matlab - 跨两种模式的交叉成对距离测量

Question

我有一个问题。我正在尝试计算向量之间的成对距离。先解释一下问题:我有两组向量X和 Y . X有三个向量 x1 , x2和 x3 . Y有三个向量 y1 , y2和 y3 .注意 X 中的向量和 Y长度为 m和 n分别。让数据集表示为这张图片:

我正在尝试计算这样的相似度矩阵:

.现在解释了不同颜色编码的部分 - 所有那些标有 0 的单元格不需要计算。我故意把它写成 100 (它可以是任何值)。必须计算灰色单元格。相似性得分计算为 L2 (xi-xj) 的规范+ L2 (yi-yj) 的规范.

这意味着条目是

M((x_i,y_j), (x_k,y_l)) := norm(x_i-x_k,2) + norm(y_j-y_l,2)

我已经写了一个基本的代码来做到这一点:

clc;clear all;close all;
%% randomly generate data
m=3; n1=4; n2=6;
train_a_mean = rand(m,n1);
train_b_mean = rand(m,n2);
p = size(train_a_mean,1)*size(train_b_mean,1);
score_mean_ab = zeros(p,p);

%% This is to store the index variables 
%% This is required for futu
idx1 = score_mean_ab;
idx2 = idx1; idx3 = idx1; idx4 = idx1;

a=1; b=1;
for i=1:size(score_mean_ab,1)
    c = 1; d = 1;
    for j=1:size(score_mean_ab,2)
        if (a==c)
            score_mean_ab(i,j) = 100;
        else            
            %% computing distances between the different modalities and
            %% summing them up
            score_mean_ab(i,j) = norm(train_a_mean(a,:)-train_a_mean(c,:),2) ...
            + norm(train_b_mean(b,:)-train_b_mean(d,:),2);
        end
        %% saving the indices
        idx1(i,j)=a; idx2(i,j)=b; idx3(i,j)=c; idx4(i,j)=d;        
        %% updating the values of c and d
        if mod(d,size(train_a_mean,1))==0
            c = c + 1;
            d = 1;
        else
            d = d+1;
        end
    end
    %% updating the values of a and b
    if mod(b,size(train_a_mean,1))==0
        a = a + 1;
        b = 1;
    else        
        b = b+1;
    end
end

对于矩阵的干样本运行:我得到这些结果 -

score_mean_ab =

  100.0000  100.0000  100.0000    0.6700    1.6548    1.5725    0.8154    1.8002    1.7179
  100.0000  100.0000  100.0000    1.6548    0.6700    1.5000    1.8002    0.8154    1.6454
  100.0000  100.0000  100.0000    1.5725    1.5000    0.6700    1.7179    1.6454    0.8154
    0.6700    1.6548    1.5725  100.0000  100.0000  100.0000    1.3174    2.3022    2.2200
    1.6548    0.6700    1.5000  100.0000  100.0000  100.0000    2.3022    1.3174    2.1475
    1.5725    1.5000    0.6700  100.0000  100.0000  100.0000    2.2200    2.1475    1.3174
    0.8154    1.8002    1.7179    1.3174    2.3022    2.2200  100.0000  100.0000  100.0000
    1.8002    0.8154    1.6454    2.3022    1.3174    2.1475  100.0000  100.0000  100.0000
    1.7179    1.6454    0.8154    2.2200    2.1475    1.3174  100.0000  100.0000  100.0000

但是我的代码很慢。我进行了很少的样本运行并得到了这些结果:

m=3; n1=3; n2=3;
Elapsed time is 0.000363 seconds.

m=10; n1=3; n2=3;
Elapsed time is 0.042015 seconds.

m=10; n1=1800; n2=1800;
Elapsed time is 0.230046 seconds.

m=20; n1=1800; n2=1800;
Elapsed time is 4.309134 seconds.

m=30; n1=1800; n2=1800;
Elapsed time is 23.058106 seconds.

我的问题:

1. 通常我的值为 m~100和 n1~2000和 n2~2000 .我自己的代码在这一点上崩溃了。有什么优化的方法可以做到这一点吗？
2. 内置的matlab函数可以吗pdist2 用于此目的？

注意:向量实际上是行向量的形式，n1的值和 n2可能不相等。

Answer 1

这是一种方法。这将计算所有条目。

m = 3;             %// number of (row) vectors in X and in Y
n1 = 3;            %// length of vectors in X
n2 = 3;            %// length of vectors in Y
X = rand(m, n1);   %// random data: X
Y = rand(m, n2);   %// random data: Y

[ii, jj] = ndgrid(1:m); 
U = reshape(sqrt(sum((X(ii,:)-X(jj,:)).^2, 2)), m, m);
V = reshape(sqrt(sum((Y(ii,:)-Y(jj,:)).^2, 2)), m, m);
result = U(ceil(1/m:1/m:m), ceil(1/m:1/m:m)) + repmat(V, m, m);

或者您可以使用 bsxfun 而不是 ndgrid:

U = sqrt(sum(bsxfun(@minus, permute(X, [1 3 2]), permute(X, [3 1 2])).^2, 3));
V = sqrt(sum(bsxfun(@minus, permute(Y, [1 3 2]), permute(Y, [3 1 2])).^2, 3));
result = U(ceil(1/m:1/m:m), ceil(1/m:1/m:m)) + repmat(V, m, m);