checking if every component of a vector is nonnegative
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What is the easiest way to check whether a vector a is >= 0?
I know you could type
a>=0 and get a vector of ones if that's true, but I am looking for something that will give me only one 1 as an answer.
For instance,
a=[2 5 3 6]; >> a>=0
ans =
1 1 1 1
But I am looking for something that will give me just
ans=
1 (as the answer)
any idea?
2 comentarios
Image Analyst
el 11 de Sept. de 2014
That is the same answer that Walter gave below almost 2 years ago.
Respuestas (2)
Walter Roberson
el 21 de Dic. de 2012
all(a >= 0)
However, what if there are NaN in the array? NaN is "nonnegative" (in some definitions) but also not >= 0 .
2 comentarios
Suraj Parasuram
el 2 de Mzo. de 2022
@Walter Roberson Is this available in Simulink? How can I implement this in Simulink?
Walter Roberson
el 2 de Mzo. de 2022
minmax block to take the min() of the input.
if-else testing whether the min was >= 0
In some contexts you would instead take sign() https://www.mathworks.com/help/simulink/slref/sign.html .
If you add 2 to the sign() then you can use that as the index into a vector of length 3 -- "some entry less than 0" "smallest entry was 0" "smallest entry was greater than 0"
Shaun VanWeelden
el 21 de Dic. de 2012
An extremely intuitive answer would just be return the minimum of the logic matrix that results from your inequality. It will only be 1 if every value turns out to be a 1.
min(A>=0) and by the way it returns a zero for NaN
2 comentarios
Teja Muppirala
el 21 de Dic. de 2012
The difference between using ALL and MIN, is that when you use ALL it will stop searching through the vector once it finds even a single location where the condition is not true. So it will work a slight bit faster than MIN in this case.
For example:
% Takes 1 sec. on my PC
R = zeros(1,1e8); R(1) = -1;
tic; for n = 1:10, all(R >= 0); end; toc;
% Takes 2 sec. on my PC
R = zeros(1,1e8); R(1) = -1;
tic; for n = 1:10, min(R >= 0); end; toc;
Walter Roberson
el 21 de Dic. de 2012
I don't think it has ever been conclusively demonstrated that all() stops at the first false, or whether it is parallelized for "big enough" matrices. My recollection is that someone created a mex version that was found to be faster than all() in ways that you would expect if all() tested all (or most) locations -- e.g., much faster for the mex version if the condition was false early.
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