不使用conv2的Matlab图像过滤 [英] Matlab image filtering without using conv2

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本文介绍了不使用conv2的Matlab图像过滤的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我被赋予了为3x3矩阵创建图像过滤功能的任务,其结果必须等于conv2的结果.我已经编写了此函数,但是它过滤的图像不正确:

I've been given a task to create image filtering function for 3x3 matrices, and its outcome must be equal to conv2's. I have written this function, but it filters image incorrectly:

function [ image ] = Func134( img,matrix )
  image=img;
  len=length(img)
  for i=2:1:len-1
    for j=2:1:len-1
      value=0;
      for g=-1:1:1
        for l=-1:1:1
          value=value+img(i+g,j+l)*matrix(g+2,l+2);
        end
      end
     image(i,j)=value;
    end
  end
i=1:1:length
image(i,1)=image(i,2)
image(i,len)=image(i,len-1)
image(1,i)=image(2,i)
image(len,i)=image(len-1,i)
end

过滤矩阵为[3,10,3; 0,0,0; ​​-3,-10,-3]

Filtration matrix is [3,10,3;0,0,0;-3,-10,-3]

请帮助找出我的代码出了什么问题.

Please help to figure out what is wrong with my code.

我在conv2和我的代码之间得到的一些示例结果如下所示.

Some sample results I get between conv2 and my code are seen below.

推荐答案

首先,这行没有意义:

i=1:1:length;

我认为您打算使用len而不是length作为结束索引:

I think you meant to use len instead of length as the ending index:

i=1:1:len;

现在引用您的代码,这是正确的,但是您正在做的是相关而不是卷积.在2D卷积中,必须对内核/蒙版进行180度旋转,然后进行加权和.因此,如果要使用conv2获得相同的结果,则必须在旋转遮罩之前对其进行预旋转.

Now referring to your code, it is correct, but what you are doing is correlation not convolution. In 2D convolution, you have to perform a 180 degree rotation of the kernel / mask and then do the weighted sum. As such, if you want to achieve the same results using conv2, you must pre-rotate the mask before calling it.

mask = [3,10,3;0,0,0;-3,-10,-3]
mask_flip = mask(end:-1:1,end:-1:1);
out = conv2(img, mask, 'same');

mask_flip包含180度旋转的内核.我们使用'same'标志来确保结果的输出大小与输入大小相同.但是,当使用conv2时,我们假定图像的边界为零填充.您的代码只需将原始图像的边界像素复制到结果图像中.这称为复制行为,但这不是conv2本身所做的. conv2假定边框像素像我之前提到的那样被零填充,所以我建议您创建两个额外的图像,一个是输出图像,该图像具有多两行和两列,另一个是输入图像大小与输出图像相同,但是您将输入图像放置在此矩阵内.接下来,对该新图像执行过滤,将得到的过滤后的像素放置在输出图像中,然后裁剪该结果.我决定创建一个新的填充输入图像,以使您的大部分代码保持完整.

mask_flip contains the 180 degree rotated kernel. We use the 'same' flag to ensure that the output size of the result is the same size as the input. However, when using conv2, we are assuming that the borders of the image are zero-padded. Your code simply copies the border pixels of the original image into the resulting image. This is known as replicating behaviour but that is not what conv2 does natively. conv2 assumes that the border pixels are zero-padded as I mentioned before, so what I would suggest you do is create two additional images, one being the output image that has 2 more rows and 2 more columns and another being the input image that is the same size as the output image but you place the input image inside this matrix. Next, perform the filtering on this new image, place the resulting filtered pixels in the output image then crop this result. I've decided to create a new padded input image in order to keep most of your code intact.

我还建议您在此处取消使用length.使用 size 来确定图像尺寸.这样的事情会起作用:

I would also recommend that you abolish the use of length here. Use size instead to determine the image dimensions. Something like this will work:

function [ image ] = Func134( img,matrix )
  [rows,cols] = size(img); %// Change

  %// New - Create a padded matrix that is the same class as the input
  new_img = zeros(rows+2,cols+2);
  new_img = cast(new_img, class(img));

  %// New -  Place original image in padded result
  new_img(2:end-1,2:end-1) = img;

  %// Also create new output image the same size as the padded result
  image = zeros(size(new_img));
  image = cast(image, class(img));

  for i=2:1:rows+1 %// Change
    for j=2:1:cols+1 %// Change
      value=0;
      for g=-1:1:1
        for l=-1:1:1
          value=value+new_img(i+g,j+l)*matrix(g+2,l+2); %// Change
        end
      end
     image(i,j)=value;
    end
  end

%// Change
%// Crop the image and remove the extra border pixels
image = image(2:end-1,2:end-1);
end


