Question

我用FFT滤镜（kernelsize = 10）卷积图像（512 * 512），看起来不错。

但是当我把它与一个我用正常方式卷曲的图像进行比较时，结果很糟糕 PSNR约为35。

67,187 / 262,144像素值的差异为1或更大（峰值在~8）（最大像素值为255）。

我的问题是，在频率空间中卷积时是否正常，或者我的卷积/转换功能是否存在问题？。因为奇怪的是当我使用double作为数据类型时我应该得到更好的结果。但它完全保持不变。

当我将图像转换为频率空间时，请勿将其旋转，然后将其转换回来，使用浮点数时PSNR大约为140.

此外，由于像素差异仅为1-10，我认为我可以排除缩放错误

编辑：~~无聊的~~感兴趣的人的更多详细信息

我使用开源kissFFT库。使用真正的二维输入（kiss_fftndr.h）

我的图像数据类型是PixelMatrix。只需一个alpha，红色，绿色和蓝色值的矩阵，从0.0到1.0浮动

我的内核也是PixelMatrix。

这里是Convolution功能的一些片段

使用的数据类型：

#define kiss_fft_scalar float
#define kiss_fft_cpx struct {
    kiss_fft_scalar r;
    kiss_fft_scalar i,
}

FFT的配置：

//parameters to kiss_fftndr_alloc:
//1st param = array with the size of the 2 dimensions (in my case dim={width, height})
//2nd param = count of the dimensions (in my case 2)
//3rd param = 0 or 1 (forward or inverse FFT)
//4th and 5th params are not relevant

kiss_fftndr_cfg stf = kiss_fftndr_alloc(dim, 2, 0, 0, 0);
kiss_fftndr_cfg sti = kiss_fftndr_alloc(dim, 2, 1, 0, 0);

填充和转换内核：

I make a new array:

kiss_fft_scalar kernel[width*height];

I fill it with 0 in a loop.

Then I fill the middle of this array with the kernel I want to use.
So if I would use a 2*2 kernel with values 1/4, 1/4, 1/4 and 1/4 it would look like

0 0 0 0 0 0
0 1/4 1/4 0
0 1/4 1/4 0
0 0 0 0 0 0

The zeros are padded until they reach the size of the image.

Then I swap the quadrants of the image diagonally. It looks like:

1/4 0 0 1/4
 0  0 0  0
 0  0 0  0
1/4 0 0 1/4

now I transform it: kiss_fftndr(stf, floatKernel, outkernel);

outkernel is declarated as 
kiss_fft_cpx outkernel= new kiss_fft_cpx[width*height]

将颜色变为数组：

kiss_fft_scalar *red = new kiss_fft_scalar[width*height];
kiss_fft_scalar *green = new kiss_fft_scalar[width*height];
kiss_fft-scalar *blue = new kiss_fft_scalar[width*height];

for(int i=0; i<height; i++) {
 for(int j=0; i<width; j++) {
  red[i*height+j] = input.get(j,i).getRed();  //input is the input image pixel matrix
  green[i*height+j] = input.get(j,i).getGreen();
  blue{i*height+j] = input.get(j,i).getBlue();
 }
}

Then I transform the arrays:

kiss_fftndr(stf, red, outred);
kiss_fftndr(stf, green, outgreen);
kiss_fftndr(stf, blue, outblue);      //the out-arrays are type kiss_fft_cpx*

卷积：

我们现在拥有的：

来自kiss_fft_cpx *
1个从kiss_fft_cpx *

它们都是复杂数组

现在出现了卷积：

for(int m=0; m<til; m++) {
 for(int n=0; n<til; n++) {
  kiss_fft_scalar real = outcolor[m*til+n].r;      //I do that for all 3 arrys in my code!
  kiss_fft_scalar imag = outcolor[m*til+n].i;      //so I have realred, realgreen, realblue
  kiss_fft_scalar realMask = outkernel[m*til+n].r; // and imagred, imaggreen, etc.
  kiss_fft_scalar imagMask = outkernel[m*til+n].i;

  outcolor[m*til+n].r = real * realMask - imag * imagMask; //Same thing here in my code i
  outcolor[m*til+n].i = real * imagMask + imag * realMask; //do it with all 3 colors
 }
}

现在我把它们变回来了：

kiss_fftndri(sti, outred, red);
kiss_fftndri(sti, outgreen, green);
kiss_fftndri(sti, outblue, blue);

and I create a new Pixel Matrix with the values from the color-arrays

PixelMatrix output;

for(int i=0; i<height; i++) {
 for(int j=0; j<width; j++) {
  Pixel p = new Pixel();
  p.setRed( red[i*height+j] / (width*height) ); //I divide through (width*height) because of the scaling happening in the FFT;
  p.setGreen( green[i*height+j] );
  p.setBlue( blue[i*height+j] );
  output.set(j , i , p);
 }
}

注意：

我已经提前注意图像的大小为2（256 * 256），（512 * 512）等。

示例：

kernelsize：10

输入：

输出：

正常卷积的输出：

我的控制台说：

142519 out of 262144 Pixels have a difference of 1 or more (maxRGB = 255)

PSNR: 32.006027221679688
MSE: 44.116752624511719

虽然我的眼睛看起来相同°。°

也许一个人感到无聊并且通过代码。这不紧急，但这是一个问题，我只是想知道我到底做错了什么^^

最后但并非最不重要的是，我的PSNR功能，虽然我不认为这是问题：D

void calculateThePSNR(const PixelMatrix first, const PixelMatrix second, float* avgpsnr, float* avgmse) {

int height = first.getHeight();
int width = first.getWidth();

BMP firstOutput;
BMP secondOutput;

firstOutput.SetSize(width, height);
secondOutput.SetSize(width, height);

double rsum=0.0, gsum=0.0, bsum=0.0;
int count = 0;
int total = 0;
for(int i=0; i<height; i++) {
    for(int j=0; j<width; j++) {
        Pixel pixOne = first.get(j,i);
        Pixel pixTwo = second.get(j,i);

        double redOne = pixOne.getRed()*255;
        double greenOne = pixOne.getGreen()*255;
        double blueOne = pixOne.getBlue()*255;

        double redTwo = pixTwo.getRed()*255;
        double greenTwo = pixTwo.getGreen()*255;
        double blueTwo = pixTwo.getBlue()*255;

        firstOutput(j,i)->Red = redOne;
        firstOutput(j,i)->Green = greenOne;
        firstOutput(j,i)->Blue = blueOne;

        secondOutput(j,i)->Red = redTwo;
        secondOutput(j,i)->Green = greenTwo;
        secondOutput(j,i)->Blue = blueTwo;

        if((redOne-redTwo) > 1.0 || (redOne-redTwo) < -1.0) {
            count++;
        }
        total++;

        rsum += (redOne - redTwo) * (redOne - redTwo);
        gsum += (greenOne - greenTwo) * (greenOne - greenTwo);
        bsum += (blueOne - blueTwo) * (blueOne - blueTwo);

    }
}
fprintf(stderr, "%d out of %d Pixels have a difference of 1 or more (maxRGB = 255)", count, total);
double rmse = rsum/(height*width);
double gmse = gsum/(height*width);
double bmse = bsum/(height*width);

double rpsnr = 20 * log10(255/sqrt(rmse));
double gpsnr = 20 * log10(255/sqrt(gmse));
double bpsnr = 20 * log10(255/sqrt(bmse));

firstOutput.WriteToFile("test.bmp");
secondOutput.WriteToFile("test2.bmp");

system("display test.bmp");
system("display test2.bmp");

*avgmse = (rmse + gmse + bmse)/3;
*avgpsnr = (rpsnr + gpsnr + bpsnr)/3;
}

Answer 1

Phonon有正确的想法。你的图像被移动了。如果您将图像移动（1,1），则MSE将近似为零（假设您相应地屏蔽或裁剪图像）。我使用下面的代码（Python + OpenCV）证实了这一点。

import cv
import sys
import math

def main():
    fname1, fname2 = sys.argv[1:]
    im1 = cv.LoadImage(fname1)
    im2 = cv.LoadImage(fname2)

    tmp = cv.CreateImage(cv.GetSize(im1), cv.IPL_DEPTH_8U, im1.nChannels)
    cv.AbsDiff(im1, im2, tmp)
    cv.Mul(tmp, tmp, tmp)
    mse = cv.Avg(tmp)
    print 'MSE:', mse

    psnr = [ 10*math.log(255**2/m, 10) for m in mse[:-1] ]
    print 'PSNR:', psnr

if __name__ == '__main__':
    main()

输出：

MSE: (0.027584912741602553, 0.026742391458366047, 0.028147870144492403, 0.0)
PSNR: [63.724087463606452, 63.858801190963192, 63.636348220531396]

Answer 2

我建议您尝试实现以下代码：

A=double(inputS(1:10:length(inputS))); %segmentation 
A(:)=-A(:);
%process the image or signal by fast fourior transformation and inverse fft
fresult=fft(inputS);
fresult(1:round(length(inputS)*2/fs))=0;
fresult(end-round(length(fresult)*2/fs):end)=0;
Y=real(ifft(fresult));

该代码可以帮助您获得相同大小的图像并且有助于删除DC分量，您可以进行卷积。

FFT卷积 - 真的很低的PSNR

2 个答案: