将Photoshop sRGB复制到LAB转换

Question

我想要实现的任务是将Photoshop RGB复制到LAB转换.
为简单起见,我将描述我只提取L通道所做的工作.

提取Photoshop的L通道

这是RGB Image,包括所有RGB颜色(请点击下载):

为了提取Photoshop的LAB,我所做的是以下内容:

1. 将图像加载到Photoshop中.
2. 将模式设置为LAB.
3. 在频道面板中选择了L频道.
4. 将模式设置为灰度.
5. 将模式设置为RGB.
6. 保存为PNG.

这是Photoshop的L通道(这正是在LAB模式中选择L通道时在屏幕上看到的):

sRGB到LAB转换

我的主要参考是Bruce Lindbloom伟大的网站.
另外一点是Photoshop在其LAB模式下使用D50 White Point(另请参阅Wikipedia的LAB Color Space Page).

假设RGB图像采用sRGB格式,则转换由下式给出:

sRGB -> XYZ (White Point D65) -> XYZ (White Point D50) -> LAB

假设数据在[0,1]范围内的Float中,则阶段由下式给出:

1. 将sRGB转换为XYZ.
   转换矩阵由RGB - > XYZ矩阵给出(参见sRGB D65).
2. 从XYZ D65
   转换为XYZ D50 使用Chromatic Adaptation Matrix完成转换.由于上一步和这是矩阵乘法,它们可以组合成一个矩阵,它来自sRGB - > XYZ D50(参见RGB底部到XYZ矩阵).请注意,Photoshop使用Bradford Adaptation Method.
3. 从XYZ D50
   转换为LAB 使用XYZ到LAB步骤完成转换.

MATLAB代码

从一开始,我只是在L频道之后,事情变得更简单了.图像被加载到MATLAB中并转换为Float [0,1]范围.

这是代码:

%% Setting Enviorment Parameters

INPUT_IMAGE_RGB             = 'RgbColors.png';
INPUT_IMAGE_L_PHOTOSHOP     = 'RgbColorsL.png';


%% Loading Data

mImageRgb   = im2double(imread(INPUT_IMAGE_RGB));
mImageLPhotoshop     = im2double(imread(INPUT_IMAGE_L_PHOTOSHOP));
mImageLPhotoshop     = mImageLPhotoshop(:, :, 1); %<! All channels are identical


%% Convert to L Channel

mImageLMatlab = ConvertRgbToL(mImageRgb, 1);


%% Display Results
figure();
imshow(mImageLPhotoshop);
title('L Channel - Photoshop');

figure();
imshow(mImageLMatlab);
title('L Channel - MATLAB');

函数ConvertRgbToL()由下式给出:

function [ mLChannel ] = ConvertRgbToL( mRgbImage, sRgbMode )

OFF = 0;
ON  = 1;

RED_CHANNEL_IDX     = 1;
GREEN_CHANNEL_IDX   = 2;
BLUE_CHANNEL_IDX    = 3;

RGB_TO_Y_MAT = [0.2225045, 0.7168786, 0.0606169]; %<! D50

Y_CHANNEL_THR = 0.008856;

% sRGB Compensation
if(sRgbMode == ON)
    vLinIdx = mRgbImage < 0.04045;

    mRgbImage(vLinIdx)  = mRgbImage(vLinIdx) ./ 12.92;
    mRgbImage(~vLinIdx) = ((mRgbImage(~vLinIdx) + 0.055) ./ 1.055) .^ 2.4;
end

% RGB to XYZ (D50)
mY = (RGB_TO_Y_MAT(1) .* mRgbImage(:, :, RED_CHANNEL_IDX)) + (RGB_TO_Y_MAT(2) .* mRgbImage(:, :, GREEN_CHANNEL_IDX)) + (RGB_TO_Y_MAT(3) .* mRgbImage(:, :, BLUE_CHANNEL_IDX));

vYThrIdx = mY > Y_CHANNEL_THR;

mY3 = mY .^ (1 / 3);

mLChannel = ((vYThrIdx .* (116 * mY3 - 16.0)) + ((~vYThrIdx) .* (903.3 * mY))) ./ 100;


end

可以看出结果是不同的.
Photoshop对于大多数颜色来说要暗得多.

有谁知道如何复制Photoshop的LAB转换？
任何人都可以在此代码中发现问题？

谢谢.

