If I understand correctly, you want to make some areas transparent inside the image. And these areas have a random shape. The easiest way (I can think of) is to create a mask and put it in the alpha channel of the image. Below is the code that shows how to do this.
If your question is: "How to create a polygon mask", I redirect you to:
SciPy Create a 2D Polygon Mask
and see the accepted answer.
w
Juha
import numpy import Image # read image as RGB and add alpha (transparency) im = Image.open("lena.png").convert("RGBA") # convert to numpy (for convenience) imArray = numpy.asarray(im) # create mask (zeros + circle with ones) center = (200,200) radius = 100 mask = numpy.zeros((imArray.shape[0],imArray.shape[1])) for i in range(imArray.shape[0]): for j in range(imArray.shape[1]): if (i-center[0])**2 + (j-center[0])**2 < radius**2: mask[i,j] = 1 # assemble new image (uint8: 0-255) newImArray = numpy.empty(imArray.shape,dtype='uint8') # colors (three first columns, RGB) newImArray[:,:,:3] = imArray[:,:,:3] # transparency (4th column) newImArray[:,:,3] = mask*255 # back to Image from numpy newIm = Image.fromarray(newImArray, "RGBA") newIm.save("lena3.png")
Edit
In fact, I could not resist ... the solution to the polygon mask was so elegant (replace the above circle with this):
# create mask polygon = [(100,100), (200,100), (150,150)] maskIm = Image.new('L', (imArray.shape[0], imArray.shape[1]), 0) ImageDraw.Draw(maskIm).polygon(polygon, outline=1, fill=1) mask = numpy.array(maskIm)
Edit2
Now that I think about it. If you have black and white svg, you can upload your svg directly as a mask (if white is your mask). I do not have any svg image samples, so I cannot verify this. I'm not sure if PIL can open svg images.