Question

我有一个由正值和负值组成的矩阵。我需要做这些事情。

让u(i,j)表示矩阵u的像素。

计算过零像素。如果u(i-1,j)和u(i+1,j)符号相反或u(i,j-1)和u(i,j+1)符号相反，则这些是网格中的像素。
然后我需要计算这些零交叉像素周围的窄带。对于每个像素，窄带的宽度为(2r+1)X(2r+1)。我正在考虑r=1因此我必须实际获得每个零交叉像素的8个邻域像素。

我在一个程序中完成了这个。请看下面的内容。

%// calculate the zero crossing pixels  
front = isfront(u);
%// calculate the narrow band of around the zero crossing pixels
band  = isband(u,front,1);

我还附加了isfront和isband函数。

function front = isfront( phi )
%// grab the size of phi
[n, m] = size(phi);

%// create an boolean matrix whose value at each pixel is 0 or 1
%// depending on whether that pixel is a front point or not
front = zeros( size(phi) );

%// A piecewise(Segmentation) linear approximation to the front is contructed by
%// checking each pixels neighbour. Do not check pixels on border.
for i = 2 : n - 1;
  for j = 2 : m - 1;
    if (phi(i-1,j)*phi(i+1,j)<0) || (phi(i,j-1)*phi(i,j+1)<0)
        front(i,j) = 100;
    else
        front(i,j) = 0;
    end
  end
end

function band = isband(phi, front, width)
%// grab size of phi
[m, n] = size(phi);

%// width=r=1;
width = 1;

[x,y] = find(front==100);

%// create an boolean matrix whose value at each pixel is 0 or 1
%// depending on whether that pixel is a band point or not
band = zeros(m, n);

%// for each pixel in phi
for ii = 1:m
  for jj = 1:n
    for k = 1:size(x,1)
        if (ii==x(k)) && (jj==y(k))
            band(ii-1,jj-1) = 100;  band(ii-1,jj) = 100; band(ii-1,jj+1) = 100;
            band(ii  ,jj-1) = 100;  band(ii  ,jj) = 100; band(ii,jj+1) = 100;
            band(ii+1,jj-1) = 100;  band(ii+1,jj) = 100; band(ii+1,jj+1) = 100;
        end
    end
  end
end

输出如下：以及计算时间：

%// Computation time

%// for isfront function
Elapsed time is 0.003413 seconds.

%// for isband function
Elapsed time is 0.026188 seconds.

当我运行代码时，我确实得到了正确的答案，但任务的计算量太大了，我不喜欢。有没有更好的方法呢？特别是isband功能？如何进一步优化我的代码？

提前致谢。

Answer 1

根据EitanT的建议，至少bwmorph已经做了你想做的事。

如果您无权访问图像处理工具箱，或只是坚持自己动手：

您可以使用矢量化

替换isfront中的三重循环

front = zeros(n,m);

zero_crossers = ...
    phi(1:end-2,2:end-1).*phi(3:end,2:end-1) < 0 | ...
    phi(2:end-1,1:end-2).*phi(2:end-1,3:end) < 0;

front([...
                   false(1,m)
    false(n-2,1)  zero_crossers  false(n-2,1)
                   false(1,m)                 ]...
) = 100;

您可以通过以下单循环替换isband：

[n,m] = size(front);
band = zeros(n,m);
[x,y] = find(front);
for ii = 1:numel(x)
    band(...
        max(x(ii)-width,1) : min(x(ii)+width,n),...
        max(y(ii)-width,1) : min(y(ii)+width,m)) = 1;
end

或者，正如米兰所建议的那样，您可以通过convolution：

应用图像扩张

kernel = ones(2*width+1);    
band = conv2(front, kernel);
band = 100 * (band(width+1:end-width, width+1:end-width) > 0);

应该更快。

您当然可以进行一些其他的小优化（isband不需要phi作为参数，您可以将front作为逻辑数组传递，以便{{1}更快，等等。）

Answer 2

如果您只对r == 1感兴趣，请查看makelut和相应的函数bwloolup。

[编辑]

% Let u(i,j) denote the pixels of the matrix u. Calculate the zero crossing
% pixels. These are the pixels in the grid if u(i-1,j) and u(i+1,j) are of
% opposite signs or u(i,j-1) and u(i,j+1) are of opposite signs.

% First, create a function which will us if a pixel is a zero crossing
% pixel, given its 8 neighbors. 

% uSign = u>0; % Note - 0 is treated as negative here. 

% uSign is 3x3, we are evaluating the center pixel uSign(2,2)
zcFun = @(uSign) (uSign(1,2)~=uSign(3,2)) || (uSign(2,1)~=uSign(2,3));

% Make a look up table which tells us what the output should be for the 2^9
% = 512 possible patterns of 3x3 arrays with 1's and 0's.
lut   = makelut(zcFun,3);

% Test image
im = double(imread('pout.tif'));
% Create positve and negative values
im = im -mean(im(:));

% Apply lookup table
imSign = im>0;
imZC   = bwlookup(imSign,lut);

imshowpair(im, imZC);

计算像素邻域的最快方法

2 个答案: