3

这是我实现这个可爱公式的尝试。

http://dl.dropbox.com/u/7348856/Picture1.png 

%WIGNER Computes Wigner-Distribution on an image (difference of two images).
function[wd] = wigner(difference)
%Image size
[M, N, ~] = size(difference);
%Window size (5 x 5)
Md = 5;
Nd = 5;
%Fourier Transform
F = fft2(difference);
%Initializing the wigner picture
wd = zeros(M, N, 'uint8');
lambda =0.02;
value = (4/(Md*Nd));
for x = 1+floor(Md/2):M - floor(Md/2)
    for y = 1+floor(Nd/2):N - floor(Nd/2)
        for l = -floor(Nd/2) : floor(Nd/2)
            for k = -floor(Md/2) : floor(Md/2)
                kernel = exp(-lambda * norm(k,l));
                kernel = kernel * value;
                theta = 4 * pi * ((real(F(x, y)) * (k/M) )+ (imag(F(x, y)) * (l/N)));
                wd(x, y) = (wd(x, y)) + (cos(theta) * difference(x + k, y + l) * difference(x - k, y - l) * (kernel));
            end
        end
    end
end
end

如您所见，外部的两个循环用于滑动窗口，而其余的内部循环用于求和的变量。

现在，我对亲爱的 stackoverflow 用户的请求是：你能帮我改进这些非常讨厌的 for 循环，这些循环花费的时间超过了它的份额，并将其变成矢量化循环吗？这种改进会带来重大变化吗？

谢谢你。

4

2 回答 2

4

这可能不是您要问的，但似乎（乍一看）求和的顺序是独立的，您可以使用 {l,k,x,y} 而不是 {x,y,l,k} . 这样做将允许您通过将内核保持在最外层循环中来更少地评估内核。

于 2012-05-25T17:42:30.147 回答
1

这四个嵌套循环基本上是以滑动邻域样式处理图像中的每个像素。我立刻想到了NLFILTER和IM2COL函数。

这是我对代码进行矢量化的尝试。请注意，我尚未对其进行彻底测试，或将性能与基于循环的解决方案进行比较：

function WD = wigner(D, Md, Nd, lambda)
    %# window size and lambda
    if nargin<2, Md = 5; end
    if nargin<3, Nd = 5; end
    if nargin<4, lambda = 5; end

    %# image size
    [M,N,~] = size(D);

    %# kernel = exp(-lambda*norm([k,l])
    [K,L] = meshgrid(-floor(Md/2):floor(Md/2), -floor(Nd/2):floor(Nd/2));
    K = K(:); L = L(:);
    kernel = exp(-lambda .* sqrt(K.^2+L.^2));

    %# frequency-domain part
    F = fft2(D);

    %# f(x+k,y+l) * f(x-k,y-l) * kernel
    C = im2col(D, [Md Nd], 'sliding');
    X1 = bsxfun(@times, C .* flipud(C), kernel);

    %# cos(theta)
    C = im2col(F, [Md Nd], 'sliding');
    C = C(round(Md*Nd/2),:);    %# take center pixels
    theta = bsxfun(@times, real(C), K/M) + bsxfun(@times, imag(C), L/N);
    X2 = cos(4*pi*theta);

    %# combine both parts for each sliding-neighborhood
    WD = col2im(sum(X1.*X2,1), [Md Nd], size(F), 'sliding') .* (4/(M*N));

    %# pad array with zeros to be of same size as input image
    WD = padarray(WD, ([Md Nd]-1)./2, 0, 'both');
end

对于它的价值，这是基于循环的版本，具有@Laurbert515建议的改进：

function WD = wigner_loop(D, Md, Nd, lambda)
    %# window size and lambda
    if nargin<2, Md = 5; end
    if nargin<3, Nd = 5; end
    if nargin<4, lambda = 5; end

    %# image size
    [M,N,~] = size(D);

    %# frequency-domain part
    F = fft2(D);

    WD = zeros([M,N]);
    for l = -floor(Nd/2):floor(Nd/2)
        for k = -floor(Md/2):floor(Md/2)
            %# kernel = exp(-lambda*norm([k,l])
            kernel = exp(-lambda * norm([k,l]));

            for x = (1+floor(Md/2)):(M-floor(Md/2))
                for y = (1+floor(Nd/2)):(N-floor(Nd/2))
                    %# cos(theta)
                    theta = 4 * pi * ( real(F(x,y))*k/M + imag(F(x,y))*l/N );

                    %# f(x+k,y+l) * f(x-k,y-l)* kernel
                    WD(x,y) = WD(x,y) + ( cos(theta) * D(x+k,y+l) * D(x-k,y-l) * kernel );
                end
            end
        end
    end
    WD = WD * ( 4/(M*N) );
end

以及我如何测试它（基于我从您之前链接到的论文中理解的内容）：

%# difference between two consecutive frames
A = imread('AT3_1m4_02.tif');
B = imread('AT3_1m4_03.tif');
D = imsubtract(A,B);
%#D = rgb2gray(D);
D = im2double(D);

%# apply Wigner-Distribution
tic, WD1 = wigner(D); toc
tic, WD2 = wigner_loop(D); toc
figure(1), imshow(WD1,[])
figure(2), imshow(WD2,[])

然后您可能需要缩放/归一化矩阵，并应用阈值...

于 2012-05-27T00:59:07.710 回答