How to find position of a new point along a series of line segments?

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Dominik Mattioli 2021-6-1

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https://ww2.mathworks.cn/matlabcentral/answers/845425-how-to-find-position-of-a-new-point-along-a-series-of-line-segments

编辑： Adam Danz 2021-6-1

Say that you know that your query point Q is colinear to one line segment in a sequence of line segments XY. If XY is an Mx2 array and you wanted to insert your 1x2 point Q into XY, how do you figure out which two points of XY to split up? Example"

XY = [1 1; 2 1; 3 1; 5 1; 9 1];
Q  = [7 1];
% Figure out where Q lies within XY
% indQ = ...
newXY = vertcat( XY( 1:indQ, : ), Q, XY( indQ+1 : end, : ) );

Sure you could test colinearity of Q to each line segment of XY. I'm wondering if there is a geometric construct for this, though.

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Answer 1

Adam Danz 2021-6-1

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/845425-how-to-find-position-of-a-new-point-along-a-series-of-line-segments#answer_714560

编辑：Adam Danz 2021-6-1

在 MATLAB Online 中打开

For collinear points,

XY = [1 1; 2 1; 3 1; 5 1; 9 1];
Q  = [7 1];
XYQ = sortrows([XY;Q])
XYQ = 6×2
     1     1
     2     1
     3     1
     5     1
     7     1
     9     1

For coordinates along a non-linear line, you can compute the distance between Q (1xn) and each point in XY (mxn). Q is inserted between the two points in XY that are the shortest distance to Q if and only if the two points are next to each other. If the two closest points are not next to each other then placement of Q cannot be determined and an error will be thrown by the assert command.

XY = [1 1; 2 1; 3 1; 5 1; 9 1];
Q  = [7 1];
dist = pdist2(Q,XY);
[~, min2idx] = mink(dist,2);
assert(diff(min2idx)==1, 'Placement cannot be determined.')
XYQ = [XY(1:min2idx(1),:); Q; XY(min2idx(2):end,:)]
XYQ = 6×2
     1     1
     2     1
     3     1
     5     1
     7     1
     9     1

Demo using non-linear coordinates

x = 0:.5:pi;

XY = [x',sin(x)'];

Q = XY(5,:)

Q = 1×2

2.0000 0.9093

XY(5,:) = []

XY = 6×2

0 0 0.5000 0.4794 1.0000 0.8415 1.5000 0.9975 2.5000 0.5985 3.0000 0.1411

dist = pdist2(Q,XY);

[~, min2idx] = mink(dist,2);

assert(diff(min2idx)==1, 'Placement cannot be determined.')

XYQ = [XY(1:min2idx(1),:); Q; XY(min2idx(2):end,:)]

XYQ = 7×2

0 0 0.5000 0.4794 1.0000 0.8415 1.5000 0.9975 2.0000 0.9093 2.5000 0.5985 3.0000 0.1411

plot(XYQ(:,1), XYQ(:,2), 'bo-')

hold on; grid on

plot(Q(1),Q(2),'r*')

The first method, using sortrows, works for all of these examples but would fail in some non-linear lines such as circles.

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Adam Danz 2021-6-1

Collinearity implies straight lines,

Three or more points are said to be collinear if they lie on a single straight line .

I'll update my answer to show how to sort nonlinear coordinates.

Dominik Mattioli 2021-6-1

Oh, thanks for catching that. I used the word because I wanted to say that Q is colinear with one of the line segments defining XY. I see your point, though.

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