I am trying to draw a tilted ellipse in image draw. However, I am not sure how to define it, since while the scheme below would move the points, I think this would just squish the ellipse, not rotate it (also I think there is something slightly wrong with the transformation in any case). I am feeding the output of this function into the ellipse command and adding it to an existing picture, so any methods that would rotate the entire image are no good. OD is just a square offset to the coordinate center I am using.
def ellipsebound(major, minor, tilt=0, offset=0, angle=0):
#creates a bound for an ellispe, defined with tilt meaning to rotate the orthogonal axis and angle corresponds to rotating the ellipse position
angle = radians(angle)
tilt = radians(tilt)
box=(
1 + int(ceil((OD+offset*cos(angle)+(major*cos(tilt)+minor*sin(tilt)))/conv)),
1 + int(ceil((OD+offset*sin(angle)+(major*sin(tilt)-minor*cos(tilt)))/conv)),
int(ceil((2*OD-(OD-offset*cos(angle)-(major*cos(tilt)+minor*sin(tilt)))/conv))),
int(ceil((2*OD-(OD-offset*sin(angle)-(major*sin(tilt)-minor*cos(tilt)))/conv)))
) #create bounding box
return box
Does anyone know how to accomplish this?
It looks like the 'box' that is being used to draw the ellipse has no rotation associated with it. It is simply defined by the (left, top, right, bottom) extents.
One possible workaround (depending on what you need to do) is to draw the ellipse (sized correctly, but without the rotation) onto an intermediary image, use the image.rotate() method, and then paste it into your target image.
I hope that helps.
Related
I'm trying to set as initial camera of a 3D volume plot where the upper left corner is the origin (x, y, z = 0). I've read the documentation about the camera controls but cannot figure out how can I accomplish this.
The initial view I want it's something like this:
I tried it and this one work on me
If you want the front upper left corner as (0,0,0)
camera = dict(
eye=dict(x=0, y=-0.5, z=-2.5)
)
fig.update_layout(scene_camera=camera, title=name)
fig.show()
what I understand from this eye is basically the position of the eye(or you) look at eyepoint(0,0,0) which is I believe the center of the 3D graph (not the coordinate)
And if you need to change the axes direction to the opposite, you can try to put it on negative on the eye position, and if it is zero you can put negative small number (in this example I used -0.5, but you can use -0.01 too)
Kivy GUI window
In the above GUI window I have a fish masked image and a fish tank which is the main background image. I want to extract the absolute lower left coordinate of the scatter to perform further cropping operation on the image and with the scatter mask. Scatter.bbox or scatter.pos is not returning absolute X,Y coordinates w.r.t image. When angle is less than 12deg it is working fine but as angle is increased new bbox rectangle translates more towards the bottom part of the window.
code for this:
`x = scatter.bbox[0][0]
y = scatter.bbox[0][1]
#I have also tried - self.to_local(*self.scatter.pos)`
For getting absolute coordinates use to_parent instead of bbox:
scatter_name.to_parent(0,0)
I want to visualize results of keypoint tracking algorithm in python. I have a sequence of (Image, Keypoint) pairs (video basically). Tracking algorithm is strong enough to give me subpixel accuracy. But i have no idea, how to visualize it properly.
I tried to round my coordinates and draw a circle by cv2.circle(image, (int(round(x)), int(round(y)))), but it leads to visual jittering of my keypoints due to small image resolution.
I checked OpenCV, Pillow, skimage, Pygame (pygame.draw.circle). All of them cannot properly draw circle with float coordinates.
DIPlib has the function DrawBandlimitedBall(), which draws a disk or a circle with smooth transitions and with floating-point origin coordinates (disclosure: I'm one of the authors). You might need to draw the circle in an empty image, then blend it in to get the effect you are looking for. Code would look something like this:
import diplib as dip
img = dip.ImageRead('/Users/cris/dip/images/flamingo.tif')
p = [366.4, 219.1]
# Create an empty image and draw a circle in it
circle = dip.Image(img.Sizes(), 1, 'SFLOAT')
circle.Fill(0)
dip.DrawBandlimitedBall(circle, diameter=22.3, origin=p, value=1, mode='empty')
circle /= dip.Maximum(circle)
# Blend: img * (1-circle) + circle * color
img *= 1 - circle
img += circle * dip.Create0D([0,255,0]) # we make the circle green here
img.Show()
dip.ImageWrite(img, 'so.jpg')
(Note that the circle actually looks better without the JPEG compression artifacts.)
You could draw the circle directly in the image, but this function adds the circle values to the image, it doesn't attempt to blend, and so you'd get a much worse look for this particular application.
I need the approximate radii of the following ellipse.
The bottom/top and left/right radii should be the same nevertheless need to be checked. Which means 4 radii should be the result of my code. I did the following in paint, the green circle should give me the top radius and red the left (the right and bottom one aren't drawn here).
The idea I'm working on is to crop the image (left/right/top/bottom side) and approximate circles to the cropped images. With the cv2.findContours-feature some white pixels get recognized as highlighted here.
Is there a way to approximate my drawn red circle from above with these given coordinates? The problems I've seen on the internet are all with a given center point or angle which I don't have. Is there a cv2 function that draws circles with only some given coordinates or something similar?
Use this function : cv2.fitEllipse(points) and pass contour points -Ziri
Yes this did the trick. I got the radii after your function with:
(x, y), radius = cv2.minEnclosingCircle(i)
I am using pygame to make a simple game. I am having issues with circle collisions. I am getting the following error:
"AttributeError: 'pygame.Rect' object has no attribute 'rect'"
Here is the particular code I am having issues with below:
if pygame.sprite.collide_circle(hero_circle, enemy_circle):
gameover()
Use pygame.mask to create a collision mesh for your objects and use the mesh to do collision detections.
In more detail:
Create an image file for both of your circles and set the bg color to something you will not use anywhere else.
Set that color to "transparent" in your image editor.
Import the images.
Create a mesh for them with pygame.mask and set it to make transparent pixels non-collidable.
Use the generated mask as your collision detection mesh.
PROFIT
(Technically this is just doing collision detection of a circle shaped area on a rectangle, but who cares!)
pygame.draw.rect()
draw a rectangle shape
rect(Surface, color, Rect, width=0) -> Rect
Draws a rectangular shape on the Surface. The given Rect is the area of the rectangle. The width argument is the thickness to draw the outer edge. If width is zero then the rectangle will be filled.
Keep in mind the Surface.fill() method works just as well for drawing filled rectangles. In fact the Surface.fill() can be hardware accelerated on some platforms with both software and hardware display modes.
The best way I've found to check circle collision detection is to calculate the distance between the center points of two circles. If the distance is less than the sum of the two circle's radii, then you've collided.
Just like how gmk said it but if your are using circles instead of rectangles, you should use this pygame function :
pygame.draw.circle(surface, color, center_point, radius, width)
This draws a circle on your surface (which would go in the surface area). Clearly the color requires a list of numbers (RGB anyone?). Your center_point decides the location of your circle since it will be the location of the center of your circle. The radius will need a number to set the radius of the circle (using the number like 25 will set your radius at 25 pixels/diameter at 50 pixels). the width section is optional as it sets the thickness of the perimeter of your circle (having 0 will have none at all). If you are not using circles, you should change your title... But anyways, I hope this helps you!