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checking if every component of a vector is nonnegative

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Del
Del 2012-12-21
评论： Walter Roberson 2022-3-2
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?

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回答（2 个）

Walter Roberson
Walter Roberson 2012-12-21
all(a >= 0)
However, what if there are NaN in the array? NaN is "nonnegative" (in some definitions) but also not >= 0 .
  2 个评论
Walter Roberson
Walter Roberson 2022-3-2
https://www.mathworks.com/help/simulink/slref/minmax.html
minmax block to take the min() of the input.
https://www.mathworks.com/help/simulink/slref/if.html
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"

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Shaun VanWeelden
Shaun VanWeelden 2012-12-21
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 个评论
Teja Muppirala
Teja Muppirala 2012-12-21
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
Walter Roberson 2012-12-21
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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