find negative values of simple summation. code efficiency
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Hello, I have the following statement which must be true: v1_before(1) == v1_transferred_vector(1) + v1_conserved_vector(1)
My goal is to find out which of the two elements on the right side are actually negative. The algebraic sign of "v1_before" is already known, as well as the absolute values of the two elements on the right side.
The indices refer to the x-value of the [x,y] vectors. So i will do exactly the same with (2) / y-values too.
I coded the following. But im wondering whether there is a more elegant and efficient way to do it. Many thanks in advance!!!!
%determine whether + / - for x-values
if v1_before(1) - v1_transferred_vector(1) - v1_conserved_vector(1) == 0 %==0 when difference is smaller then 1e-15/eps
%do nothing, leave them positiv
elseif v1_before(1) + v1_transferred_vector(1)- v1_conserved_vector(1) == 0
v1_transferred_vector(1) = - v1_transferred_vector(1);
elseif v1_before(1) - v1_transferred_vector(1) + v1_conserved_vector(1) == 0
v1_conserved_vector(1) = - v1_conserved_vector(1);
else %v1_before(1) + v1_transferred_vector(1) + v1_conserved_vector(1) == 0
v1_transferred_vector(1) = - v1_transferred_vector(1);
v1_conserved_vector(1) = - v1_conserved_vector(1);
end
6 个评论
Rik
2021-1-3
I don't think there is an elegant way to do this (your code is already fairly clear and efficient), but you could consider storing the three values in temporary variables and using a nested loop (or ndgrid) to make sure you don't miss any combination and it is still easy to adapt the code.
Ansgar Elles
2021-1-4
Ansgar Elles
2021-1-5
Rik
2021-1-5
If this code only runs once it doesn't matter if it takes 1 ms or 100 ms, but if this code is executed thousands of times in a time-sensitive context, then it does start to matter.
The main point is to spend time optimizing the code that matters. You can spend a few hours here to save a few miliseconds, or spend your time on code that runs often and takes longer to complete.
Ansgar Elles
2021-1-5
回答（1 个）
Walter Roberson
2021-1-5
M = [1 1; 1 -1; -1 1; -1 -1];
m = M(v1_before(1) == M*[v1_transferred_vector(1); v1_conserved_vector(1)],:);
v1_transfered_vector(1) = m(1)*v1_transfered_vector(1);
v1_conserved_vector(1) = m(2)*v1_conserved_vector(1);
This can be vectorized with a slight bit of work.
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2021-1-3
2021-1-5
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