I'm trying to create a code that doing something like this ......

조회 수: 1 (최근 30일)
Javier
Javier 2013년 7월 4일
Hello........ I'm trying to create a code that doing something like this
data=[1 1 1 -1 1 1 -1 -1 -1 1 1 1]
take the sequence of positive and negative data
= 3 -1 2 -3 3
and, take the position of the value when it changes from positive to negative
= 1 4 5 6 7 10
%----------------------------
I made this code, but only works for positive numbers....... =(
data=[1 1 1 -1 1 1 -1 -1 -1 1 1 1]
z = v > 0;
id = find([true;diff(v.') ~= 0]);
k = diff([id;numel(v)+1]);
out = k(z(id));
Thanks for your help

채택된 답변

Guru
Guru 2013년 7월 4일
You had the answer, just had to create a variable of the negative values. I am assuming that
v = data
So just change your code to this
data=[1 1 1 -1 1 1 -1 -1 -1 1 1 1]
z = v > 0;
id = find([true;diff(v.') ~= 0]);
k = diff([id;numel(v)+1]);
% Set all values as if they were negative 1, then assign those that are not
out = -k;
out(z(id)) = k(z(id));

추가 답변 (1개)

Jan
Jan 2013년 7월 5일
data = [1 1 1 -1 1 1 -1 -1 -1 1 1 1];
[v, n] = RunLength(data);
neg = v < 0;
n(neg) = -n(neg);
  댓글 수: 1
Javier
Javier 2013년 7월 5일
Thanks Jan Simon, but I think that my MEX version not support your code "RunLength =( " Warning: You are using gcc version "4.6.3-1ubuntu5)". The version currently supported with MEX is "4.3.4". For a list of currently supported compilers see: http://www.mathworks.com/support/compilers/current_release/

댓글을 달려면 로그인하십시오.

카테고리

Help CenterFile Exchange에서 Logical에 대해 자세히 알아보기

태그

Community Treasure Hunt

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

Start Hunting!

Translated by