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I have a matrix which consists of positive and negative values. I need to do these things.

Let u(i,j) denote the pixels of the matrix u.

  1. 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.
  2. Then I need to calculate the narrow band around these zero crossing pixels. The width of the narrow band is (2r+1)X(2r+1) for each pixel. I am taking r=1 so for it I have to actually get the 8 neighborhood pixels of each zero crossing pixels.

I have done this in a program. Please see it below. 

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

I am also attaching the isfront and isband functions.

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

The outputs are given below:, as well as the computation time:

Figure

%// Computation time

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

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

When I run the code I do get the correct answers but the computation for the tasks is too much to my liking. Is there a better way to do it ? Especially the isband function? How can I optimize my code further ?

Thanks in advance.

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2 に答える 2

1

r==1 だけに関心がある場合は、makelutと対応する関数 bwloolup を見てください。

[編集]

% 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);
于 2013-09-03T13:03:57.307 に答える