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c++ - Opencv average filter gives different output than Matlab average filter

问题描述

The following OpenCV code is written to serve as a diagnostic to compare average filter implementation in Matlab and OpenCV. The OpenCV code is 

Mat P(Size(5,5),CV_64FC1,Scalar(0));

for(int i = 0; i < 5; i++)
{
    for (int j = 0; j < 5 ; j++)
        P.at<double>(i,j) = i;
}


cout<<"Original Matrix is :"<<endl;
cout<<P<<endl;

Mat averageFilter(2,2,CV_64FC1,Scalar(0)),U;
averageFilter = cv::Scalar::all(1.0/(2*2));


filter2D(P, U, -1 , averageFilter, Point( -1, -1 ), 0, BORDER_REPLICATE );
cout<<"Filtered Matrix is :"<<endl;
cout<<U<<endl;

The output is 

Original Matrix is :
[0, 0, 0, 0, 0;
  1, 1, 1, 1, 1;
  2, 2, 2, 2, 2;
  3, 3, 3, 3, 3;
  4, 4, 4, 4, 4]
Filtered Matrix is :
[0, 0, 0, 0, 0;
  0.5, 0.5, 0.5, 0.5, 0.5;
  1.5, 1.5, 1.5, 1.5, 1.5;
  2.5, 2.5, 2.5, 2.5, 2.5;
  3.5, 3.5, 3.5, 3.5, 3.5]

Matlab code to replicate the same operation is : 

ma = [0 0 0 0 0;1 1 1 1 1;2 2 2 2 2;3 3 3 3 3;4 4 4 4 4];
MEANF = fspecial('average',[2 2]);
U = imfilter(ma, MEANF, 'replicate');

The output is 

U =

    0.5000    0.5000    0.5000    0.5000    0.5000
    1.5000    1.5000    1.5000    1.5000    1.5000
    2.5000    2.5000    2.5000    2.5000    2.5000
    3.5000    3.5000    3.5000    3.5000    3.5000
    4.0000    4.0000    4.0000    4.0000    4.0000

What is the reason for discrepancy between the two outputs ?

标签： c++matlabopencvimage-processing

解决方案

您看到的差异是由于对均匀大小过滤器中原点位置的不同选择。对于奇数大小的内核，软件倾向于一致并选择中间像素作为原点。但对于偶数大小的内核，有两种选择同样有意义。

我可以通过更改滤波器的原点来复制 MATLAB 中的 OpenCV 输出：

ma = repmat((0:4).',1,5);
filt = zeros(3);
filt(2:3,2:3) = 1/4;
U = imfilter(ma,filt,'replicate')
filt = rot90(filt,2);
V = imfilter(ma,filt,'replicate')

这给了我U和你一样的东西：

U =
    0.5000    0.5000    0.5000    0.5000    0.5000
    1.5000    1.5000    1.5000    1.5000    1.5000
    2.5000    2.5000    2.5000    2.5000    2.5000
    3.5000    3.5000    3.5000    3.5000    3.5000
    4.0000    4.0000    4.0000    4.0000    4.0000

这V就是你在 OpenCV 中看到的：

V =
         0         0         0         0         0
    0.5000    0.5000    0.5000    0.5000    0.5000
    1.5000    1.5000    1.5000    1.5000    1.5000
    2.5000    2.5000    2.5000    2.5000    2.5000
    3.5000    3.5000    3.5000    3.5000    3.5000

更有指导意义的是创建一个除中间一个值外全为零的输入：

ma = zeros(5);
ma(3,3) = 1;
filt = zeros(3);
filt(2:3,2:3) = 1/4;
U = imfilter(ma,filt,'replicate')
filt = rot90(filt,2);
V = imfilter(ma,filt,'replicate')

现在我明白了：

U =
         0         0         0         0         0
         0    0.2500    0.2500         0         0
         0    0.2500    0.2500         0         0
         0         0         0         0         0
         0         0         0         0         0

V =
         0         0         0         0         0
         0         0         0         0         0
         0         0    0.2500    0.2500         0
         0         0    0.2500    0.2500         0
         0         0         0         0         0

很明显，内核移动了一个像素。同样，fspecial('average',[2 2])你会得到 的结果U，如果你在 OpenCV 中复制它，你会看到类似的输出V。

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