将OpenCV BGR 8位图像转换为CIE L * a * b * [英] Converting an OpenCV BGR 8-bit Image to CIE L*a*b*

问题描述

我试图将表示具有8位深度的 RGB 图像的给定 Mat 转换为 Lab 文档中提供的函数：

I am trying to convert a given Mat representing an RGB image with 8-bit depth to Lab using the function provided in the documentation:

cvtColor(source, destination, <conversion code>);

我尝试过以下转换代码：

I have tried the following conversion codes:

CV_RGB2Lab
CV_BGR2Lab
CV_LBGR2Lab

我每次都收到奇怪的结果，对于一些样品，L值大于100，字面上<107,125,130>。

I have received bizarre results each time around, with an "L" value of greater than 100 for some samples, literally <107, 125, 130>.

I我也使用Photoshop检查结果 - 但考虑到107超出了接受范围0≤L≤100，我不能理解我的错误是什么。

I am also using Photoshop to check the results - but given that 107 is beyond the accepted range of 0 ≤ L ≤ 100, I can not comprehend what my error is.

更新：
我将在这里发布我的整体结果：
给定由8位BGR表示的图像（Mat），图像可以通过以下转换：

Update: I'll post my overall results here: Given an image (Mat) represented by 8-bit BGR, the image can be converted by the following:

cvtColor(source, destination, CV_BGR2Lab);

然后可以通过以下方式访问像素值：

The pixel values can then be accessed in the following manner:

int step = destination.step;
int channels = destination.channels();
for (int i = 0; i < destination.rows(); i++) {
    for (int j = 0; j < destination.cols(); j++) {
        Point3_<uchar> pixelData;
        //L*: 0-255 (elsewhere is represented by 0 to 100)
        pixelData.x = destination.data[step*i + channels*j + 0];
        //a*: 0-255 (elsewhere is represented by -127 to 127)
        pixelData.y = destination.data[step*i + channels*j + 1];
        //b*: 0-255 (elsewhere is represented by -127 to 127)
        pixelData.z = destination.data[step*i + channels*j + 2];
    }
}

推荐答案

因为 L 值在OpenCV中的范围 [0..255] 。您可以简单地将此值缩放到所需的间隔（ [0..100] ）。

That's because L value is in range [0..255] in OpenCV. You can simply scale this value to needed interval ([0..100] in your case).

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