Python PIL - function to divide blend two images?

EDIT: Code is working now, thanks to Mark and zephyr. zephyr also has two alternate w开发者_开发百科orking solutions below.

I want to divide blend two images with PIL. I found ImageChops.multiply(image1, image2) but I couldn't find a similar divide(image, image2) function.

Divide Blend Mode Explained (I used the first two images here as my test sources.)

Is there a built-in divide blend function that I missed (PIL or otherwise)?

My test code below runs and is getting close to what I'm looking for. The resulting image output is similar to the divide blend example image here: Divide Blend Mode Explained.

Is there a more efficient way to do this divide blend operation (less steps and faster)? At first, I tried using lambda functions in Image.eval and ImageMath.eval to check for black pixels and flip them to white during the division process, but I couldn't get either to produce the correct result.

EDIT: Fixed code and shortened thanks to Mark and zephyr. The resulting image output matches the output from zephyr's numpy and scipy solutions below.

# PIL Divide Blend test

import Image, os, ImageMath

imgA = Image.open('01background.jpg')
imgA.load()
imgB = Image.open('02testgray.jpg')
imgB.load()

# split RGB images into 3 channels
rA, gA, bA = imgA.split()
rB, gB, bB = imgB.split()

# divide each channel (image1/image2)
rTmp = ImageMath.eval("int(a/((float(b)+1)/256))", a=rA, b=rB).convert('L')
gTmp = ImageMath.eval("int(a/((float(b)+1)/256))", a=gA, b=gB).convert('L')
bTmp = ImageMath.eval("int(a/((float(b)+1)/256))", a=bA, b=bB).convert('L')

# merge channels into RGB image
imgOut = Image.merge("RGB", (rTmp, gTmp, bTmp))

imgOut.save('PILdiv0.png', 'PNG')

os.system('start PILdiv0.png')

You are asking:

Is there a more efficient way to do this divide blend operation (less steps and faster)?

You could also use the python package blend modes. It is written with vectorized Numpy math and generally fast. Install it via pip install blend_modes. I have written the commands in a more verbose way to improve readability, it would be shorter to chain them. Use blend_modes like this to divide your images:

from PIL import Image
import numpy
import os
from blend_modes import blend_modes

# Load images
imgA = Image.open('01background.jpg')
imgA = numpy.array(imgA)
# append alpha channel
imgA = numpy.dstack((imgA, numpy.ones((imgA.shape[0], imgA.shape[1], 1))*255))
imgA = imgA.astype(float)

imgB = Image.open('02testgray.jpg')
imgB = numpy.array(imgB)
# append alpha channel
imgB = numpy.dstack((imgB, numpy.ones((imgB.shape[0], imgB.shape[1], 1))*255))
imgB = imgB.astype(float)

# Divide images
imgOut = blend_modes.divide(imgA, imgB, 1.0)

# Save images
imgOut = numpy.uint8(imgOut)
imgOut = Image.fromarray(imgOut)
imgOut.save('PILdiv0.png', 'PNG')

os.system('start PILdiv0.png')

Be aware that for this to work, both images need to have the same dimensions, e.g. imgA.shape == (240,320,3) and imgB.shape == (240,320,3).

There is a mathematical definition for the divide function here: http://www.linuxtopia.org/online_books/graphics_tools/gimp_advanced_guide/gimp_guide_node55_002.html

Here's an implementation with scipy/matplotlib:

import numpy as np
import scipy.misc as mpl

a = mpl.imread('01background.jpg')
b = mpl.imread('02testgray.jpg')

c = a/((b.astype('float')+1)/256)
d = c*(c < 255)+255*np.ones(np.shape(c))*(c > 255)

e = d.astype('uint8')

mpl.imshow(e)
mpl.imsave('output.png', e)

If you don't want to use matplotlib, you can do it like this (I assume you have numpy):

imgA = Image.open('01background.jpg')
imgA.load()
imgB = Image.open('02testgray.jpg')
imgB.load()

a = asarray(imgA)
b = asarray(imgB)
c = a/((b.astype('float')+1)/256)
d = c*(c < 255)+255*ones(shape(c))*(c > 255)
e = d.astype('uint8')

imgOut = Image.fromarray(e)
imgOut.save('PILdiv0.png', 'PNG')

The problem you're having is when you have a zero in image B - it causes a divide by zero. If you convert all of those values to one instead I think you'll get the desired result. That will eliminate the need to check for zeros and fix them in the result.