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。计算完全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实时优化diag(A*B)，或者我最好在这种情况下使用for循环？

Answer 1

还可以将diag(A*B)实施为sum(A.*B',2)。让我们根据此问题的建议对此进行基准测试以及所有其他实现/解决方案。

下面列出了作为函数实现的不同方法，用于基准测试：

求和方法-1

function out = sum_mult_method1(A,B)

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

Sum-multiplication method-2

function out = sum_mult_method2(A,B)

out = sum(A.'.*B).';

For-loop方法

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

完整/直接乘法

function out = direct_mult_method(A,B)

out = diag(A*B);

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

<强>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.

中级结论

看起来sum-multiplication方法是最好的方法，但bsxfun方法似乎是在M从100增加到1000时赶上它们。

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.

替代基准测试代码（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) ->')

<强>剧情

enter image description here

最终结论

似乎sum-multiplication方法在某个阶段很好，大约是M=5000标记，之后bsxfun似乎有轻微的上风。

未来工作

可以研究不同的N并研究这里提到的实现的性能。

Answer 2

是的，这是for循环更好的罕见情况之一。

我通过探查器运行以下脚本：

M = 5000;
N = 5000;

A = rand(M, N); B = rand(N, M);
product = A*B;
diag1 = diag(product);

A = rand(M, N); B = rand(N, M);
diag2 = diag(A*B);

A = rand(M, N); B = rand(N, M);
diag3 = zeros(M,1);
for i=1:M
    diag3(i) = A(i,:) * B(:,i);
end

我在每次测试之间重置A和B，以防MATLAB试图通过缓存加速任何事情。

结果（为简洁起见编辑）：

  time   calls  line
  6.29       1    5 product = A*B; 
< 0.01       1    6 diag1 = diag(product); 

  5.46       1    9 diag2 = diag(A*B); 

             1   12 diag3 = zeros(M,1); 
             1   13 for i=1:M 
  0.52    5000   14     diag3(i) = A(i,:) * B(:,i); 
< 0.01    5000   15 end

正如我们所看到的，在这种情况下，for循环变量比其他两个变量快一个数量级。虽然diag(A*B)变体实际上比diag(product)变体更快，但它最多只是边缘。

我尝试了一些不同的M和N值，在我的测试中，只有当M = 1时，for循环变量才会变慢。

Answer 3

实际上，可以<{1}}循环使用bsxfun的奇迹更快地执行此操作：

for

这大约是我的机器上显式diag4 = sum(bsxfun(@times,A,B.'),2)循环的两倍，对于大型矩阵（2,000乘2,000和更大），对于大于500 x 500的矩阵来说速度更快。

请注意，由于求和和乘法的顺序不同，所有这些方法都会产生数值上不同的结果。

Answer 4

您只能计算没有循环的对角元素：只需使用

sum(A.'.*B).'

或

sum(A.*B.',2)

MATLAB是否优化了诊断（A * B）？

4 个答案: