Crop non symmetric area of an image with Python/PIL

Question

Is there a way to cut out non rectangular areas of an image with Python PIL?

e.g. in this picture I want to exclude all black areas as well as tower

Answer 1

If I understood correctly, you want to make some areas transparent within the image. And these areas are random shaped. Easiest way (that I can think of) is to create a mask and put it to the alpha channel of the image. Below is a code that shows how to do this.

If your question was "How to create a polygon mask" I will redirect you to:

SciPy Create 2D Polygon Mask

and look the accepted answer.

br,

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

Actually, I could not resist... the polygon mask solution 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 when I think of it. If you have a black and white svg, you can load your svg directly as mask (assuming white is your mask). I have no sample svg images, so I cannot test this. I am not sure if PIL can open svg images.