Answer 1

最新答案（现在知道是错误的，等待正确答案）

Photoshop 是一个非常古老且混乱的软件。没有明确的文档说明当您执行从一种模式到另一种模式的转换时，为什么像素值会发生这样或那样的情况。

出现问题的原因是，当您在 Adob​​e Photoshop 中将选定的 L* 通道转换为灰度时，伽玛发生了变化。本质上，转换使用 1.74 的伽马值进行单通道到灰度转换。不要问我为什么，我猜这与旧激光打印机有关（？）。

无论如何，这是我发现的最好的方法：

打开文件，将其转为 LAB 模式，仅选择 L 通道

然后转到：

编辑 > 转换为配置文件

您将选择“自定义伽玛”并输入值2.0（不要问我为什么2.0效果更好，我不知道Adobe的软件制造商在想什么......）此操作会将您的图片变成灰度图片只有一个频道

然后你可以将其转换为RGB模式。

如果将结果与您的结果进行比较，您将看到高达 4 点的差异 - 全部位于最暗的区域。

我怀疑这是因为伽玛曲线应用程序不适用于暗值中的 LAB 模式（参见，如您所知，所有低于 0.008856 的 XYZ 值在 LAB 中都是线性的）

结论：

据我所知，Adobe Photoshop中没有正确的实现方法来将L通道从LAB模式提取到灰度模式！

之前的回答

这是我用自己的方法得到的结果：

看起来和Adobe Photoshop 的结果一模一样。

我不确定您这边出了什么问题，因为您描述的步骤与我遵循的步骤完全相同，并且我会建议您遵循。我没有 Matlab，所以我使用了 python：

import cv2, Syn

# your file
fn = "EASA2.png"

#reading the file
im = cv2.imread(fn,-1)

#openCV works in BGR, i'm switching to RGB
im = im[:,:,::-1]

#conversion to XYZ
XYZ = Syn.sRGB2XYZ(im)

#white points D65 and D50
WP_D65 = Syn.Yxy2XYZ((100,0.31271, 0.32902))
WP_D50 = Syn.Yxy2XYZ((100,0.34567, 0.35850))

#bradford
XYZ2 = Syn.bradford_adaptation(XYZ, WP_D65, WP_D50) 

#conversion to L*a*b*
LAB = Syn.XYZ2Lab(XYZ2, WP_D50)

#picking the L channel only
L = LAB[:,:,0] /100. * 255.

#image output
cv2.imwrite("result.png", L)

Syn 库是我自己的东西，以下是函数（抱歉搞得一团糟）：

def sRGB2XYZ(sRGB):

    sRGB = np.array(sRGB)
    aShape = np.array([1,1,1]).shape
    anotherShape = np.array([[1,1,1],[1,1,1]]).shape
    origShape = sRGB.shape

    if sRGB.shape == aShape:
        sRGB = np.reshape(sRGB, (1,1,3))

    elif len(sRGB.shape) == len(anotherShape):
        h,d = sRGB.shape
        sRGB = np.reshape(sRGB, (1,h,d))

    w,h,d = sRGB.shape

    sRGB = np.reshape(sRGB, (w*h,d)).astype("float") / 255.

    m1 = sRGB[:,0] > 0.04045
    m1b = sRGB[:,0] <= 0.04045
    m2 = sRGB[:,1] > 0.04045
    m2b = sRGB[:,1] <= 0.04045
    m3 = sRGB[:,2] > 0.04045
    m3b = sRGB[:,2] <= 0.04045

    sRGB[:,0][m1] = ((sRGB[:,0][m1] + 0.055 ) / 1.055 ) ** 2.4
    sRGB[:,0][m1b] = sRGB[:,0][m1b] / 12.92

    sRGB[:,1][m2] = ((sRGB[:,1][m2] + 0.055 ) / 1.055 ) ** 2.4
    sRGB[:,1][m2b] = sRGB[:,1][m2b] / 12.92

    sRGB[:,2][m3] = ((sRGB[:,2][m3] + 0.055 ) / 1.055 ) ** 2.4
    sRGB[:,2][m3b] = sRGB[:,2][m3b] / 12.92

    sRGB *= 100. 

    X = sRGB[:,0] * 0.4124 + sRGB[:,1] * 0.3576 + sRGB[:,2] * 0.1805
    Y = sRGB[:,0] * 0.2126 + sRGB[:,1] * 0.7152 + sRGB[:,2] * 0.0722
    Z = sRGB[:,0] * 0.0193 + sRGB[:,1] * 0.1192 + sRGB[:,2] * 0.9505

    XYZ = np.zeros_like(sRGB)

    XYZ[:,0] = X
    XYZ[:,1] = Y
    XYZ[:,2] = Z

    XYZ = np.reshape(XYZ, origShape)

    return XYZ

def Yxy2XYZ(Yxy):

    Yxy = np.array(Yxy)
    aShape = np.array([1,1,1]).shape
    anotherShape = np.array([[1,1,1],[1,1,1]]).shape
    origShape = Yxy.shape

    if Yxy.shape == aShape:
        Yxy = np.reshape(Yxy, (1,1,3))

    elif len(Yxy.shape) == len(anotherShape):
        h,d = Yxy.shape
        Yxy = np.reshape(Yxy, (1,h,d))

    w,h,d = Yxy.shape

    Yxy = np.reshape(Yxy, (w*h,d)).astype("float")

