Question

我需要计算百万粒子的归一化互相关的最大值。 normxcorr2的两个参数的大小是56 * 56。我不能将计算并行化。是否有任何建议加速代码，特别是我不需要所有结果但只需要每个互相关的最大值（知道位移）？

算法示例

%The choice of 170 particles is because in each time 
%the code detects 170 particles, so over 10000 images it's 1 700 000 particles
particle_1=rand(54,54,170);
particle_2=rand(56,56,170);
for i=1:170
    C=normxcorr2(particle_1(:,:,i),particle_2(:,:,i));
    L(i)=max(C(:));
end

Answer 1

我没有MATLAB所以我在这个网站上运行了以下代码：https://www.tutorialspoint.com/execute_matlab_online.php实际上是八度。所以我实现了“天真”规范化的互相关，实际上对于这些小图像尺寸，天真的表现更好：

经过的时间是2.62645秒 - 对于normxcorr2
经过的时间是0.199034秒 - 对于我的naive_normxcorr2

该代码基于文章http://scribblethink.org/Work/nvisionInterface/nip.pdf，该文章介绍了如何使用integral image以有效的方式计算规范化所需的标准偏差，这是box_corr函数。

此外，MATLAB的normxcorr2返回一个填充图像，所以我在未填充的部分上取最大值。

pkg load image

function [N] = naive_corr(pat,img)

[n,m] = size(img);
[np,mp] = size(pat);

 N = zeros(n-np+1,m-mp+1);

 for i = 1:n-np+1
    for j = 1:m-mp+1
        N(i,j) = sum(dot(pat,img(i:i+np-1,j:j+mp-1)));
    end
 end

end


%w_arr the array of coefficients for the boxes
%box_arr of size [k,4] where k is the number boxes, each box represented by
%4 something ... 
function [C] = box_corr2(img,box_arr,w_arr,n_p,m_p)

% construct integral image + zeros pad (for boundary problems)
I = cumsum(cumsum(img,2),1);
I = [zeros(1,size(I,2)+2); [zeros(size(I,1),1) I zeros(size(I,1),1)]; zeros(1,size(I,2)+2)];

% initialize result matrix
[n,m] = size(img);
C = zeros(n-n_p+1,m-m_p+1);
%C = zeros(n,m);

jump_x = 1;
jump_y = 1;

x_start = ceil(n_p/2);
x_end = n-x_start+mod(n_p,2);
x_span = x_start:jump_x:x_end;

y_start = ceil(m_p/2);
y_end = m-y_start+mod(m_p,2);
y_span = y_start:jump_y:y_end;

arr_a = box_arr(:,1) - x_start;
arr_b = box_arr(:,2) - x_start+1;
arr_c = box_arr(:,3) - y_start;
arr_d = box_arr(:,4) - y_start+1;

% cumulate box responses
k = size(box_arr,1); % == numel(w_arr)
for i = 1:k
    a = arr_a(i);
    b = arr_b(i);
    c = arr_c(i);
    d = arr_d(i);

    C = C ...
        + w_arr(i) * ( I(x_span+b,y_span+d) ...
                       - I(x_span+b,y_span+c) ...
                       - I(x_span+a,y_span+d) ...
                       + I(x_span+a,y_span+c) );
end

end



function [NCC]  = naive_normxcorr2(temp,img)

    [n_p,m_p]=size(temp);

    M = n_p*m_p;

    % compute template mean & std
    temp_mean = mean(temp(:));
    temp = temp - temp_mean;

    temp_std = sqrt(sum(temp(:).^2)/M);

    % compute windows' mean & std
    wins_mean =  box_corr2(img,[1,n_p,1,m_p],1/M,  n_p,m_p);
    wins_mean2 = box_corr2(img.^2,[1,n_p,1,m_p],1/M,n_p,m_p);


    wins_std = real(sqrt(wins_mean2 - wins_mean.^2));
    NCC_naive = naive_corr(temp,img);

    NCC = NCC_naive ./ (M .* temp_std .* wins_std);
end 

n = 170;

particle_1=rand(54,54,n);
particle_2=rand(56,56,n);

[n_p1,m_p1,c_p1]=size(particle_1);
[n_p2,m_p2,c_p2]=size(particle_2);

L1 = zeros(n,1);
L2 = zeros (n,1);

tic
for i=1:n
    C1=normxcorr2(particle_1(:,:,i),particle_2(:,:,i));

    C1_unpadded = C1(n_p1:n_p2 , m_p1:m_p2);
    L1(i)=max(C1_unpadded(:));

end

toc

tic
for i=1:n

    C2=naive_normxcorr2(particle_1(:,:,i),particle_2(:,:,i));
    L2(i)=max(C2(:));
end

toc

加快normxcorr2最大值的计算

1 个答案: