Mapping 1D vector to 2D area

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Prasanna Routray
Prasanna Routray il 27 Set 2024
Commentato: Prasanna Routray il 2 Nov 2024 alle 6:43

load xPoints; load yPoints; j=boundary(xPoints,yPoints,0.1); Plot(xPoints(j),yPoints(j))

%How do I map the x-values to y-values here?

  4 Commenti
Prasanna Routray
Prasanna Routray il 27 Set 2024
Hi Rahul,
I want to get a set of points that map to a particular x-value.
The boundary actually forms an area.
I want to get a function or method so that whenever I get an 'x', I should be able to get a set of y-values.
Hope that make the question clear. The forum was having an issue with server so, posted it from my phone.
Image Analyst
Image Analyst il 27 Set 2024
What if, for a given vertical line (like you specified x with some specific value), there are no y values for that exact x value? Maybe some are close but not exact. Do you want to find all y values within a certain tolerance of your specified x? If so use ismembertol().

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Cris LaPierre
Cris LaPierre il 27 Set 2024
Are you wanting all the corresponding yPoints, or just those on the boundary?
load xPoints;
load yPoints;
j=boundary(xPoints,yPoints,0.1);
To me, the simplest approach is to find the indices of the desired X value, and use that the extract the corresponding Y values.
idx = xPoints==2;
yPoints(idx)
ans = 5×1
0.0016 0.0015 0.0015 0.0014 0.0014
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That will return all points. If you just want them from the boundary, try this.
ids = xPoints(j)==2;
yPoints(j(ids))
ans = 0.0014
Here, only one value is returned because only one X value in boundary exactly equals 2. In that case, you could use ismembertol.
LIA = ismembertol(xPoints(j),2,0.01);
yPoints(j(LIA))
ans = 2×1
0.0013 0.0014
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  9 Commenti
Cris LaPierre
Cris LaPierre il 1 Nov 2024 alle 14:38
Modificato: Cris LaPierre il 1 Nov 2024 alle 18:02
The difference approach doesn't appear to be working here. Assuming the boundary shape is always like this, consider using the location of the max x value to separate your data.
load xPointsNew;
load yPointsNew;
j=boundary(xPointsNew,yPointsNew,0.1);
plot(xPointsNew(j),yPointsNew(j))
[~,ind] = max(xPointsNew(j))
ind = 64
% Define x value
x = 5;
% Can only use interp on unique X values, so split j into increasing and
% decreasing x values
idx1 = 1:ind-1;
idx2 = ind:length(j);
% interpolate to find corresponding y values when increasing and decreasing
y1 = interp1(xPointsNew(j(idx1)),yPointsNew(j(idx1)),x);
y2 = interp1(xPointsNew(j(idx2)),yPointsNew(j(idx2)),x);
y = [y1,y2]
y = 1×2
0.0036 0.0048
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hold on
plot(x*ones(length(y),1),y)
hold off
Prasanna Routray
Prasanna Routray il 2 Nov 2024 alle 6:43
Hi Cris, great that you could found another way.
I too found a different way that uses polyshape and intersect function.
load xPointsNew;
load yPointsNew;
j=boundary(xPointsNew,yPointsNew,0.01);
plot(xPointsNew(j),yPointsNew(j))
pgon = polyshape(xPointsNew(j),yPointsNew(j));
x= 6;
lineseg = [x 0; x 0.01]; % [x1 y1; x2 y2]
[in,out] = intersect(pgon,lineseg);
hold on
plot(in(:,1),in(:,2),'b',out(:,1),out(:,2),'r')
xlabel( 'X [m]', 'Interpreter', 'latex', FontSize=18 );
ylabel( 'Y [m]', 'Interpreter', 'latex', FontSize=18 );
set(gca, fontsize=22, fontname='Times')
ax = gca
ax.YAxis.Exponent = 0;
axis padded
hold on
Cheers!

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Rahul
Rahul il 27 Set 2024
I believe that you're trying to want to obtain a reverse mapping, from xPoints data to yPoints data, using a MATLAB function. Here's how you can code the same:
function [res_x, res_y] = getYs(x, xPoints, yPoints)
x = 5;
n = size(xPoints, 1);
res_y = [];
res_x = [];
for i=1:n
if xPoints(i) == x
res_y = [res_y yPoints(i)];
res_x = [res_x x];
end
end
end
Use the above function to get yPoints values corresponding to a given 'x', plot the resultant values on the figure, and display the resultant array 'res_y':
load xPoints; load yPoints;
j=boundary(xPoints,yPoints,0.1);
plot(xPoints(j),yPoints(j), 'Color','black');
hold on;
% Call getYs to get corresponding y values for a given x = 5
x = 5;
[res_x, res_y] = getYs(x, xPoints, yPoints);
% Plot returned data using dotted red line on same graph
plot(res_x, res_y, 'r.');
disp(res_y);
hold off;
  1 Commento
Prasanna Routray
Prasanna Routray il 30 Set 2024
Hi Rahul,
I have been trying to do reverse mapping. however, I'm looking for a set of points and not just the boundary points as I posted in one of the replays.
Thanks

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