c - 离散余弦变换DCT实现C

标签 c dct

我试图在 C 中实现正向和反向离散余弦变换 (DCT)。代码是通过 dct() 函数将单个输入像素 block 转换为转换矩阵,然后通过 dct() 函数返回原始像素值idct() 函数。请参阅随附的代码。我的 idct 输出是 244、116、244、116 等的连续值。从 idct 值的外观来看,我的程序似乎没有运行。有人可以帮我看看我的结果是什么吗应该期待每个功能后?显然在 idct 之后,我应该非常接近原始输入矩阵。

谢谢

 # include <stdio.h>
 # define PI 3.14

void dct(float [][]);       // Function prototypes
void idct(float [][]);     // Function prototypes

void dct(float inMatrix[8][8]){

    double dct,
    Cu,
    sum,
    Cv;

    int i,
    j,
    u,
    h = 0,
    v;

    FILE * fp = fopen("mydata.csv", "w");

    float dctMatrix[8][8],
    greyLevel;                       

    for (u = 0; u < 8; ++u) {
        for (v = 0; v < 8; ++v) {

            if (u == 0) {
                Cu = 1.0 / sqrt(2.0);
            } else {
                Cu = 1.0;
            }

            if (v == 0) {
                Cv = 1.0 / sqrt(2.0);
            } else {
                Cu = (1.0);
            }   

            sum = 0.0;  

            for (i = 0; i < 8; i++) {
                for (j = 0; j < 8; j++) {

                    // Level around 0
                    greyLevel = inMatrix[i][j];

                    dct = greyLevel * cos((2 * i + 1) * u * PI / 16.0) *
                        cos((2 * j + 1) * v * PI / 16.0);

                    sum += dct;

                }               
            }
            dctMatrix[u][v] = 0.25 * Cu * Cv * sum;
            fprintf(fp, "\n %f", dctMatrix[u][v]);          
        }
        fprintf(fp, "\n");
    }  
    idct(dctMatrix);  
 }

void idct(float dctMatrix[8][8]){

    double idct,
    Cu,
    sum,
    Cv;

    int i,
    j,
    u,
    v;

    float idctMatrix[8][8],
    greyLevel;

    FILE * fp = fopen("mydata.csv", "a");

    fprintf(fp, "\n Inverse DCT");                     

    for (i = 0; i < 8; ++i) {
        for (j = 0; j < 8; ++j) { 

            sum = 0.0;  

        for (u = 0; u < 8; u++) {
            for (v = 0; v < 8; v++) {

            if (u == 0) {
                Cu = 1.0 / sqrt(2.0);
            } else {
                Cu = 1.0;
              }

            if (v == 0) {
                Cv = 1.0 / sqrt(2.0);
            } else {
                Cu = (1.0);
              }   

                    // Level around 0
                greyLevel = dctMatrix[u][v];

                idct = (greyLevel * cos((2 * i + 1) * u * PI / 16.0) *
                        cos((2 * j + 1) * v * PI / 16.0));

                sum += idct;

                }               
            }
            idctMatrix[i][j] = 0.25 * Cu * Cv * sum;
            fprintf(fp, "\n %f", idctMatrix[i][j]);         
        }
        fprintf(fp, "\n");
    }    
 }


int main() {

   float    
    testBlockA[8][8] = { {255, 255, 255, 255, 255, 255, 255, 255},
                         {255, 255, 255, 255, 255, 255, 255, 255},
                         {255, 255, 255, 255, 255, 255, 255, 255},
                         {255, 255, 255, 255, 255, 255, 255, 255},
                         {255, 255, 255, 255, 255, 255, 255, 255},
                         {255, 255, 255, 255, 255, 255, 255, 255},
                         {255, 255, 255, 255, 255, 255, 255, 255},
                         {255, 255, 255, 255, 255, 255, 255, 255} },

    testBlockB[8][8] = {{255, 0, 255, 0, 255, 0, 255, 0},
                        {0, 255, 0, 255, 0, 255, 0, 255},
                        {255, 0, 255, 0, 255, 0, 255, 0},
                        {0, 255, 0, 255, 0, 255, 0, 255},
                        {255, 0, 255, 0, 255, 0, 255, 0},
                        {0, 255, 0, 255, 0, 255, 0, 255},
                        {255, 0, 255, 0, 255, 0, 255, 0},
                        {0, 255, 0, 255, 0, 255, 0, 255} };

    dct(testBlockB);
}

最佳答案

if 语句中的 Cv 常量赋值至少有两个拼写错误:

    if (v == 0) {
        Cv = 1.0 / sqrt(2.0);
    } else {
        Cu = (1.0); // << this should be Cv = 1.0
    }   

虽然没有检查得太正确。使用 german wikipedia关于余弦变换,以下代码有效... 我不想花时间弄清楚你是如何定义转换常量的。 我想您需要确保使用正确的常量和反函数:

