我想在此paper中实现颜色转移算法,并引用此tutorial在OpenCV C++中转移该算法。
但是我得到了一些奇怪的结果,例如:
This是源图像,this是目标图像,但是合并后的结果看起来像this。
结果的某些部分看起来很奇怪。
这是我的源代码
Mat src; Mat tar; Mat result;
class imageInfo{
public:
double lMean, lStd, aMean, aStd, bMean, bStd;
};
/// Function header
void image_stats(Mat img,imageInfo *info);
/** @function main */
int main(int argc, char** argv)
{
vector<Mat> mv;
imageInfo srcInfo, tarInfo;
src = imread("images/autumn.jpg");
tar = imread("images/fallingwater.jpg");
imshow("src", src);
imshow("tar", tar);
cvtColor(src, src, CV_BGR2Lab);
cvtColor(tar, tar, CV_BGR2Lab);
image_stats(src, &srcInfo);
image_stats(tar, &tarInfo);
split(tar, mv);
Mat l = mv[0];
Mat a = mv[1];
Mat b = mv[2];
/*pixel color modify*/
for (int i = 0; i<l.rows; i++){
for (int j = 0; j<l.cols; j++){
double li = l.data[l.step[0] * i + l.step[1] * j];
if (i == 426 && j == 467)
cout << "i:" << i << "j:" << j << " " << li << endl;
li -= tarInfo.lMean;
li = (tarInfo.lStd / srcInfo.lStd)*li;
li += srcInfo.lMean;
li = (int)li % 256;
l.data[l.step[0] * i + l.step[1] * j] = li;
}
}
for (int i = 0; i<a.rows; i++){
for (int j = 0; j<a.cols; j++){
double ai = a.data[a.step[0] * i + a.step[1] * j];
ai -= tarInfo.aMean;
ai = (tarInfo.aStd / srcInfo.aStd)*ai;
ai += srcInfo.aMean;
ai = (int)ai % 256;
a.data[a.step[0] * i + a.step[1] * j] = ai;
}
}
for (int i = 0; i<b.rows; i++){
for (int j = 0; j<b.cols; j++){
double bi = b.data[b.step[0] * i + b.step[1] * j];
bi -= tarInfo.bMean;
bi = (tarInfo.bStd / srcInfo.bStd)*bi;
bi += srcInfo.bMean;
bi = (int)bi % 256;
b.data[b.step[0] * i + b.step[1] * j] = bi;
}
}
mv.clear();
mv.push_back(l);
mv.push_back(a);
mv.push_back(b);
merge(mv, result);
cvtColor(result, result, CV_Lab2BGR);
imshow("result", result);
imwrite("result.png", result);
waitKey(0);
return(0);
}
image_stats函数:
void image_stats(Mat img, imageInfo *info){
int Max=0;
vector<Mat> mv;
vector<int> vl, va, vb;
split(img, mv);
Mat l = mv[0];
Mat a = mv[1];
Mat b = mv[2];
/*statistics L space*/
for (int i = 0; i<l.rows; i++){
for (int j = 0; j<l.cols; j++){
int li = l.data[l.step[0] * i + l.step[1] * j];
vl.push_back(li);
}
}
double sum_l = std::accumulate(vl.begin(), vl.end(), 0.0);
double mean_l = sum_l / vl.size();
std::vector<double> diff_l(vl.size());
std::transform(vl.begin(), vl.end(), diff_l.begin(),
std::bind2nd(std::minus<double>(), mean_l));
double sq_sum_l = std::inner_product(diff_l.begin(), diff_l.end(), diff_l.begin(), 0.0);
double stdev_l = std::sqrt(sq_sum_l / vl.size());
info->lMean = mean_l;
info->lStd = stdev_l;
/*statistics A space*/
for (int i = 0; i<a.rows; i++){
for (int j = 0; j<a.cols; j++){
int ai = a.data[a.step[0] * i + a.step[1] * j];
va.push_back(ai);
}
}
double sum_a = std::accumulate(va.begin(), va.end(), 0.0);
double mean_a = sum_a / va.size();
std::vector<double> diff_a(va.size());
std::transform(va.begin(), va.end(), diff_a.begin(),
std::bind2nd(std::minus<double>(), mean_a));
double sq_sum_a = std::inner_product(diff_a.begin(), diff_a.end(), diff_a.begin(), 0.0);
double stdev_a = std::sqrt(sq_sum_a / va.size());
info->aMean = mean_a;
info->aStd = stdev_a;
/*statistics B space*/
for (int i = 0; i<b.rows; i++){
for (int j = 0; j<b.cols; j++){
int bi = b.data[b.step[0] * i + b.step[1] * j];
vb.push_back(bi);
}
}
double sum_b = std::accumulate(vb.begin(), vb.end(), 0.0);
double mean_b = sum_b / vb.size();
std::vector<double> diff_b(vb.size());
std::transform(vb.begin(), vb.end(), diff_b.begin(),
std::bind2nd(std::minus<double>(), mean_b));
double sq_sum_b = std::inner_product(diff_b.begin(), diff_b.end(), diff_b.begin(), 0.0);
double stdev_b = std::sqrt(sq_sum_b / vb.size());
info->bMean = mean_b;
info->bStd = stdev_b;
}
最佳答案
在主要功能中,您可能应将结果值限制为上限和下限[0; 255],而不是取模。如果是li = 256;
,则代码li = (int)li % 256;
将其设为零。
关于c++ - 颜色转移:我在OpenCV C++中的代码有什么问题,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/30570628/