通过 python 算法将 Square Hald clut 转换为 Classic Hald clut

Question

我的问题是：我有一个 Square Hald 8 512x512px，我想把它转换成 .CUBE 文件

将 Square Hald 转换为 Classic Hald，然后轻松转换为 CUBE（参见第二个算法）图片：Square Hald 8 - Classic Hald 8

我正在尝试反转这个简单的 python 算法，该算法可以很好地转换 Classic Hald -> Square Hald:

import numpy, cv2

hald = cv2.imread("classic_hald_8.png")

size = int(hald.shape[0] ** (1.0/3.0) + .5)

clut = numpy.concatenate([
    numpy.concatenate(hald.reshape((size,size,size**2,size**2, 3))[row], axis=1)
    for row in range(size)
])

cv2.imwrite("square_hald_8.png", clut)

import imageio as iio, numpy as np
imagen=iio.imread("Classic_Hald_8.png")

r,g,b=(imagen[:,:,0]).reshape(-1), (imagen[:,:,1]).reshape(-1), (imagen[:,:,2]).reshape(-1)
np.savetxt("Classic_Hald_8.cube",X=np.column_stack((r/255,g/255,b/255)),fmt='%1.6f', header="LUT_3D_SIZE 64", comments="")

我们需要重塑价值观。一些想法或建议？谢谢

UPDATE2: 感谢 GrossGrade 作者 Eugene Vdovin，我已经解决了。如果有人建议实施 3D 结构（也许使用 numpy？），则可以很好地接受。我对 python

很菜鸟

from PIL import Image

im = Image.open('test.png','r')
values = im.load()
hald_side_in_pixels = im.size[0]
hald_in_pixels = im.size[0]*im.size[0]
lutSize = int(hald_in_pixels ** (1.0/3.0) + .5) 

fr = [0.0]*hald_in_pixels
fg = [0.0]*hald_in_pixels
fb = [0.0]*hald_in_pixels

cubeIndex = 0
for y in range(hald_side_in_pixels):
 for x in range(hald_side_in_pixels):
  iR = cubeIndex % lutSize
  iG = y % lutSize
  iB = int(x/lutSize)+(int(y/lutSize)*int(hald_side_in_pixels/lutSize))
  idx = iR * lutSize * lutSize + iG * lutSize + iB
  fr[idx],fg[idx],fb[idx] = values[x,y]  
  cubeIndex+=1


with open("test.cube", "w") as output:
 output.write("DOMAIN_MIN 0 0 0\nDOMAIN_MAX 1 1 1\nLUT_3D_SIZE " + str(lutSize) + '\n')
 for iB in range(lutSize):
  for iG in range(lutSize):
   for iR in range(lutSize):
    idx = iR * lutSize * lutSize + iG * lutSize + iB
    output.write((str("%.9f" % (fr[idx]/255)) + ' ' +  str("%.9f" % (fg[idx]/255)) + ' ' + str("%.9f" % (fb[idx]/255)))+ '\n')
output.close()

UPDATE3: 我听从了建议，我用 numpy 创建了一个 3D 数组，现在更干净了，但它比 3x 1D 数组慢了大约 150ms，我post代码

from PIL import Image
import numpy as np

im = Image.open('test.png','r')
values = im.load()
hald_side_in_pixels = im.size[0]
lutSize = int((hald_side_in_pixels*hald_side_in_pixels) ** (1.0/3.0) + .5)

LUT = np.empty((lutSize,lutSize,lutSize), dtype=bytearray)
cubeIndex = 0
for y in range(hald_side_in_pixels):
 for x in range(hald_side_in_pixels):
  iR = cubeIndex % lutSize
  iG = y % lutSize
  iB = int(x/lutSize)+(int(y/lutSize)*int(hald_side_in_pixels/lutSize))
  LUT[iR,iG,iB]=values[x,y]
  cubeIndex+=1


with open("test1.cube", "w") as output:
 output.write("DOMAIN_MIN 0 0 0\nDOMAIN_MAX 1 1 1\nLUT_3D_SIZE " + str(lutSize) + '\n')
 for iB in range(lutSize):
  for iG in range(lutSize):
   for iR in range(lutSize):
    output.write((str("%.9f" % (LUT[iR,iG,iB][0]/255)) + ' ' +  str("%.9f" % (LUT[iR,iG,iB][1]/255)) + ' ' + str("%.9f" % (LUT[iR,iG,iB][2]/255)))+ '\n')
output.close()

Answer 1

这里是解析HALD图像的每个像素并将其值放入3DLUT的C++代码，随后以CUBE格式保存到文件中。代码缺少一些东西来保持一切紧凑。希望这会帮助您理解索引算法，以便您可以自己在 python 中实现它。

std::string formatstring(const char *Format, ...)
{
    if(Format == NULL || Format[0] == '[=10=]') return "";
    static std::string Res;
    va_list args;
    va_start(args, Format);
    int len = _vscprintf(Format, args);
    Res.resize(len);
    vsprintf((char*)Res.data(), Format, args);
    va_end(args);
    return Res;
}

void convert_hald_to_cube()
{
    CLUT LUT; // you must implement a class to store 3DLUT data of any size

    // Here you getting your HALD image data into a 2D-array hald_colors
    // ...

    // Getting 3DLUT values from a HALD 2D-array
    int hald_side_in_pixels = 512;
    int lutSize = (int)(powf((float)hald_side_in_pixels*hald_side_in_pixels,1.f/3.f)+0.5f); // +0.5f is for rounding a positive value
    int iR,iG,iB;
    int cubeIndex = 0;
    for(int y=0; y<hald_side_in_pixels; ++y) 
    {
        for(int x=0; x<hald_side_in_pixels; ++x)
        {
            iR = cubeIndex % lutSize;
            iG = y % lutSize;
            iB = (x/lutSize)+((y/lutSize)*(hald_side_in_pixels/lutSize));
                        
            // Here you copy the hald_colors[x][y] color value to the 3DLUT voxel value at indexes {iR,iG,iB}
                        
            cubeIndex++;
        }
    }

    // Putting 3DLUT values to a CUBE file
    FILE *f = fopen("OutputCubeFile.cube", "w+");
    fputs("TITLE my cube file\n",f);                        
    fputs("DOMAIN_MIN 0 0 0\n",f);
    fputs("DOMAIN_MAX 1 1 1\n",f);
    std::string s = "LUT_3D_SIZE " + std::to_string((_ULonglong)lutSize) + "\n";
    fputs(s.c_str(),f);
    for(int iB=0; iB<lutSize; ++iB) 
    {
        for(int iG=0; iG<lutSize; ++iG) 
        {
            for(int iR=0; iR<lutSize; ++iR) 
            {
                float fr,fg,fb;
                
                // Here you copy the 3DLUT voxel value at indexes {iR,iG,iB} to the values fr,fg,fb
                
                std::string outputvalues = formatstring("%.9f %.9f %.9f\n",fr,fg,fb); 
                fputs(outputvalues.c_str(),f);
            }
        }
    }
    fclose(f);
}

通过 python 算法将 Square Hald clut 转换为 Classic Hald clut

Convert Square Hald clut to Classic Hald clut through a python alghoritm

python

android

opencv

numpy

cube