negative continuous position on vector

Hi I have a question, i know how to solve with the code I show in the bottom of the page but i think that it has to be an easyer way. What I want to do is: starting with a vector row AA containing only one one in one position, for example:

AA = [0 0 1 0 0 0 0];

I want to get a BB vector that's next:

BB = [-2 -1 0 1 2 3 4];

A vector in which the numbers to the left of 1 are negative crescent in which the first component is the most negative position possible, the point 1 becomes 0 and the right of 1 is positive crescent until the end of the vector.

The code I know how to do is this:

AA = [0 0 1 0 0 0 0]
[m,n] = size(AA);
x = find(AA(1,:),1)
neg = AA(1:x-1)
BB(1,x) = 0
BB(1,1:length(neg)) = -length(neg):1:-1
BB(1,x+1:n) = 1:1:n-x

But I don't see that this is a very good way to get the solution...

Answer 1

[Bruno Luong]

BB = (1:length(AA))-find(AA==1,1)

[Alejandro Fernández]

Well, you could make it a lot shorter than I did... thank you very much.

Answer 2

[hosein Javan]

how about:

k = find(logical(A));
n = length(A);
B = 1-k:n-k

A =

0 0 0 0 1 0 0 0

B =

-4 -3 -2 -1 0 1 2 3

[Alejandro Fernández]

Yes thank you very much, I have to give the MVP to Bruno because he did it earlier and in less steps but I appreciate it very much.

[hosein Javan]

Ofcourse. thank you.

Answer 3

[Sara Boznik]

AA=[0 0 1 0 0 0 0]
ne=-1;
po=1;
[n,m]=size(AA)
b=find(AA(1,:)==1)
BB=zeros(n,m)
for i=1:b-1
    BB(1,i)=ne
    ne=ne-1
end
for j=b+1:m
    BB(1,j)=po
    po=po+1
end

[Alejandro Fernández]

Thank you so much, you can see the previous coments, they make it with just 1 step, thank you so much by the way.