Python PIL-函数划分混合两个图像? [英] Python PIL - function to divide blend two images?

问题描述

编辑:感谢Mark和zephyr，代码现在可以正常工作了.zephyr在下面还有两个替代的工作解决方案.

Code is working now, thanks to Mark and zephyr. zephyr also has two alternate working solutions below.

我想用PIL划分两个图像.我找到了 ImageChops.multiply(image1，image2)，但找不到类似的 divide(image，image2)函数.

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.

说明了混合混合模式" (我在这里使用前两个图像作为测试源.)

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

是否有我错过的内置除法混合功能(PIL或其他方式)?

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

我下面的测试代码正在运行，并且已经接近我要寻找的代码.生成的图像输出类似于此处的除法混合示例图像:

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.

是否有一种更有效的方法来执行除法混合操作(步骤更少，速度更快)?最初，我尝试在 Image.eval 和 ImageMath.eval 中使用lambda函数检查黑色像素并将其在分割过程中翻转为白色，但是我无法都能得到正确的结果.

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.

固定代码，由于Mark和zephyr而缩短了代码.最终的图像输出与以下zephyr的numpy和scipy解决方案的输出相匹配.

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')

推荐答案

除法函数在此处有数学定义: http://www.linuxtopia.org/online_books/graphics_tools/gimp_advanced_guide/gimp_guide_node55_002html.

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

这是一个使用scipy/matplotlib的实现:

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)

如果您不想使用matplotlib，则可以这样做(我假设您有numpy):

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')

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