negative continuous position on vector

2 views (last 30 days)
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...

Accepted Answer

Bruno Luong
Bruno Luong on 16 Aug 2020
Edited: Bruno Luong on 17 Aug 2020
BB = (1:length(AA))-find(AA==1,1)
  1 Comment
Alejandro Fernández
Alejandro Fernández on 17 Aug 2020
Well, you could make it a lot shorter than I did... thank you very much.

Sign in to comment.

More Answers (2)

hosein Javan
hosein Javan on 16 Aug 2020
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
  2 Comments
Alejandro Fernández
Alejandro Fernández on 17 Aug 2020
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.

Sign in to comment.


Sara Boznik
Sara Boznik on 16 Aug 2020
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
  1 Comment
Alejandro Fernández
Alejandro Fernández on 17 Aug 2020
Thank you so much, you can see the previous coments, they make it with just 1 step, thank you so much by the way.

Sign in to comment.

Categories

Find more on Statistics and Machine Learning Toolbox in Help Center and File Exchange

Products


Release

R2020a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!