寻找对图像进行按位XOR的方法 [英] Searching for a way to do Bitwise XOR on images

问题描述

我正在寻找一种方法来获取两个图像的按位异或命令行(或以可以在程序或脚本中实现的另一种方式).

I am looking for a way to get the Bitwise XOR of two images on the command line(or in another way that can be implemented in a program or script).

这将产生与在支持它的图片编辑器(Paint.NET，Photoshop等)中使用异或混合"模式相同的最终图片.

This should result in the same final picture as using the XOR Blending mode in picture editors that support it (Paint.NET, Photoshop, etc)

举例来说，假设我有图片A:

和图像B:

然后结果应如下所示:

其中最有趣的部分是，当您再次将图像C与图像B异或时，您将获得图像A的精确副本.

The fun part of this is of course, that when you XOR image C with image B again, you will get an exact copy of image A.

现在，我一直在互联网上寻找以编程方式执行此操作的方法，但是我什么也没发现.甚至ImageMagick也不支持对图像进行按位XOR.

Now, I have been looking all over the internet for a way to do this programmatically, but I have found nothing. Even ImageMagick does not support doing a bitwise XOR on images.

某人知道这样做的方法吗?

Does sombebody know a way to do this?

推荐答案

ImageMagick可以做到，尽管有点令人费解.一种方法是:

ImageMagick can do it, although it's a bit convoluted. One way is:

convert img1 img2 -fx "(((255*u)&(255*(1-v)))|((255*(1-u))&(255*v)))/255" img_out

(img1，img2，img_out分别是两个输入和单个输出文件名).

(img1,img2,img_out are the two input and single output file names respectively).

这有点丑陋(我敢肯定有人拥有比我更多的ImageMagick-fu可以清理它，但是它的工作原理是这样的:

It's a bit ugly (I'm sure someone with more ImageMagick-fu than me could clean it up but it works like this:

1. -fx "xxx"基本上说对图像执行操作xxx". 在上面的表达式中，u和v分别代表第一和第二输入图像.

1. -fx "xxx" basically says "perform the operation xxx on the image". In the expression above, u and v stand for the first and second input images respectively.

现在，-fx仅以按位运算符的方式具有按位与&和按位OR |. 要重构按位异或，我们需要

Now, -fx only has bitwise AND & and bitwise OR | in the way of bitwise operators. To reconstruct bitwise XOR, we need

convert img1 img2 -fx "(u & NOT v) | (NOT u & v)" img_out

要获取NOT(存在逻辑NOT，但没有按位的NOT)，请记住，NOT x = 255-x if x是8位. 因此，要获得NOT u，我们可以做255-u，假设图像u是8位的. 因此，ImageMagick命令将为:

To get the NOT (there is a logical NOT but no bitwise NOT), we remember that NOT x = 255-x if x is 8-bit. So to get NOT u we can just do 255-u, assuming image u is 8-bit. Hence, the ImageMagick command would be:

convert img1.png img2.img -fx "((255-u)&v)|(u&(255-v))" image_xor.png

• 这里的一个问题是，当ImageMagick执行fx时，它会标准化[0,1]范围内的u和v中的所有像素，而不是我们期望的[0,255]，并按位进行非整数螺钉塞满.

• • The one problem here is that when ImageMagick does fx it normalises all the pixels in u and v in the range [0,1] instead of [0,255] as we expect, and doing bitwise on non-integers screws stuff up.

    因此，我们必须将上述表达式中u和v的所有出现次数乘以255(因此按位运算有效)，然后将255除以最后返回ImageMagick期望的范围[0,1].

    Hence, we have to multiply all occurrences of u and v in the above expression by 255 (so the bitwise operations work), and divide by 255 at the very end to get back in the range [0,1] that ImageMagick expects.

    这给了我们原始的命令，

    This gives us the original command,

    convert img1 img2 -fx "(((255*u)&(255*(1-v)))|((255*(1-u))&(255*v)))/255" img_out
    

    Voila！

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