Mapping 1D vector to 2D area

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Prasanna Routray
Prasanna Routray 2024-9-27
评论： Prasanna Routray 2024-11-2，6:43

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?

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Prasanna Routray
Prasanna Routray 2024-9-27
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 2024-9-27
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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采纳的回答

Cris LaPierre
Cris LaPierre 2024-9-27
Ran in:
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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Cris LaPierre
Cris LaPierre 2024-11-1，14:38
编辑：Cris LaPierre 2024-11-1，18:02
Ran in:
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 2024-11-2，6:43
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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更多回答（1 个）

Rahul
Rahul 2024-9-27
Hi @Prasanna Routray,
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 个评论
Prasanna Routray
Prasanna Routray 2024-9-30
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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