필터 지우기
필터 지우기

How to trim a ponit cloud

조회 수: 12 (최근 30일)
Hall marokey
Hall marokey 2017년 2월 19일
답변: Hongxu Guo 2020년 10월 13일
I have generated pointcloud -- ptCloud using pcfromkinect function ( as in the Matlab example) . Now I want to trim the pointcloud using
x = ptCloud.Location(100:200 ,100:200,1);
y = ptCloud.Location(100:200 ,100:200,2);
z = ptCloud.Location(100:200 ,100:200,3);
R = ptCloud.Location(100:200 ,100:200,1);
G = ptCloud.Location(100:200 ,100:200,2);
B = ptCloud.Location(100:200 ,100:200,3);
PTC.Location(:,:,1) =x;
PTC.Location(:,:,2) =y;
PTC.Location(:,:,3) =z;
PTC.Color(:,:,1) = R;
PTC.Color(:,:,2) = G;
PTC.Color(:,:,3) = B;
then pcshow(PTC)
but not working any sugesion or matlab file exchange files wiill be highy appreciated
  댓글 수: 3
Image Analyst
Image Analyst 2017년 2월 19일
I don't even know what you want to do. What does it mean to "trim a point cloud"? Do you mean crop? Remove outliers? "Trim" does not seem to be a precisely defined term in your context. Like for strings, trim means to remove leading and trailing white space. But I have no idea what you mean. Well, 2 ideas that I just gave but I still am not sure if it's one of those or something else.
Hall marokey
Hall marokey 2017년 3월 10일
In fact I have a point cloud obtained from kinect v2. Now I want to identify objects using color information . Once identified I would like to reduce the whole point cloud to identified object only. So I search the color information in the point cloud ptCloud.Color with RGB values. I dont know how use a find command for a 3d array for color or like [i,j] = find(ptCloud.color ==[ R G B]) Thanks for help

댓글을 달려면 로그인하십시오.

답변 (1개)

Hongxu Guo
Hongxu Guo 2020년 10월 13일
Just in case someone is still looking for solving similar issue
Pls look here.
https://www.mathworks.com/help/vision/ref/pointcloud.findpointsinroi.html

카테고리

Help CenterFile Exchange에서 Point Cloud Processing에 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