Create a vector of vector exponents

Question

Jean Dupin 2019-5-17

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/462720-create-a-vector-of-vector-exponents

编辑： Adam Danz 2019-5-20

Hi,

Assume I have a vector v and a square matrix A

v = [1, 2, 3, 4, 5];
A = randn(5);

I want to create a matrix

m = [v*A; v*A^2; v*A^3; ..; v*A^n]

Where n is a user input and ^ results in the matrix product operator (not dot product .^)

I can do that easily with a loop. Is there a quicker way of doing this without using a loop?

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Matt J 2019-5-17

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/462720-create-a-vector-of-vector-exponents#answer_375557

编辑：Matt J 2019-5-17

No, a for-loop is fastest, but you want to implement it the right way, with a recursive update,

m=repmat(v*A,n,1); %pre-allocate
for i=2:n
    m(i,:) = m(i-1,:)*A;
end

That way, the process takes only n matrix multiplications.

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

Answer 2

Adam Danz 2019-5-17

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/462720-create-a-vector-of-vector-exponents#answer_375531

编辑：Adam Danz 2019-5-20

No loop method:

m = cell2mat(arrayfun(@(x)v*A.^x,1:n,'UniformOutput',false)')

Test result with loop version

mm = zeros(6,5);
for i = 1:n
    mm(i,:) = v*A.^i; 
end
isequal(m,mm) % = 1, same result

Which is faster?

To test speed, I ran 100,000 iterations of the one-liner and the same number of iterations of the for-loop (just the loop, the pre-allocation was done before the timer started). The variable n=20. The median speed of the for-loop was 2.7 times faster than the arrayfun() method (p<0.001 wilcox signed rank test). The for-loop is the clear winner for speed.