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我要实现在C + + DCT算法,这是我本次代码:问题,以DCT实施
// dct: computes the discrete cosinus tranform of a 8x8 block
template<typename Tin=uchar,typename Tout=float>
inline cv::Mat_<Tout> dct(const cv::Mat_<Tin>& oBlock) {
int indexNumber;
float pi = 3.14159265359;
float fcoscos, fxy, cos1, cos2, forCos1, forCos2;
cv::Mat_<Tout> resultBloc(8, 8);
for (int u = 0; u < oBlock.rows; u++){
for (int v = 0; v < oBlock.cols; v++){
float cu=0, cv=0, Result=0;
// calcul c(u)
if (u == 0){
cu = (float)sqrt((float)1/(float)oBlock.rows);
}
else {
cu = (float)sqrt((float)2/(float)oBlock.rows);
}
// calcul c(v)
if (v == 0){
cv = (float)sqrt((float)1/(float)oBlock.cols);
}
else {
cv = (float)sqrt((float)2/(float)oBlock.cols);
}
float sums = 0;
for (int x = 0; x < oBlock.rows; x++){
for (int y = 0; y < oBlock.cols; y++){
indexNumber = x * oBlock.rows + y;
fxy = (int)oBlock.data[indexNumber];
forCos1 = (pi*((2 * x) + 1)*u)/(2 * oBlock.rows);
forCos2 = (pi*((2 * y) + 1)*v)/(2 * oBlock.cols);
cos1 = cos(forCos1);
cos2 = cos(forCos2);
fcoscos = fxy * cos1 * cos2;
sums += fcoscos;
}
}
// calcul total
Result = sums*cu*cv;
indexNumber = u * oBlock.rows + v;
resultBloc.data[indexNumber] = Result;
}
}
return resultBloc;
}
我比较了CV DCT算法如下结果:
cv::Mat_<float> tempImage(8,8);
for (int i = 0; i < vecImageCut[0].cols*vecImageCut[0].rows; i++){
tempImage.data[i] = (int)vecImageCut[0].data[i];
}
cv::Mat_<float> dctCV;
cv::dct(tempImage, dctCV);
for (int i = 0; i < blocksAfterDCT[0].cols*blocksAfterDCT[0].rows; i++){
std::cerr << "Difference DCT for pixel " << i << " : " << dctCV.data[i] - blocksAfterDCT[0].data[i] << std::endl;
}
DCT和cv DCT之间的结果非常不同,所以我认为我的DCT算法是错误的,但我搜索了几个小时,找不到我的错误,谁能告诉我我在哪里做错了什么?
的OpenCV的DCT使用DFT计算,如下所述: HTTP: //www.ece.utexas.edu/~bevans/courses/ee381k/lectures/09_DCT – sturkmen
其中** DCT **?其中有更多** DCT-I,II,III,IV **请确保你比较相同的东西。另请参阅[快速DCT](http://stackoverflow.com/a/22779268/2521214)。它也使用** FFT **,但您可以通过在行上使用** 1D DCT **来执行** 2D DCT **转置...执行列...转置...正常化。 1D DCT更容易这个过程也比直接2D计算更快,而且更容易调试。如果您需要** FFT **帮助,请参见[如何计算离散傅立叶变换?](http://stackoverflow.com/a/26355569/2521214)。 – Spektre