Question

在我的matlab程序中，我有几个实例需要创建一个矩阵，这些条目取决于它的索引并用它执行矩阵向量运算。我想知道如何最有效地实现这一目标。

例如，我需要加快速度：

N = 1e4;
x = rand(N,1);

% Option 1
tic
I = 1:N;
J = 1:N;
S = zeros(N,N);
for i = 1:N
    for j = 1:N
        S(i,j) = (i+j)/(abs(i-j)+1);
    end
end
a = x'*S*x
fprintf('Option 1 takes %.4f sec\n',toc)
clearvars -except x N

我尝试加快速度，所以我尝试了以下选项：

% Option 2
tic
I = 1:N;
J = 1:N;
Sx = zeros(N,1);
for i = 1:N
    Srow_i = (i+J)./(abs(i-J)+1);
    Sx(i)= Srow_i*x;
end
a = x'*Sx
fprintf('Option 2 takes %.4f sec\n',toc)
clearvars -except x N

和

% Option 3
tic
I = 1:N;
J = 1:N;
S = bsxfun(@plus,I',J)./(abs(bsxfun(@minus,I',J))+1);
a = x'*S*x
fprintf('Option 3 takes %.4f sec\n',toc)
clearvars -except x N

和（感谢其中一个答案）

% options 4
tic
[I , J] = meshgrid(1:N,1:N);
S = (I+J) ./ (abs(I-J) + 1);
a = x' * S * x;
fprintf('Option 4 takes %.4f sec\n',toc)
clearvars -except x N

Otion 2是最有效的。是否有更快的选择来执行此操作？

更新

我也尝试过Abhinav的选项：

% Option 5 using Tony's Trick
tic
i = 1:N;
j = (1:N)';
I = i(ones(N,1),:);
J = j(:,ones(N,1));
S = (I+J)./(abs(I-J)+1);
a = x'*S*x;
fprintf('Option 5 takes %.4f sec\n',toc)
clearvars -except x N

似乎最有效的程序取决于N的大小。对于不同的N，我得到以下输出：

N = 100：

Option 1 takes 0.00233 sec
Option 2 takes 0.00276 sec
Option 3 takes 0.00183 sec
Option 4 takes 0.00145 sec
Option 5 takes 0.00185 sec

N = 10000：

Option 1 takes 3.29824 sec
Option 2 takes 0.41597 sec
Option 3 takes 0.72224 sec
Option 4 takes 1.23450 sec
Option 5 takes 1.27717 sec

因此，对于小N，选项2是最慢的，但对于较大的N，它变得最有效。可能是因为内存？有人可以解释一下吗？

Answer 1

您可以使用meshgrid创建索引，无需循环：

N = 1e4;
[I , J] = meshgrid(1:N,1:N);
x = rand(N,1);
S = (I+J) ./ (abs(I-J) + 1);
a = x' * S * x;

<强>更新

由于@Optimist显示此代码的性能低于Option2和Option3，我决定略微改进Option2：

N = 1e4;
x = rand(N,1);
Sx = zeros(N,1);
for i = 1:N
    Srow_i = (i+1:i+N)./[i:-1:2,1:N-i+1] ;
    Sx(i)= Srow_i*x;
end
a = x'*Sx;

Answer 2

You should try using the Tony's trick to do vector stacking/tiling in Matlab the fastest way. I have answered a similar question here. Here is the Tony's Trick option.

% Option using Tony's Trick
tic
i = 1:N;
j = (1:N)';
I = i(ones(N,1),:);
J = j(:,ones(N,1));
S = (I+J)./(abs(I-J)+1);

a = x'*S*x
fprintf('Option 1 takes %.4f sec\n',toc)

Edit 1: I ran a few tests and found the following. Up to N=1000, the Tony's trick option is slightly faster than the Option 2. Beyond that, Option 2 again catches up and becomes faster.

Possible Reason : This should be so because, up until the size of the array could fit in the cache, the fully vectorized code (Tony's Trick option) is faster BUT as soon as the array sizes grow (N>1000), it spills into memory caches away from the CPU and then Matlab uses some internal optimization to breakdown the Tony's Trick code into piecemeal code so that it no-longer enjoys the benefit of complete vectorization.

如何在matlab中有效地构造一个依赖于索引的矩阵

2 个答案: