【C++】使用离散余弦变换滤波算法（DCT）实现
最编程
2024-08-14 13:44:01
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const int height = 128, width = 128, channel = 3;
// DCT hyper-parameter
int T = 8;
int K = 8;
// DCT coefficient
struct dct_str {
double coef[height][width][channel];
};
// Discrete Cosine transformation
dct_str dct(cv::Mat img, dct_str dct_s){
double I;
double F;
double Cu, Cv;
for (int ys = 0; ys < height; ys += T){
for (int xs = 0; xs < width; xs += T){
for (int c = 0; c < channel; c++){
for (int v = 0; v < T; v ++){
for (int u = 0; u < T; u ++){
F = 0;
if (u == 0){
Cu = 1. / sqrt(2);
} else{
Cu = 1;
}
if (v == 0){
Cv = 1. / sqrt(2);
}else {
Cv = 1;
}
for (int y = 0; y < T; y++){
for(int x = 0; x < T; x++){
I = (double)img.at<cv::Vec3b>(ys + y, xs + x)[c];
F += 2. / T * Cu * Cv * I * cos((2. * x + 1) * u * M_PI / 2. / T) * cos((2. * y + 1) * v * M_PI / 2. / T);
}
}
dct_s.coef[ys + v][xs + u][c] = F;
}
}
}
}
}
return dct_s;
}
// Inverse Discrete Cosine transformation
cv::Mat idct(cv::Mat out, dct_str dct_s){
double f;
double Cu, Cv;
for(int ys = 0; ys < height; ys += T){
for(int xs = 0; xs < width; xs += T){
for(int c = 0; c < channel; c++){
for(int y = 0; y < T; y++){
for(int x = 0; x < T; x++){
f = 0;
for (int v = 0; v < K; v++){
for (int u = 0; u < K; u++){
if (u == 0){
Cu = 1. / sqrt(2);
} else {
Cu = 1;
}
if (v == 0){
Cv = 1. / sqrt(2);
} else {
Cv = 1;
}
f += 2. / T * Cu * Cv * dct_s.coef[ys + v][xs + u][c] * cos((2. * x + 1) * u * M_PI / 2. / T) * cos((2. * y + 1) * v * M_PI / 2. / T);
}
}
f = fmin(fmax(f, 0), 255);
out.at<cv::Vec3b>(ys + y, xs + x)[c] = (uchar)f;
}
}
}
}
}
return out;
}
// Main
int main(int argc, const char* argv[]){
// read original image
cv::Mat img = cv::imread("lena.jpg", cv::IMREAD_COLOR);
// DCT coefficient
dct_str dct_s;
// output image
cv::Mat out = cv::Mat::zeros(height, width, CV_8UC3);
// DCT
dct_s = dct(img, dct_s);
// IDCT
out = idct(out, dct_s);
cv::imwrite("out.jpg", out);
//cv::imshow("answer", out);
//cv::waitKey(0);
cv::destroyAllWindows();
```
Python代码如下：
```python
import cv2
import numpy as np
import matplotlib.pyplot as plt
T = 8
K = 8
channel = 3
def w(x, y, u, v):
cu = 1.
cv = 1.
if u == 0:
cu /= np.sqrt(2)
if v == 0:
cv /= np.sqrt(2)
theta = np.pi / (2 * T)
return (( 2 * cu * cv / T) * np.cos((2*x+1)*u*theta) * np.cos((2*y+1)*v*theta))
def dct(img):
H, W, _ = img.shape
F = np.zeros((H, W, channel), dtype=np.float32)
for c in range(channel):
for yi in range(0, H, T):
for xi in range(0, W, T):
for v in range(T):
for u in range(T):
for y in range(T):
for x in range(T):
F[v+yi, u+xi, c] += img[y+yi, x+xi, c] * w(x,y,u,v)
return F
def idct(F):
H, W, _ = F.shape
out = np.zeros((H, W, channel), dtype=np.float32)
for c in range(channel):
for yi in range(0, H, T):
for xi in range(0, W, T):
for y in range(T):
for x in range(T):
for v in range(K):
for u in range(K):
out[y+yi, x+xi, c] += F[v+yi, u+xi, c] * w(x,y,u,v)
out = np.clip(out, 0, 255)
out = np.round(out).astype(np.uint8)
return out
img = cv2.imread("imori.jpg").astype(np.float32)
F = dct(img)
out = idct(F)
cv2.imshow("result", out)
cv2.waitKey(0)
cv2.imwrite("out.jpg", out)