Looking for where line intersects on itself

I am looking to see where a line that is plotted by x and y intersects on itself. I have tried using unique() with the index given to see where the same combination of x and y occur, but based on the data that index did not make sense. I have also tried linexline() which was for two lines intersecting not a line intersecting itself. I have also tried to just zoom in on the graph and find the point. Although the amount of detail required for that was not working on my graph.

4 Kommentare

Debadipto
Debadipto am 13 Jul. 2022
Please refer to this answer to a similar question.
Caitlin Bemis
Caitlin Bemis am 13 Jul. 2022
This does answer the question partly, but I also need the index to create new subsets of data.
My end goal is to find the AUC of the new subset of data to calculate work.
Star Strider
Star Strider am 15 Jul. 2022
This answer was posted while I was sleeping, and since I cannot improve on it, I deleted my answer.
Caitlin Bemis
Caitlin Bemis am 18 Jul. 2022
Thank you, anyways!

Melden Sie sich an, um zu kommentieren.

 Akzeptierte Antwort

Bruno Luong
Bruno Luong am 15 Jul. 2022
Bearbeitet: Bruno Luong am 17 Jul. 2022
If you interpolate your data by line segments it cross it selft at least 135 times
data = readtable('data.csv');
xy=table2array(data)';
X=polyselfx(xy);
fprintf('Number of intersections = %d\n', size(X,2))
Number of intersections = 135
figure
plot(xy(1,:),xy(2,:))
hold on
plot(X(1,:),X(2,:),'rx')
function X = polyselfx(P)
% Empty buffer
X = zeros(2,0);
filled = 0;
sizec = 0;
for n=2:size(P,2)-1
cn = seg2poly(P(:,n:n+1), P(:,1:n-1));
m = size(cn,2);
filled = filled+m;
% Buffer too small
if sizec < filled
sizec = max(filled, 2*sizec);
X(2,sizec) = 0;
end
% Store the result
X(:,filled+(-m+1:0)) = cn;
end
% remove the tails
X(:,filled+1:end) = [];
end % polyselfx
%%
function [cross_X, cross_t1, cross_t2, cross_i2] = seg2poly(s1, P)
a = s1(:,1);
M = P-a;
b = s1(:,2)-a;
nb2 = b(1)^2+b(2)^2;
% Check if the points are on the left/right side
x = [b(2) -b(1)]*M; % 1 x n
sx = sign(x);
% x -coordinates has opposite signs
crossx = sx(1:end-1).*sx(2:end) <= 0;
ix = find(crossx);
% cross point to the y-axis (along the segment)
x1 = x(ix);
x2 = x(ix+1);
d = b.'/nb2;
y1 = d*M(:,ix);
y2 = d*M(:,ix+1);
dx = x2-x1;
t1 = (y1.*x2-y2.*x1)./dx;
% Check if the cross point is inside the segment
ind = t1>0 & t1<=1;
if any(ind)
cross_t1 = t1(ind);
cross_t2= -x1(ind)./dx(ind);
cross_X = a + b*cross_t1;
cross_i2 = ix(ind);
else
cross_X = zeros(2,0);
cross_t1 = zeros(1,0);
cross_t2 = zeros(1,0);
cross_i2 = [];
end
end % seg2poly

5 Kommentare

Caitlin Bemis
Caitlin Bemis am 17 Jul. 2022
Thank you! This worked perfectly
Caitlin Bemis
Caitlin Bemis am 17 Jul. 2022
Is there a way to see where the index is for each spot? That way I can segment the data into sections.
Bruno Luong
Bruno Luong am 17 Jul. 2022
Inside the loop when cn is not impty the indexes are:
  • n and cross_i2, the 4th output argument of seg2poly which is not retreived in my code. If you need it just put them in array.
  • cross_t1 and cross_t2 are the fractional parts of the index where the intersection occurs.
Caitlin Bemis
Caitlin Bemis am 18 Jul. 2022
Okay. I am not sure how to add things to the array. I attempted to add them to the seg2poly, but the response is either "too many input arguments" or "won't be used'. Unforunately, I am very new to Matlab trying something much more complex than my skills, obviously.
Bruno Luong
Bruno Luong am 18 Jul. 2022
Bearbeitet: Bruno Luong am 18 Jul. 2022
Here we go I modify the code and you get in idx the (first) indexes of the line segments and the fractional parametriec coordinates t of the intersections
data = readtable('data.csv');
xy=table2array(data)';
[X,idx,t]=polyselfx(xy);
fprintf('Number of intersections = %d\n', size(X,2))
figure
plot(xy(1,:),xy(2,:))
hold on
plot(X(1,:),X(2,:),'rx')
function [X, loc, t] = polyselfx(P)
% Crossing points polygons P
%
% INPUTS:
% P: two-row arrays, each column is a vertices
% OUTPUTS:
% X is two-row array, each column is an crossing point
% loc: two-row array, which edges the crossing point belong?
% first row corresponds to P1, second row to P2
% edge#1 is P(:,[1 2]), edge#2 is P(:,[2 3]), ... etc
% t: floating parametric of the crossing point
% Empty buffer
X = zeros(2,0);
loc = zeros(2,0);
t = zeros(2,0);
filled = 0;
sizec = 0;
for n=2:size(P,2)-1
[cn, t1, t2, i2] = seg2poly(P(:,n:n+1), P(:,1:n-1));
m = size(cn,2);
filled = filled+m;
% Buffer too small
if sizec < filled
sizec = max(filled, 2*sizec);
X(2,sizec) = 0;
loc(2,sizec) = 0;
t(2,sizec) = 0;
end
% Store the result
ifill = filled+(-m+1:0);
X(:,ifill) = cn;
loc(1,ifill) = n;
loc(2,ifill) = i2;
t(1,ifill) = t1;
t(2,ifill) = t2;
end
% remove the tails
X(:,filled+1:end) = [];
loc(:,filled+1:end) = [];
t(:,filled+1:end) = [];
end % polyselfx
%%
function [cross_X, cross_t1, cross_t2, cross_i2] = seg2poly(s1, P)
a = s1(:,1);
M = P-a;
b = s1(:,2)-a;
nb2 = b(1)^2+b(2)^2;
% Check if the points are on the left/right side
x = [b(2) -b(1)]*M; % 1 x n
sx = sign(x);
% x -coordinates has opposite signs
crossx = sx(1:end-1).*sx(2:end) <= 0;
ix = find(crossx);
% cross point to the y-axis (along the segment)
x1 = x(ix);
x2 = x(ix+1);
d = b.'/nb2;
y1 = d*M(:,ix);
y2 = d*M(:,ix+1);
dx = x2-x1;
t1 = (y1.*x2-y2.*x1)./dx;
% Check if the cross point is inside the segment
ind = t1>0 & t1<=1;
if any(ind)
cross_t1 = t1(ind);
cross_t2= -x1(ind)./dx(ind);
cross_X = bsxfun(@plus, a, b*cross_t1);
cross_i2 = ix(ind);
else
cross_X = zeros(2,0);
cross_t1 = zeros(1,0);
cross_t2 = zeros(1,0);
cross_i2 = [];
end
end % seg2poly

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Kategorien

Produkte

Version

R2020a

Community Treasure Hunt

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

Start Hunting!

Translated by