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Help to Improve this short code (intersection point between two straight lines given end points)

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Aaron Garcia
Aaron Garcia 2020-4-29
关闭： MATLAB Answer Bot 2021-8-20
Hello there!!
I have write this code to find the intersection point given initial and final point of two straight lines, its to simple but in implementation calling this function to many times is not fast as another functions (InterX.m) . Im not shure why is too slow. Please help !!!! Thank you
function [pX] = pInter(A,B,C,D)
%intersection point between two straight lines given end points
%A : initial point of first straight line
%B : final point of first straight line
%C: initial point of second straight line
%D : final point of second straight line
a = B(2) - A(2);
b = A(1) - B(1);
c = a*A(1) + b*A(2);
a1 = D(2) - C(2);
b1 = C(1) - D(1);
c1 = a1*C(1)+ b1*C(2);
det = a*b1 - a1*b;
if det == 0
pX= NaN; %are parallel
else
X = (b1*c - b*c1)/det;
Y = (a*c1 - a1*c)/det;
pX=[X,Y];
end
end
  1 个评论
Aaron Garcia
Aaron Garcia 2020-4-29
编辑：Aaron Garcia 2020-4-29
I find the solution !!! by calling it 1000 times in another code, total time decrease from 15s to 3s !!!
The problem was how i call for the function(variables) , this is how i was doing:
Pi1 = [x(1) L1(1)];
Pi2 = [x(end) L2(end)];
Pf1 = [x(1) L1(1)];
Pf2 = [x(end) L2(end)];
pX=pInter(Pi1,Pf1,Pi2,Pf2);
First solution:
Call
pX = pInterNew1(x,L1,L2);
Function
function [pX] = pInterNew1(x,L1,L2)
A = [x(1) L1(1)];
B = [x(end) L1(end)];
C = [x(1) L1(1)];
D = [x(end) L2(end)];
%====Same code of original function
end
Anoter solution:
Call
pX = pInterNew2(x(1),x(end),L1(1),L1(end),L2(1),L2(end));
function
function [pX] = pInterNew2(xin,xEnd,Pi1,Pf1,Pi2,Pf2)
a = Pf1 - Pi1;
b = xin - xEnd;
c = a*xin + b*Pi1;
a1 = Pf2 - Pi2;
b1 = xin - xEnd;
c1 = a1*xin+ b1*Pi2;
det = a*b1 - a1*b;
if det == 0
pX= NaN; %are parallel
else
X = (b1*c - b*c1)/det;
Y = (a*c1 - a1*c)/det;
pX=[X,Y];
end
end

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回答（1 个）

David Hill
David Hill 2020-4-29
function ans = pInter(A,B,C,D)
a=B-A;
b=D-C;
if isequal(a(2)/a(1),b(2)/b(1))
NaN;
else
[-a(2)/a(1),1;-b(2)/b(1),1]\[A(2)-A(1)*a(2)/a(1);C(2)-C(1)*b(2)/b(1)];
end
  1 个评论
David Hill
David Hill 2020-4-29
I would just load up all your points in a matrices A,B,C,D and only execute the function once.
function p = pInter(A,B,C,D)
a=B-A;
b=D-C;
p=zeros(size(A));
for k=1:length(A)
if isequal(a(k,2)/a(k,1),b(k,2)/b(k,1))
p(k,:)=[NaN,NaN];
else
p(k,:)=([-a(k,2)/a(k,1),1;-b(k,2)/b(k,1),1]\[A(k,2)-A(k,1)*a(k,2)/a(k,1);C(k,2)-C(k,1)*b(k,2)/b(k,1)])';
end
end

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