Improving speed in construction of a matrix

2024 1 16
1 个回答
回答已采纳
更新时间：2024 1 20
11 次查看（30 天）

0 个投票

Hi.
Im currently implementing a code, where part of the code constructs a matrix several million times. Currently, the construction of this matrix takes much more time, than the actual computing invovled with the matrix. Its a rather simple matrix, but takes a third of the total runtime. I want to construct it faster, but cant seem to bring the time down. Ive tried two approaches.
One implementation:
B = zeros(9,9);
B(1:3,1) = A(1,:);
B(4:6,2) = A(1,:);
B(7:9,3) = A(1,:);
B(1:3,4) = A(2,:);
B(4:6,5) = A(2,:);
B(7:9,6) = A(2,:);
B(1:3,7) = A(3,:);
B(4:6,8) = A(3,:);
B(7:9,9) = A(3,:);
And the other implementation:
Z = zeros(3,1);
B =[A(:,1) Z Z A(:,2) Z Z A(:,3) Z Z
Z A(:,1) Z Z A(:,2) Z Z A(:,3) Z
Z Z A(:,1) Z Z A(:,2) Z Z A(:,3)];
Both giving a inaduqate execution time. Is there a faster way to construct this type of matrix?

6 个评论
显示 4更早的评论 隐藏 4更早的评论

Torsten
Torsten 2024-1-16
编辑：Torsten 2024-1-16
Ran in:
Ran in:
How long is "inadequate" ?
A = rand(3,3);
B = zeros(9,9);
Z = zeros(3,1);
tic
for i = 1:100000
B(1:3,1) = A(1,:);
B(4:6,2) = A(1,:);
B(7:9,3) = A(1,:);
B(1:3,4) = A(2,:);
B(4:6,5) = A(2,:);
B(7:9,6) = A(2,:);
B(1:3,7) = A(3,:);
B(4:6,8) = A(3,:);
B(7:9,9) = A(3,:);
end
toc
Elapsed time is 0.979855 seconds.
tic
for i = 1:100000
B =[A(:,1) Z Z A(:,2) Z Z A(:,3) Z Z
Z A(:,1) Z Z A(:,2) Z Z A(:,3) Z
Z Z A(:,1) Z Z A(:,2) Z Z A(:,3)];
end
toc
Elapsed time is 0.620697 seconds.
Walter Roberson
Walter Roberson 2024-1-16
Z = zeros(3,1);
A1 = A(:,1); A2 = A(:,2); A3 = A(:,3);
B =[A1 Z Z A2 Z Z A3 Z Z
Z A1 Z Z A2 Z Z A3 Z
Z Z A1 Z Z A2 Z Z A3];
Matt J
Matt J 2024-1-16
Ran in:
Ran in:
Your two implementations give different results:
A=reshape(1:9,3,3)
A = 3×3
1 4 7 2 5 8 3 6 9
B = zeros(9,9);
B(1:3,1) = A(1,:);
B(4:6,2) = A(1,:);
B(7:9,3) = A(1,:);
B(1:3,4) = A(2,:);
B(4:6,5) = A(2,:);
B(7:9,6) = A(2,:);
B(1:3,7) = A(3,:);
B(4:6,8) = A(3,:);
B(7:9,9) = A(3,:)
B = 9×9
1 0 0 2 0 0 3 0 0 4 0 0 5 0 0 6 0 0 7 0 0 8 0 0 9 0 0 0 1 0 0 2 0 0 3 0 0 4 0 0 5 0 0 6 0 0 7 0 0 8 0 0 9 0 0 0 1 0 0 2 0 0 3 0 0 4 0 0 5 0 0 6 0 0 7 0 0 8 0 0 9
Z = zeros(3,1);
B =[A(:,1) Z Z A(:,2) Z Z A(:,3) Z Z
Z A(:,1) Z Z A(:,2) Z Z A(:,3) Z
Z Z A(:,1) Z Z A(:,2) Z Z A(:,3)]
B = 9×9
1 0 0 4 0 0 7 0 0 2 0 0 5 0 0 8 0 0 3 0 0 6 0 0 9 0 0 0 1 0 0 4 0 0 7 0 0 2 0 0 5 0 0 8 0 0 3 0 0 6 0 0 9 0 0 0 1 0 0 4 0 0 7 0 0 2 0 0 5 0 0 8 0 0 3 0 0 6 0 0 9
cTroels
cTroels 2024-1-16
The function is called 1-10 million times, dependent on the problem, taking 40 seconds of 120 seconds total runtime.
Matt J
Matt J 2024-1-16
编辑：Matt J 2024-1-16
I think it unlikely you are going to be able to reliably optimize such an infinitessimal task in MCode. Different computers will give you different relative performance. You need to try to reorganize your computation in large vectorized batches.
cTroels
cTroels 2024-1-16
I was able to bring the execution time of the function down to 4s by just writing out the whole matrix explicitly. Should have done that initially i guess. But thanks for the suggestions! Appreciate the inputs.

请先登录，再进行评论。

请先登录，再回答此问题。

 采纳的回答

Benjamin Thompson
Benjamin Thompson 2024-1-20

0 个投票

Not sure where your data for A is coming from, or where the output of B is going to. if each A -> B mapping is independent try parfor if you have a multicore processor or using a gpuArray.

更多回答（0 个）

类别

在 帮助中心 和 File Exchange 中查找有关 Linear Algebra 的更多信息

产品

版本

R2023a

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by