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我正在用 FFT 滤波器(内核大小 = 10)对图像(512*512)进行卷积,看起来不错。

但是当我将它与我以正常方式进行复杂处理的图像进行比较时,结果是可怕的。
PSNR 约为 35。

67,187/262,144 像素值相差 1 或更多(峰值约为 8)(最大像素值为 255)。

我的问题是,在频率空间中进行卷积时是否正常,或者我的卷积/变换函数可能存在问题?. 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。但它完全一样。

当我将图像转换为频率空间时,不要对其进行卷积,然后将其转换回来就可以了,使用浮点数时 PSNR 约为 140。

此外,由于像素差异只有 1-10,我想我可以排除缩放错误

编辑:无聊感兴趣的人的更多详细信息

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

我的图像数据类型是 PixelMatrix。简单的 alpha、red、green 和 blue 值从 0.0 到 1.0 float 的矩阵

我的内核也是一个 PixelMatrix。

这是卷积函数的一些片段

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]

将颜色放入数组:

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:

  • 3个来自kiss_fft_cpx类型的转换颜色数组*
  • 1个来自kiss_fft_cpx类型的转换内核数组*

它们都是复杂的数组

现在是卷积:

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
 }
}

现在我将它们转换回来:

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);
 }
}

笔记:

  • 我已经提前注意图像的大小为 2 (256*256)、(512*512) 等的幂。

例子:

内核大小: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

虽然在我看来它们看起来是一样的°.°

也许一个人很无聊,并通过代码。不是很紧急,但这是一种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一点是我的 PSNR 功能,虽然我并不认为这是问题所在: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;
}
4

2 回答 2

2

Phonon 的想法是正确的。你的图像被转移了。如果您将图像移动 (1,1),则 MSE 将近似为零(前提是您相应地屏蔽或裁剪图像)。我使用下面的代码(Python + OpenCV)确认了这一点。

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]
于 2011-09-06T03:45:14.553 回答
0

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));

该代码可帮助您获得相同大小的图像并有利于去除 DC 分量,您可以进行卷积。

于 2013-08-28T23:22:27.020 回答