为了进行比较,我生成了这个随机矩阵:


To compare, I've generated this random matrix:

>> rng(123);
>> A = rand(10,10)

A =

    0.6965    0.3432    0.6344    0.0921    0.6240    0.1206    0.6693    0.0957    0.3188    0.7050
    0.2861    0.7290    0.8494    0.4337    0.1156    0.8263    0.5859    0.8853    0.6920    0.9954
    0.2269    0.4386    0.7245    0.4309    0.3173    0.6031    0.6249    0.6272    0.5544    0.3559
    0.5513    0.0597    0.6110    0.4937    0.4148    0.5451    0.6747    0.7234    0.3890    0.7625
    0.7195    0.3980    0.7224    0.4258    0.8663    0.3428    0.8423    0.0161    0.9251    0.5932
    0.4231    0.7380    0.3230    0.3123    0.2505    0.3041    0.0832    0.5944    0.8417    0.6917
    0.9808    0.1825    0.3618    0.4264    0.4830    0.4170    0.7637    0.5568    0.3574    0.1511
    0.6848    0.1755    0.2283    0.8934    0.9856    0.6813    0.2437    0.1590    0.0436    0.3989
    0.4809    0.5316    0.2937    0.9442    0.5195    0.8755    0.1942    0.1531    0.3048    0.2409
    0.3921    0.5318    0.6310    0.5018    0.6129    0.5104    0.5725    0.6955    0.3982    0.3435

现在运行上面讨论的内容:

Now running with what we talked about above:

mask = [3,10,3;0,0,0;-3,-10,-3];
mask_flip = mask(end:-1:1,end:-1:1);
B = Func134(A,mask);
C = conv2(A, mask_flip,'same');

对于您的函数和conv2的输出,我们得到以下信息:

We get the following for your function and the output of conv2:

>> B

B =

   -5.0485  -10.6972  -11.9826   -7.2322   -4.9363  -10.3681  -10.9944  -12.6870  -12.5618  -12.0295
    4.4100    0.1847   -2.2030   -2.7377    0.6031   -3.7711   -2.5978   -5.8890   -2.9036    2.7836
   -0.6436    6.6134    4.2122   -0.7822   -2.3282    1.6488    0.4420    2.2619    4.2144    3.2372
   -4.8046   -1.0665    0.1568   -1.5907   -4.6943    0.3036    0.4399    4.3466   -2.5859   -3.4849
   -0.7529   -5.5344    1.3900    3.1715    2.9108    4.6771    7.0247    1.7062   -3.9277   -0.6497
   -1.9663    2.4536    4.2516    2.2266    3.6084    0.6432   -1.0581   -3.4674    5.3815    6.1237
   -0.9296    5.1244    0.8912   -7.7325  -10.2260   -6.4585   -1.4298    6.2675   10.1657    5.3225
    3.9511   -1.7869   -1.9199   -5.0832   -3.2932   -2.9853    5.5304    5.9034    1.4683   -0.7394
    1.8580   -3.8938   -3.9216    3.8254    5.4139    1.8404   -4.3850   -7.4159   -4.9894   -0.5096
    6.4040    7.6395    7.3643   11.8812   10.6537   10.8957    5.0278    3.0277    4.2295    3.3229

>> C

C =

   -5.0485  -10.6972  -11.9826   -7.2322   -4.9363  -10.3681  -10.9944  -12.6870  -12.5618  -12.0295
    4.4100    0.1847   -2.2030   -2.7377    0.6031   -3.7711   -2.5978   -5.8890   -2.9036    2.7836
   -0.6436    6.6134    4.2122   -0.7822   -2.3282    1.6488    0.4420    2.2619    4.2144    3.2372
   -4.8046   -1.0665    0.1568   -1.5907   -4.6943    0.3036    0.4399    4.3466   -2.5859   -3.4849
   -0.7529   -5.5344    1.3900    3.1715    2.9108    4.6771    7.0247    1.7062   -3.9277   -0.6497
   -1.9663    2.4536    4.2516    2.2266    3.6084    0.6432   -1.0581   -3.4674    5.3815    6.1237
   -0.9296    5.1244    0.8912   -7.7325  -10.2260   -6.4585   -1.4298    6.2675   10.1657    5.3225
    3.9511   -1.7869   -1.9199   -5.0832   -3.2932   -2.9853    5.5304    5.9034    1.4683   -0.7394
    1.8580   -3.8938   -3.9216    3.8254    5.4139    1.8404   -4.3850   -7.4159   -4.9894   -0.5096
    6.4040    7.6395    7.3643   11.8812   10.6537   10.8957    5.0278    3.0277    4.2295    3.3229

这篇关于不使用conv2的Matlab图像过滤的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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