Question

这是我尝试实施这个可爱的公式。

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，而剩下的内部循环用于求和的变量。

现在，我对我心爱的stackoverflow用户的请求是：你能帮助我改进这些非常讨厌的循环，花费超过它的时间，并将其转换为矢量化循环吗？这种改善会有重大改变吗？

谢谢。

Answer 1

这可能不是你要问的，但似乎（乍一看）总结的顺序是独立的，而不是{x，y，l，k}你可以去{l，k，x ，Y}。执行此操作将允许您通过将内核保留在最外层循环中来评估内核的次数。

Answer 2

这四个嵌套循环基本上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。

这是我尝试矢量化代码。请注意，我没有对其进行全面测试，或者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

值得一提的是，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

以及我如何对其进行测试（根据我对您paper所关联的previously的理解）：

%# 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/标准化矩阵，并应用阈值...

MATLAB和矢量化循环中的双重求和

2 个答案: