Using Logical Indexing with Matrices

조회 수: 3 (최근 30일)
Andrew
Andrew 2014년 2월 6일
편집: Matt J 2014년 2월 6일
Hello,
I am currently working on something where I have a large number of three dimensional points (8.9e7 points) expressed in a matrix like X=[x;y;z] where x, y, and z are row vectors containing the Cartesian coordinates of each point.
I need to search these points and pull out sets of points that fall on a given plane with a tolerance and store these points to a new matrix like X2=[x2;y2;z2] where x2, y2, and z2 are points from X that fall on the defined plane.
I have tried something like this:
%generating spherical point cloud
n=8937*10000;
L=sqrt(n*pi());
k=1:n;
z=ones(size(k))-(2*k-ones(size(k)))./n;
phi=acos(z);
theta=L*phi;
x=sin(phi).*cos(theta);
y=sin(phi).*sin(theta);
rbody=1737; alt=50;
X=[x*rbody;y*rbody;z*rbody];
%defining plane
ns=[1,1,1]; %normal vector
planeloc=rbody^2/(alt+rbody); %distance from origin
%Storing values that fit on the plane
X2=X(ns*(X-planeloc*ns'*ones(1,length(x)))>=-0.0001 & ns*(X-planeloc*ns'*ones(1,length(x)))<=0.0001);
max(max(X2)) %testing to see if any values fit
but it does not perform what I want. When I use the logical index
X(ns*(X-planeloc*ns'*ones(1,length(x)))>=-0.0001 & ns*(X-planeloc*ns'*ones(1,length(x)))<=0.0001)
because the index returns values for each element of the matrix. Is there some way that I can use logical indices to select the columns of X that satisfy
(ns*(X-planeloc*ns'*ones(1,length(x)))>=-0.0001 & ns*(X-planeloc*ns'*ones(1,length(x)))<=0.0001)
without running a for loop (which would obviously take forever due to the huge number of points).
Thanks, Andrew

채택된 답변

Matt J
Matt J 2014년 2월 6일
편집: Matt J 2014년 2월 6일
tmp=ns*X-planeloc*norm(ns)^2;
X2=X(:,abs(tmp)<.0001);

추가 답변 (0개)

카테고리

Help CenterFile Exchange에서 Resizing and Reshaping Matrices에 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