假设我有两个非常大的矩阵A（M×N）和B（N×M）。我需要的对角线A*B。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位于对角线之外的元素。
MATLAB 是否实现了这一点并进行即时优化diag(A*B)自动地，还是在这种情况下我最好使用 for 循环？
还可以实施diag(A*B) as sum(A.*B',2)。让我们将其与针对此问题建议的所有其他实现/解决方案进行基准测试。
为了进行基准测试，下面列出了作为函数实现的不同方法：
和乘法-1
function out = sum_mult_method1(A,B)
out = sum(A.*B',2);
和乘法2
function out = sum_mult_method2(A,B)
out = sum(A.'.*B).';
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
完全/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);
基准测试代码
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
Results
*** 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
好像sum-multiplication方法是最好的方法，但是bsxfun方法似乎是赶上他们M从 100 增加到 1000。
接下来，仅使用以下参数测试了更高的基准尺寸sum-multiplication and bsxfun方法。尺寸为 -
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) ->')
Plot
最终结论
它似乎sum-multiplication方法在某个阶段是很好的，大约是M=5000标记，然后bsxfun似乎略占上风。
未来的工作
人们可以研究不同的N并研究此处提到的实现的性能。