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

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0 个投票

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

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Aaron Garcia 2020-4-29

编辑：Aaron Garcia 2020-4-29

在 MATLAB Online 中打开

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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Answer 1

David Hill 2020-4-29

在 MATLAB Online 中打开

0 个投票

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

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David Hill 2020-4-29

在 MATLAB Online 中打开

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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