matlab - MATLAB 是否优化 diag(A*B)？

Question

假设我有两个非常大的矩阵 A (M-by-N) 和 B (N-by-M)。我需要 A*B 的对角线。计算完整的 A*B 需要 M*M*N 次乘法，而计算它的对角线只需要 M*N 次乘法，因为不需要计算最终会在对角线之外的元素。

MATLAB 是否自动实现了这一点并即时优化 diag(A*B)，还是在这种情况下我最好使用 for 循环？

Answer 1

也可以将 diag(A*B) 实现为 sum(A.*B',2)。让我们将其与针对此问题建议的所有其他实现/解决方案一起进行基准测试。

下面列出了作为基准实现的不同方法:

1. 和乘法-1

   function out = sum_mult_method1(A,B)
   
   out = sum(A.*B',2);
   

2. 和乘法-2

   function out = sum_mult_method2(A,B)
   
   out = sum(A.'.*B).';
   

3. For循环法

   function out = for_loop_method(A,B)
   
   M = size(A,1);
   out = zeros(M,1);
   for i=1:M
       out(i) = A(i,:) * B(:,i);
   end
   

4. 全/直乘法

   function out = direct_mult_method(A,B)
   
   out = diag(A*B);
   

5. Bsxfun-方法

   function out = bsxfun_method(A,B)
   
   out = sum(bsxfun(@times,A,B.'),2);
   

基准代码

num_runs = 1000;
M_arr = [100 200 500 1000];
N = 4;

%// Warm up tic/toc.
tic();
elapsed = toc();
tic();
elapsed = toc();

for k2 = 1:numel(M_arr)
    M = M_arr(k2);

    fprintf('\n')
    disp(strcat('*** Benchmarking sizes are M =',num2str(M),' and N = ',num2str(N)));

    A = randi(9,M,N);
    B = randi(9,N,M);

    disp('1. Sum-multiplication method-1');
    tic
    for k = 1:num_runs
        out1 = sum_mult_method1(A,B);
    end
    toc
    clear out1

    disp('2. Sum-multiplication method-2');
    tic
    for k = 1:num_runs
        out2 = sum_mult_method2(A,B);
    end
    toc
    clear out2

    disp('3. For-loop method');
    tic
    for k = 1:num_runs
        out3 = for_loop_method(A,B);
    end
    toc
    clear out3

    disp('4. Direct-multiplication method');
    tic
    for k = 1:num_runs
        out4 = direct_mult_method(A,B);
    end
    toc
    clear out4

    disp('5. Bsxfun method');
    tic
    for k = 1:num_runs
        out5 = bsxfun_method(A,B);
    end
    toc
    clear out5

end

结果

*** Benchmarking sizes are M =100 and N =4
1. Sum-multiplication method-1
Elapsed time is 0.015242 seconds.
2. Sum-multiplication method-2
Elapsed time is 0.015180 seconds.
3. For-loop method
Elapsed time is 0.192021 seconds.
4. Direct-multiplication method
Elapsed time is 0.065543 seconds.
5. Bsxfun method
Elapsed time is 0.054149 seconds.

*** Benchmarking sizes are M =200 and N =4
1. Sum-multiplication method-1
Elapsed time is 0.009138 seconds.
2. Sum-multiplication method-2
Elapsed time is 0.009428 seconds.
3. For-loop method
Elapsed time is 0.435735 seconds.
4. Direct-multiplication method
Elapsed time is 0.148908 seconds.
5. Bsxfun method
Elapsed time is 0.030946 seconds.

*** Benchmarking sizes are M =500 and N =4
1. Sum-multiplication method-1
Elapsed time is 0.033287 seconds.
2. Sum-multiplication method-2
Elapsed time is 0.026405 seconds.
3. For-loop method
Elapsed time is 0.965260 seconds.
4. Direct-multiplication method
Elapsed time is 2.832855 seconds.
5. Bsxfun method
Elapsed time is 0.034923 seconds.

*** Benchmarking sizes are M =1000 and N =4
1. Sum-multiplication method-1
Elapsed time is 0.026068 seconds.
2. Sum-multiplication method-2
Elapsed time is 0.032850 seconds.
3. For-loop method
Elapsed time is 1.775382 seconds.
4. Direct-multiplication method
Elapsed time is 13.764870 seconds.
5. Bsxfun method
Elapsed time is 0.044931 seconds.

中间结论

看起来 sum-multiplication 方法是最好的方法，尽管 bsxfun 方法似乎正在 catch 它们，因为 M 从 100 增加到 1000。

接下来，仅使用 sum-multiplication 和 bsxfun 方法测试了更高的基准测试大小。尺寸是 -

M_arr = [1000 2000 5000 10000 20000 50000];

结果是——

*** Benchmarking sizes are M =1000 and N =4
1. Sum-multiplication method-1
Elapsed time is 0.030390 seconds.
2. Sum-multiplication method-2
Elapsed time is 0.032334 seconds.
5. Bsxfun method
Elapsed time is 0.047377 seconds.

*** Benchmarking sizes are M =2000 and N =4
1. Sum-multiplication method-1
Elapsed time is 0.040111 seconds.
2. Sum-multiplication method-2
Elapsed time is 0.045132 seconds.
5. Bsxfun method
Elapsed time is 0.060762 seconds.

*** Benchmarking sizes are M =5000 and N =4
1. Sum-multiplication method-1
Elapsed time is 0.099986 seconds.
2. Sum-multiplication method-2
Elapsed time is 0.103213 seconds.
5. Bsxfun method
Elapsed time is 0.117650 seconds.

*** Benchmarking sizes are M =10000 and N =4
1. Sum-multiplication method-1
Elapsed time is 0.375604 seconds.
2. Sum-multiplication method-2
Elapsed time is 0.273726 seconds.
5. Bsxfun method
Elapsed time is 0.226791 seconds.

*** Benchmarking sizes are M =20000 and N =4
1. Sum-multiplication method-1
Elapsed time is 1.906839 seconds.
2. Sum-multiplication method-2
Elapsed time is 1.849166 seconds.
5. Bsxfun method
Elapsed time is 1.344905 seconds.

*** Benchmarking sizes are M =50000 and N =4
1. Sum-multiplication method-1
Elapsed time is 5.159177 seconds.
2. Sum-multiplication method-2
Elapsed time is 5.081211 seconds.
5. Bsxfun method
Elapsed time is 3.866018 seconds.

备用基准测试代码(带有 `timeit)

num_runs = 1000;
M_arr = [1000 2000 5000 10000 20000 50000 100000 200000 500000 1000000];
N = 4;

timeall = zeros(5,numel(M_arr));
for k2 = 1:numel(M_arr)
    M = M_arr(k2);

    A = rand(M,N);
    B = rand(N,M);

    f = @() sum_mult_method1(A,B);
    timeall(1,k2) = timeit(f);
    clear f

    f = @() sum_mult_method2(A,B);
    timeall(2,k2) = timeit(f);
    clear f

    f = @() bsxfun_method(A,B);
    timeall(5,k2) = timeit(f);
    clear f

end

figure,
hold on
plot(M_arr,timeall(1,:),'-ro')
plot(M_arr,timeall(2,:),'-ko')
plot(M_arr,timeall(5,:),'-.b')
legend('sum-method1','sum-method2','bsxfun-method')
xlabel('M ->')
ylabel('Time(sec) ->')

绘图

最终结论

看起来 sum-multiplication 方法在某些阶段之前很棒，大约在 M=5000 标记之后 bsxfun 似乎有略占上风。

future 的工作

可以查看不同的 N 并研究此处提到的实现的性能。