I want to use drawRect( [0 0 60 60] ) to draw a square.
But the step size of x- and y-axis change when I resize the window:
And both of them don't "look like" square.
So, is there any method to set the x- and y-axis step width consistent? No matter how I resize the window?
Here is the result using #Dan's solution: axis( ..., "equal" )
try axis("equal"). From the docs this will
Force x distance to equal y-distance.
Related
I have a dataset that provides bounding box coordinates in the following format.
height- 84 width- 81 x - 343 y - 510. Now, I want to normalize these values (0-1) to train them using the yolov5 model. I have looked online and found that I can normalize these values in 2 ways. Way 1:
Normalized(Xmin) = (Xmin+w/2)/Image_Width
Normalized(Ymin) = (Ymin+h/2)/Image_Height
Normalized(w) = w/Image_Width
Normalized(h) = h/Image_Height
Way 2: divide x_center and width by image width, and y_center and height by image height.
Now, I am not sure which way I should follow to normalize the values in the given dataset. Can anyone suggest me any solution? Also, the size of the given images in my dataset is 1024 x 1024. Now, if I convert the images in 512 x 512 size, how do I figure the new bounding box coordinates i.e what will be the value of height widht x and y?
First, Yolov5 will resize your images and bounding boxes for you, so you don't have to worry about that. By default, it will resize the longest side to 640px and the shortest side will be resized to a length that preserves the proportion of the original image.
About the normalization [0-1]. Yolov5 expects the center points of the bbox, not the minimum points, so if your box dimensions areheight = 84px and width = 81px and those x and y are the minimum points of the bbox (i'm not sure from your post), your formula works, because you're computing the center points:
Normalized(**x_center**) = (Xmin+w/2)/Image_Width
Normalized(**y_center**) = (Ymin+h/2)/Image_Height
...
About the resizing:
https://github.com/ultralytics/yolov5/discussions/7126#discussioncomment-2429260
I've implemented a zoom and crop on the HTML5 Canvas. Zoom is actually increasing the height and width of the Canvas so that it looks zoomed. For crop, I wrote an algorithm to select a rectangular area using mouse and then crop it. Now, if I want to crop when the image is zoomed in or out, while selecting the crop area I have to consider the top and left position displacement caused due to the zoom , which works fine.
So I'm now implementing a rotate (using css3 transform: rotate). The problem is, when I rotate the image by a certain angle, the selection appears a little away from the actual mouse position. This used to happen for the zoom effect as well, but since I used to subtract the added left and top distance from the x and y position resp., I was able to draw the selection even when the image was zoomed. I don't understand how I should do it for a rotated image!
The following image might help you understand my problem a little more clearly:
There's a div around the canvas, reflecting the canvas. It'll have the same width, height, top, left properties as the canvas. This is done on purpose since I can't add the selection, which is absolute, as the child of the canvas. Now this cover, when selected in FireBug, still shows as a rectangle with increased width and height and changed top and left positions.
I understand I have to calculate the displacement like I'm already doing for zoom, but I don't know how to do it! I have spent a lot of time trying out stuff like Pythagoras algorithm and rotational matrix and blah blah!
Please help me out!
You can rotate each of the vertices using this function where "pnt" is a vertex, "pivot" is the point you're rotating around and "angle" is the angle in radians
function rotatePoint(pnt, pivot, angle){
var data = figureAngle(pivot, pnt),
theta = data.angle + angle,
rise = Math.sin(theta) * data.length,
run = Math.cos(theta) * data.length;
return {
x: pivot.x + run,
y: pivot.y + rise
}
}
function figureAngle(start, end){
var rise = (end.y - start.y),
run = (end.x - start.x),
length = Math.sqrt(Math.pow(rise, 2) + Math.pow(run, 2));
return {length: length, angle: Math.atan2(-rise, -run) + Math.PI};
}
Then your horizontal shift is going to be the smallest x of the 4 new vertices, and your vertical shift is going to be the smallest y.
EDIT: this assumes your top left coordinate is [0, 0] before you rotate. If not you need to subtract your starting coordinates from the results i.e. if your top-left corner starts at [50, 100], your horizontal shift would be xMin - 50, and your vertical would be yMin - 100
I want to decrease an 480 X 480 bitmap image size to 30 X 30 pixel size but keeping the whole height and width intact. (I do not want to scale or use height/width property! )
So if i divide 480/16 = 30. So i need to take average pixel values of 30 pixel elements and put it into new image.
How to take the average in actionscript 3.0? I looked at getpixels() method, is their any simple way/methods to achieve this?
Let me put in more simple way - I am trying to reduce pixels in an bitmap image from 480 X 480 to 30 X 30, the height and width remain same and i expect some amount of distortion after converting image to 30 X 30.
I did scaling but it reduces width and height, if i again increase width and height it just regains normal pixels. Thanks!
Why don't you simply then make a copy of the whole image in code, but use the simple scaling to scale the copy, and only present that to the user. Also look at this from Stack Overflow
How to resize dynamically loaded image into flash (as3)
Is it possible to set the width and height of a canvas element and have the existing content scale to fit these new dimensions?
Right now the user uploads an image and my page creates a canvas element containing this image and whose dimensions are the same size as the image. I want to limit the size that I'm working with, however, so here is an example of what I want to have happen:
The user uploads an image that is 1600 x 1200 pixels (Not saved to server)
The data goes right to an html5 canvas object
The canvas height and width are set to 800 x 600 and the image content scales appropriately and then is displayed.
Right now if I set the canvas width and height it just crops the image at those dimension, not resizing as I would like. Thanks!
There are many ways to call drawImage
One is:
ctx.drawImage(image, sx, sy, sw, sh, dx, dy, dw, dh)
Where dx,dy,dw,dh are the destination x, y, width, and height.
If you make dw and dh always 800x600, the image will drawn will always automatically be scaled to 800x600.
Here's a tiny example that will always draw any size image to 800x600 http://jsfiddle.net/jAT8Y/
Another alternative is to use context.scale function. Your image quality will remain better if you use scale.
Having trouble scaling with . It seems to make sense to code up a drawing in canvas to a fixed size (ie 800x600) then scale it for specific locations - but sizing occurs in 4 places: 1) in the context definition (ie ctx.width = 800 2) with ctx.scale; 3) in html with
I can scale it with ctx.scale(0.25,0.25) and use but this doesn't appear right - it seems to want the scale to be proportional.
css sizing simply makes it fuzzy so not a good way to go. Any ideas?
Actually, you can resize a canvas using stylesheets. The results may vary across browsers as HTML5 is still in the process of being finalized.
There is no width or height property for a drawing context, only for canvas. A context's scale is used to resize the unit step size in x or y dimensions and it doesn't have to be proportional. For example,
context.scale(5, 1);
changes the x unit size to 5, and y's to 1. If we draw a 30x30 square now, it will actually come out to be 150x30 as x has been scaled 5 times while y remains the same. If you want the logo to be larger, increase the context scale before drawing your logo.
Mozilla has a good tutorial on scaling and transformations in general.
Edit: In response to your comment, the logo's size and canvas dimensions will determine what should be the scaling factor for enlarging the image. If the logo is 100x100 px in size and the canvas is 800x600, then you are limited by canvas height (600) as its smaller. So the maximum scaling that you can do without clipping part of the logo outside canvas will be 600/100 = 6
context.scale(6, 6)
These numbers will vary and you can do your own calculations to find the optimal size.
You could convert the logo to svg and let the browser do the scaling for you, with or without adding css mediaqueries.
Check out Andreas Bovens' presentation and examples.
You can resize the image when you draw it
imageobject=new Image();
imageobject.src="imagefile";
imageobject.onload=function(){
context.drawImage(imageobject,0,0,imageobject.width,imageobject.height,0,0,800,600);
}
The last 2 arguments are the width an height to resize the image
http://www.w3.org/TR/html5/the-canvas-element.html#dom-context-2d-drawimage
If you set the element.style.width and element.style.height attributes (assuming element is a canvas element) you are stretching the contents of the canvas. If you set the element.width and element.height you are resizing the canvas itself not the content. The ctx.scale is for dynamic resizing whenever you drawing something with javascript and gives you the same stretching effect as element.style.