    XYZ = np.zeros_like(Yxy)

    XYZ[:,0] = Yxy[:,1] * ( Yxy[:,0] / Yxy[:,2] )
    XYZ[:,1] = Yxy[:,0]
    XYZ[:,2] = ( 1 - Yxy[:,1] - Yxy[:,2] ) * ( Yxy[:,0] / Yxy[:,2] )

    return np.reshape(XYZ, origShape)

def bradford_adaptation(XYZ, Neutral_source, Neutral_destination):
    """should be checked if it works properly, but it seems OK"""

    XYZ = np.array(XYZ)
    ashape = np.array([1,1,1]).shape
    siVal = False

    if XYZ.shape == ashape:


        XYZ = np.reshape(XYZ, (1,1,3))
        siVal = True


    bradford = np.array(((0.8951000, 0.2664000, -0.1614000),
                          (-0.750200, 1.7135000,  0.0367000),
                          (0.0389000, -0.068500,  1.0296000)))

    inv_bradford = np.array(((0.9869929, -0.1470543, 0.1599627),
                              (0.4323053,  0.5183603, 0.0492912),
                              (-.0085287,  0.0400428, 0.9684867)))

    Xs,Ys,Zs = Neutral_source
    s = np.array(((Xs),
                   (Ys),
                   (Zs)))

    Xd,Yd,Zd = Neutral_destination
    d = np.array(((Xd),
                   (Yd),
                   (Zd)))


    source = np.dot(bradford, s)
    Us,Vs,Ws = source[0], source[1], source[2]

    destination = np.dot(bradford, d)
    Ud,Vd,Wd = destination[0], destination[1], destination[2]

    transformation = np.array(((Ud/Us, 0, 0),
                                (0, Vd/Vs, 0),
                                (0, 0, Wd/Ws)))

    M = np.mat(inv_bradford)*np.mat(transformation)*np.mat(bradford)

    w,h,d = XYZ.shape
    result = np.dot(M,np.rot90(np.reshape(XYZ, (w*h,d)),-1))
    result = np.rot90(result, 1)
    result = np.reshape(np.array(result), (w,h,d))

    if siVal == False:
        return result
    else:
        return result[0,0]

def XYZ2Lab(XYZ, neutral):
    """transforms XYZ to CIE Lab
    Neutral should be normalized to Y = 100"""

    XYZ = np.array(XYZ)
    aShape = np.array([1,1,1]).shape
    anotherShape = np.array([[1,1,1],[1,1,1]]).shape
    origShape = XYZ.shape

    if XYZ.shape == aShape:
        XYZ = np.reshape(XYZ, (1,1,3))

    elif len(XYZ.shape) == len(anotherShape):
        h,d = XYZ.shape
        XYZ = np.reshape(XYZ, (1,h,d))

    N_x, N_y, N_z = neutral
    w,h,d = XYZ.shape

    XYZ = np.reshape(XYZ, (w*h,d)).astype("float")

    XYZ[:,0] = XYZ[:,0]/N_x
    XYZ[:,1] = XYZ[:,1]/N_y
    XYZ[:,2] = XYZ[:,2]/N_z

    m1 = XYZ[:,0] > 0.008856
    m1b = XYZ[:,0] <= 0.008856
    m2 = XYZ[:,1] > 0.008856 
    m2b = XYZ[:,1] <= 0.008856
    m3 = XYZ[:,2] > 0.008856
    m3b = XYZ[:,2] <= 0.008856

    XYZ[:,0][m1] = XYZ[:,0][XYZ[:,0] > 0.008856] ** (1/3.0)
    XYZ[:,0][m1b] = ( 7.787 * XYZ[:,0][m1b] ) + ( 16 / 116.0 )

    XYZ[:,1][m2] = XYZ[:,1][XYZ[:,1] > 0.008856] ** (1/3.0)
    XYZ[:,1][m2b] = ( 7.787 * XYZ[:,1][m2b] ) + ( 16 / 116.0 )

    XYZ[:,2][m3] = XYZ[:,2][XYZ[:,2] > 0.008856] ** (1/3.0)
    XYZ[:,2][m3b] = ( 7.787 * XYZ[:,2][m3b] ) + ( 16 / 116.0 )

    Lab = np.zeros_like(XYZ)

    Lab[:,0] = (116. * XYZ[:,1] ) - 16.
    Lab[:,1] = 500. * ( XYZ[:,0] - XYZ[:,1] )
    Lab[:,2] = 200. * ( XYZ[:,1] - XYZ[:,2] )

    return np.reshape(Lab, origShape)