颜色深度还原的OpenCV和LUT(color depth reduction with openc

我想通过执行颜色深度缩放颜色减少。

像这样的例子：

第一图像是CGA分辨率，第二个是EGA，第三是HAM。我想和CV :: LUT这样做，因为我认为这是做的佰。我可以用这个代码灰阶做：

Mat img = imread("test1.jpg", 0);
uchar* p;
Mat lookUpTable(1, 256, CV_8U);
p = lookUpTable.data;
for( int i = 0; i < 256; ++i)
    p[i] = 16 * (i/16)
LUT(img, lookUpTable, reduced);

原版的：

彩色简化：

但是，如果我试图用颜色做我得到了奇怪的结果..

与此代码：

imgColor = imread("test1.jpg");
Mat reducedColor;
int n = 16;
for (int i=0; i<256; i++) {
    uchar value = floor(i/n) * n;
    cout << (int)value << endl;
    lut.at<Vec3b>(i)[2]= (value >> 16) & 0xff;
    lut.at<Vec3b>(i)[1]= (value >> 8) & 0xff;
    lut.at<Vec3b>(i)[0]= value & 0xff;
} 
LUT(imgColor, lut, reducedColor);

Answer 1:

你可能已经被搬到现在的，但问题的根源是，你正在做一个16位移位uchar value ，这仅仅是8位长。即使在这种情况下，8位移位实在是太多了，因为你会擦除在所有位uchar 。再有是，这样的事实cv::LUT文档明确指出src必须是一个，这显然是不能在代码的情况下“的8位元素的输入数组”。净结果是，只有彩色图像（蓝色通道）的所述第一信道是由转化的cv::LUT 。

解决这些限制的最佳方式是跨渠道分离的彩色图像，分别变换每个信道，然后合并转化通道进入一个新的彩色图像。请参见下面的代码：

/*
Calculates a table of 256 assignments with the given number of distinct values.

Values are taken at equal intervals from the ranges [0, 128) and [128, 256),
such that both 0 and 255 are always included in the range.
*/
cv::Mat lookupTable(int levels) {
    int factor = 256 / levels;
    cv::Mat table(1, 256, CV_8U);
    uchar *p = table.data;

    for(int i = 0; i < 128; ++i) {
        p[i] = factor * (i / factor);
    }

    for(int i = 128; i < 256; ++i) {
        p[i] = factor * (1 + (i / factor)) - 1;
    }

    return table;
}

/*
Truncates channel levels in the given image to the given number of
equally-spaced values.

Arguments:

image
    Input multi-channel image. The specific color space is not
    important, as long as all channels are encoded from 0 to 255.

levels
    The number of distinct values for the channels of the output
    image. Output values are drawn from the range [0, 255] from
    the extremes inwards, resulting in a nearly equally-spaced scale
    where the smallest and largest values are always 0 and 255.

Returns:

Multi-channel images with values truncated to the specified number of
distinct levels.
*/
cv::Mat colorReduce(const cv::Mat &image, int levels) {
    cv::Mat table = lookupTable(levels);

    std::vector<cv::Mat> c;
    cv::split(image, c);
    for (std::vector<cv::Mat>::iterator i = c.begin(), n = c.end(); i != n; ++i) {
        cv::Mat &channel = *i;
        cv::LUT(channel.clone(), table, channel);
    }

    cv::Mat reduced;
    cv::merge(c, reduced);
    return reduced;
}

Answer 2:

既i和n是整数，因此i/n是一个整数。也许你想它转换成（DOUBLE (double)i/n发言并乘以之前） n ？