Creating a matrix that is more time efficient

2018 3 16
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更新时间：2018 3 17
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Hey guys,
My problem is as follows.
I want to create a matrix that has certain logic like so.
% code
n = 50; % Just an arbitrary number
B = rand(1,n);
A = rand(1,n);
% The components of the vectors A and B are not necessary a number between 0 and 1. But a calculated number, but for the purpose of this problem I'd rather use this random generated vector.
for i = 1:n
for j = 1:n
if i == j
C(i, j) = 5*B(1, i) + A(1,j);
else
C(i, j) = -B(1, i) + 3*A(1, j);
end
end
end
The code up above does the works, but it takes quite a long time, since it makes each component individually.
So my question is, is there a more efficient way to create this matrix?

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James Tursa
James Tursa 2018-3-16
编辑：James Tursa 2018-3-16

1 个投票

C = 3*A - B(:); % or C = bsxfun(@minus,3*A,B(:))
C(1:n+1:end) = 5*B + A;

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Nicholas Ootani
Nicholas Ootani 2018-3-16
Really liked your answer, but if you allow me, lets increase a little bit the difficult. The equation that i'm working with is exactly as follows - when you aren't working with the main diagonal , in another words i =~ j.
C(i,j) = sin(A(j))/((cos(A(j))-cos(A(i)))^2)*(((1-(-1)^(j-i))/(2*(n+1))));
Do you know how to solve that little part that uses "(j-i)"?
And for the main diagonal the equation is as follows.
B(i,j) = -((n+1)/(4*sin(A(i)))+B(i));
But with your solution, I've figured it out.
Thank you very much.
James Tursa
James Tursa 2018-3-16
编辑：James Tursa 2018-3-16
There's no B in your C off-diagonal expression? Or is that a typo? I assume your diagonal line is a typo and should have had C(i,i) on the lhs.
Nicholas Ootani
Nicholas Ootani 2018-3-17
That is correct, there isn't any B in the off diagonal. The problem are those indexes that i don't want to use a for loop to cover than all.
And the main diagonal was a mistake
C(i,j) = -((n+1)/(4*sin(A(i)))+B(i));
James Tursa
James Tursa 2018-3-17
What about this:
N = 1:n;
C = sin(A)./((cos(A)-cos(A(:))).^2).*(((1-(-1).^(N-N(:)))/(2*(n+1))));
C(1:n+1:end) = -((n+1)./(4*sin(A))+B);
This assumes you are running with a later version of MATLAB that has implicit expansion for the element-wise operators. If not, then you would have to use bsxfun( ) for a few of those operations.
Nicholas Ootani
Nicholas Ootani 2018-3-17
I'm running the R2016a version. And probably it doesn't work, i'm getting a "Matrix dimension must agree".
So probably somehow bsxfun could a way to do it.
But anyway Thank you very much, at least i've learned some new logic to use in future programs =D
James Tursa
James Tursa 2018-3-17
The bsxfun( ) version:
N = 1:n;
% D = sin(A)./((cos(A)-cos(A(:))).^2).*(((1-(-1).^(N-N(:)))/(2*(n+1))));
cosA = cos(A);
sinA = sin(A);
costemp = bsxfun(@minus,cosA,cosA(:));
sincostemp = bsxfun(@rdivide,sinA,costemp.^2);
Ntemp = bsxfun(@minus,N,N(:));
D = sincostemp.*(((1-(-1).^Ntemp)/(2*(n+1))));
D(1:n+1:end) = -((n+1)./(4*sinA)+B);
Nicholas Ootani
Nicholas Ootani 2018-3-17
It worked, man, you are awesome.
Never thought that you could use bsxfun() like that.
once more, thanks, problem solved

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