FillholeFilterType::Pointer fillHoleFilter=FillholeFilterType::New();
fillHoleFilter->SetInput(someFilter->GetOutput());
fillHoleFilter->SetForegroundValue(itk::NumericTraits<PixelType>::max());
fillHoleFilter->Update();
However, it does not fill the holes. ImageType is float and its dimension is 3. How to solve this issue?
I ran SimpleITK.BinaryFillhole on your image, setting the foreground value to 1.0. Basically the same thing you did except the function version of the filter. It worked as expected.
The BinaryFillhole filter fills in holes that are contained completely within the object. In your case, there were some tiny, 1 pixel, holes that were filled in.
Maybe you were expecting some of the holes/junk along the object boundary to be filled. Because those holes are in some way (perhaps very small way) connected to the boundary, the filter does not fill those holes.
Perhaps you could try some smoothing of the image first.
BinaryFillholeImageFilter expects a binary image. Convert your image to itk::Image<unsigned char, 3> first, and then apply hole filling.
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.
I'm developing a simple a graphical editor for my flash-based app. In my editor there's a posibility of scaling, range of scaling is big (maximum scale is 16.0, minimum scale is 0.001 and default scale is 0.2). So it's quite possible that a user can draw a line with thickness 0.1 or 300.0, and it looks that line possible thickness (in Graphics.lineStyle()) has upper border. As I found out from livedocs maximum value is 255. So if thickness is greater then 255.0 there'is drawn a line of thickness 255.0. Whether mentioned upper border exists and how big is it. Here're my questions:
Right now I'm drawing lines with drawPath() or lineTo() methods. Natural walkarround if thickness is greater then 255.0 is to draw a rectange instead of segment and two circles on the ends of segment (instead of lineTo()). Or even to draw two thin segments and two half-circles and fill interior. Maybe there's more elegant/quick solution?
Another question is if the thickness of line is big but less then 255.0 (e.g. 100.0), what is faster drawing a line with lineTo() or drawing two thin segments and two half-circles and fill interior?
And finally, maybe someone knows a good article/book where I can read what's inside all methods of flash.display.Graphics class (or even not flash specific article/book on graphics)?
Any thoughts are appreciated. Thank you in advance!
I agree with f-a that putting the line in a container would probably be better and more efficient than drawing a rectangle and extra circles.
I don't think that the math would be too difficult to work out. For efficiency you should probably only do this if the line style is going to be over 255.
To setup the display object to hold your line I would start by halving the width of your line (the length can stay the same). Then create a new sprite and draw the line in the sprite at half size (e.g. if you wanted 300, just draw it at 150). It would be most simple to just start at (0,0) and draw the segment straight so that all of your transformations can be applied to the new sprite.
From here you can just double the scaleY of the sprite to get the desired line weight. It should keep the same length and the ends should also be rounded correctly.
Hope this helped out!
A cool resource for working with the graphics class is Flash and Math. This site has several cool effects and working examples and source code.
http://www.flashandmath.com/
I want to test each pixel in an image and check it's color if it is white then I must display
the corresponding (x,Y) for that pixel but I didn't find until now a function that help me
do something like that. So please if you have such function tell me.
thanks at all
If you are using matlab for image editting, the function for this purpose would be imread, i.e. if you have a MN image them imread would split it in a matrix of [MN*3], i.e. a 3 dimensional array representing RGB components of the image. Now if at a given X,Y coordinate, all the three i.e. Red, Green and Blue components have equal value, then that pixel is not color. If they have different values then it is coloured pixel. Using an if condition check and generate a new picture matrix accordingly. Display the new picture matrix.
I don't know if there is such a function. maybe in some deep graphics toolkit? it'll certainly not be part of a basic language framework (i.e. core stuff in objc or system in c#).
you might have to bite the bullet and just work your way through all of the pixels manually in o(n^2).
foreach horizontal pixel
foreach vertical pixel
if(pixel at (horizontal,vertical) is white)
return (horizontal, vertical)