필터 지우기
필터 지우기

Comparing vector elements by percentage

조회 수: 2 (최근 30일)
yousef Yousef
yousef Yousef 2015년 11월 1일
답변: Image Analyst 2015년 11월 2일
Angle=[5 10 20 60 180 190 195 300 310], these angles determine the possible directions of robots. If there are 2 robots starting with the same angle or plus minus 5%(5 degrees) they will make collision. I'd like to have the result in on vector such that:[2 2 1 1 1 2 2 1 1]. Any help?
  댓글 수: 5
Stefan Raab
Stefan Raab 2015년 11월 2일
편집: Stefan Raab 2015년 11월 2일
Hello,
the codeline
collision(i)= Angle(inToleranceIndexes)
won't work, if inToleranceIndexes has more than one value==1. Then Angle(inToleranceIndexes) i.e. is [300 310] and you can't assign a vector to a single element as you do with collision(i).
But a question in general: You say tolerance is 0.05*targetValue, that means your tolerance increases with increasing start-angle. Is this really how it is supposed to be?
Best regards, Stefan
yousef Yousef
yousef Yousef 2015년 11월 2일
Hi,the tolerance is plus minus 5 degrees.The tolerance is fixed and can not be changed.I have another try,please check it.
  • Angle=[5 10 20 60 180 190 200 300 310];
  • collision=[];
  • for i=1:9
  • c=[];
  • targetValue = Angle(i);
  • tolerance = 0.5 * targetValue;
  • differences = abs(Angle-targetValue) ;
  • inToleranceIndexes = differences < tolerance;
  • withinTolerance = Angle(inToleranceIndexes);
  • c=[c;inToleranceIndexes];
  • end
  • collision=[collision;c];

댓글을 달려면 로그인하십시오.

답변 (1개)

Image Analyst
Image Analyst 2015년 11월 2일
As mentioned in your duplicate question there will be no collision unless you have different starting points. And all your code does is tell you that each angle interferes/collides with no other angle except itself, which is pretty useless.

카테고리

Help CenterFile Exchange에서 Programming에 대해 자세히 알아보기

제품

Community Treasure Hunt

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

Start Hunting!

Translated by