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# negative continuous position on vector

조회 수: 3(최근 30일)
Alejandro Fernández 2020년 8월 16일
댓글: hosein Javan 2020년 8월 17일
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...

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### 채택된 답변

Bruno Luong 2020년 8월 16일
편집: Bruno Luong 2020년 8월 17일
BB = (1:length(AA))-find(AA==1,1)
##### 댓글 수: 1표시숨기기 없음
Alejandro Fernández 2020년 8월 17일
Well, you could make it a lot shorter than I did... thank you very much.

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### 추가 답변(2개)

hosein Javan 2020년 8월 16일
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
##### 댓글 수: 2표시숨기기 이전 댓글 수: 1
hosein Javan 2020년 8월 17일
Ofcourse. thank you.

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Sara Boznik 2020년 8월 16일
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
##### 댓글 수: 1표시숨기기 없음
Alejandro Fernández 2020년 8월 17일
Thank you so much, you can see the previous coments, they make it with just 1 step, thank you so much by the way.

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R2020a

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