#include <stdio.h>
#include <math.h>
#include <stdlib.h>

void dct(float **DCTMatrix, float **Matrix, int N, int M);
void write_mat(FILE *fp, float **testRes, int N, int M);
void idct(float **Matrix, float **DCTMatrix, int N, int M);
float **calloc_mat(int dimX, int dimY);
void free_mat(float **p);


float **calloc_mat(int dimX, int dimY){
    float **m = calloc(dimX, sizeof(float*));
    float *p = calloc(dimX*dimY, sizeof(float));
    int i;
    for(i=0; i <dimX;i++){
    m[i] = &p[i*dimY];

    }
   return m;
}

void free_mat(float **m){
  free(m[0]);
  free(m);
}

void write_mat(FILE *fp, float **m, int N, int M){

   int i, j;
   for(i =0; i< N; i++){
    fprintf(fp, "%f", m[i][0]);
    for(j = 1; j < M; j++){
       fprintf(fp, "\t%f", m[i][j]);
        }   
    fprintf(fp, "\n");
   }
   fprintf(fp, "\n");
}

void dct(float **DCTMatrix, float **Matrix, int N, int M){

    int i, j, u, v;
    for (u = 0; u < N; ++u) {
        for (v = 0; v < M; ++v) {
        DCTMatrix[u][v] = 0;
            for (i = 0; i < N; i++) {
                for (j = 0; j < M; j++) {
                    DCTMatrix[u][v] += Matrix[i][j] * cos(M_PI/((float)N)*(i+1./2.)*u)*cos(M_PI/((float)M)*(j+1./2.)*v);
                }               
            }
        }
    }  
 }

void idct(float **Matrix, float **DCTMatrix, int N, int M){
    int i, j, u, v;

    for (u = 0; u < N; ++u) {
        for (v = 0; v < M; ++v) {
          Matrix[u][v] = 1/4.*DCTMatrix[0][0];
          for(i = 1; i < N; i++){
          Matrix[u][v] += 1/2.*DCTMatrix[i][0];
           }
           for(j = 1; j < M; j++){
          Matrix[u][v] += 1/2.*DCTMatrix[0][j];
           }

           for (i = 1; i < N; i++) {
                for (j = 1; j < M; j++) {
                    Matrix[u][v] += DCTMatrix[i][j] * cos(M_PI/((float)N)*(u+1./2.)*i)*cos(M_PI/((float)M)*(v+1./2.)*j);
                }               
            }
        Matrix[u][v] *= 2./((float)N)*2./((float)M);
        }
    }  
 }



int main() {

   float    
    testBlockA[8][8] = { {255, 255, 255, 255, 255, 255, 255, 255},
                         {255, 255, 255, 255, 255, 255, 255, 255},
                         {255, 255, 255, 255, 255, 255, 255, 255},
                         {255, 255, 255, 255, 255, 255, 255, 255},
                         {255, 255, 255, 255, 255, 255, 255, 255},
                         {255, 255, 255, 255, 255, 255, 255, 255},
                         {255, 255, 255, 255, 255, 255, 255, 255},
                         {255, 255, 255, 255, 255, 255, 255, 255} },

    testBlockB[8][8] = {{255, 0, 255, 0, 255, 0, 255, 0},
                        {0, 255, 0, 255, 0, 255, 0, 255},
                        {255, 0, 255, 0, 255, 0, 255, 0},
                        {0, 255, 0, 255, 0, 255, 0, 255},
                        {255, 0, 255, 0, 255, 0, 255, 0},
                        {0, 255, 0, 255, 0, 255, 0, 255},
                        {255, 0, 255, 0, 255, 0, 255, 0},
                        {0, 255, 0, 255, 0, 255, 0, 255} };

    FILE * fp = fopen("mydata.csv", "w");
    int dimX = 8, dimY = 8;
    int i, j;

    float **testBlock = calloc_mat(dimX, dimY);
    float **testDCT = calloc_mat(dimX, dimY);
    float **testiDCT = calloc_mat(dimX, dimY);

    for(i = 0; i<dimX; i++){
      for(j = 0; j<dimY; j++){
        testBlock[i][j] = testBlockB[i][j];
      }
    }

    dct(testDCT, testBlock, dimX, dimY);
    write_mat(fp, testDCT, dimX, dimY);

    idct(testiDCT, testDCT, dimX, dimY);
    write_mat(fp, testiDCT, dimX, dimY);

    fclose(fp);
    free_mat(testBlock);
    free_mat(testDCT);
    free_mat(testiDCT);

    return 0;
}

编辑 dct 基于 wiki 中公式 DCT-II 的叉积。 idct 基于公式 DCT-III 与每个维度的归一化因子 2/N 的叉积(因为这是文本中提到的 DCT-II 的逆)。 编辑 我很确定在您的版本中,逆 dct 中的因子应该是 sqrt(2) 而不是 1/sqrt(2)。

关于c - 离散余弦变换DCT实现C,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/8310749/

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