Question

我已将图像分成多个子图像。现在我想比较每个子图像与其8个邻居的平均强度。但在某些方面，不到8个邻居。例如，对于第一个块（i = 1，J = 1），上左块（i-1，j-1）不存在。我该如何检查并跳到下一个有效的？

file='myimg.bmp';
I=imread(file);

blockSizeR = 128; % Rows in block.
blockSizeC = 128; % Columns in block.

wholeBlockRows = floor(rows / blockSizeR);
blockVectorR = [blockSizeR * ones(1, wholeBlockRows), rem(rows, blockSizeR)];

wholeBlockCols = floor(columns / blockSizeC);
blockVectorC = [blockSizeC * ones(1, wholeBlockCols), rem(columns, blockSizeC)];

ca = mat2cell(I, blockVectorR, blockVectorC);

%get the mean value of each cell
meanValues = cellfun(@(x) mean(x(:)),ca);


for j=1:size(ca(2))
    for i=1:size(ca(1))
        currentSlice = ca(i,j);
        MeanOfCurrentSlice =  cellfun(@(x) mean(x(:)),currentSlice);

        %here I want to minus the 8 neighbors average grayscale intensity from the currentSlice average grayscale inensity and take the absolute sum 

    end

end

Answer 1

为您提供每个元素的最近邻居索引的解决方案：

%creation of the index matrix (here a 3x3 matrix)
M = reshape([1:9],3,3);

%subdivide the matrix into 3x3 array
IND = nlfilter(M,[3 3],@(x){x(:)});

%elimination of the value where IND == 0 or IND == index value of the element
for ii = 1:size(M,1)
    for jj = 1:size(M,2)
        IND{ii,jj}(IND{ii,jj}==0|IND{ii,jj}==sub2ind(size(M),ii,jj)) = [];
    end
end

PS：nlfilter是图片处理工具箱的一部分，但您可以轻松创建自己的类似功能。

第1步：

第2步：

IND = 
{
  [1,1] = 
     0   0   0   0   1   2   0   4   5
  [2,1] =
     0   0   0   1   2   3   4   5   6
  [3,1] =
     0   0   0   2   3   0   5   6   0
  [1,2] =
     0   1   2   0   4   5   0   7   8
  [2,2] =
     1   2   3   4   5   6   7   8   9
  [3,2] =
     2   3   0   5   6   0   8   9   0
  [1,3] =
     0   4   5   0   7   8   0   0   0
  [2,3] =
     4   5   6   7   8   9   0   0   0
  [3,3] =
     5   6   0   8   9   0   0   0   0
}

第3步：

IND =
{
  [1,1] = %neighbors of the value M[1,1] 
     2   4   5
  [2,1] =
     1   3   4   5   6
  [3,1] =
     2   5   6
  [1,2] =
     1   2   5   7   8
  [2,2] =
     1   2   3   4   6   7   8   9
  [3,2] =
     2   3   5   8   9
  [1,3] =
     4   5   8
  [2,3] =
     4   5   6   7   9
  [3,3] =
     5   6   8
}

检查最近的8个邻居中的有效指数

1 个答案: