Circle in pixel grid

Question

Given the center (x,y)and radius r, how can I draw a circle C((x,y),r)in a pixel grid using python? It’s good to assume that the pixel grid is large enough.

Answer 1

Here's the RosettaCode Midpoint Circle Algorithm in Python

def circle(self, x0, y0, radius, colour=black):
    f = 1 - radius
    ddf_x = 1
    ddf_y = -2 * radius
    x = 0
    y = radius
    self.set(x0, y0 + radius, colour)
    self.set(x0, y0 - radius, colour)
    self.set(x0 + radius, y0, colour)
    self.set(x0 - radius, y0, colour)

    while x < y:
      if f >= 0: 
        y -= 1
        ddf_y += 2
        f += ddf_y
        x += 1
        ddf_x += 2
        f += ddf_x    
        self.set(x0 + x, y0 + y, colour)
        self.set(x0 - x, y0 + y, colour)
        self.set(x0 + x, y0 - y, colour)
        self.set(x0 - x, y0 - y, colour)
        self.set(x0 + y, y0 + x, colour)
        self.set(x0 - y, y0 + x, colour)
        self.set(x0 + y, y0 - x, colour)
        self.set(x0 - y, y0 - x, colour)
        Bitmap.circle = circle

        bitmap = Bitmap(25,25)
        bitmap.circle(x0=12, y0=12, radius=12)
        bitmap.chardisplay()

Answer 2

Assuming you want to achieve a result (as opposed to learning bitmap algorithms), just use Pillow :

from PIL import Image, ImageDraw
image = Image.new('1', (10, 10)) #create new image, 10x10 pixels, 1 bit per pixel
draw = ImageDraw.Draw(image)
draw.ellipse((2, 2, 8, 8), outline ='white')
print list(image.getdata())

output:

[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 255, 255, 255, 255, 255, 0, 0, 0, 0, 255, 255, 0, 0, 0, 255, 255, 0, 0, 0, 255, 0, 0, 0, 0, 0, 255, 0, 0, 0, 255, 0, 0, 0, 0, 0, 255, 0, 0, 0, 255, 0, 0, 0, 0, 0, 255, 0, 0, 0, 255, 255, 0, 0, 0, 255, 255, 0, 0, 0, 0, 255, 255, 255, 255, 255, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

The range is 0..255, because Pillow stores bytes per pixel even for an image with 1 bit per pixel (as more efficient).

If you need a range of 0..1, you can divide by 255:

[x/255 for x in image.getdata()]