How to Interpolate and Find X given Y

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Hello,
I have a problem where I need to get the x-values corresponding to where a horizontal line intersects with the data below.
I've tried using interp1 but it seems like it only returns to upper bound value, when I would like for it to return the location of both interception points.
Attached is the graphical output to the code below.
x = [50 43 35 22 15 12 11 10 8.5 6.5 5.75 5 4 3 2.75 2.5 2.25 2 1.67 1.37 1.2 1 0.7 0.55 0.3 0.25];
y = [0.3 0.35 0.43 0.52 0.6 0.67 0.79 0.91 1.02 1.1 1.2 1.35 1.23 1.23 1.20 1.17 1.15 1.14 1.07 1.02 0.95 0.83 0.73 0.31 0.21 0.18];
y_target = 0.8;
figure(1)
semilogx(x,y*0.75)
grid on
title('x y reverse interpolation')
xlabel('x')
ylabel('y')
hold on
y1 = yline(y_target,'-');
x1 = xline(1.67,'-'); %would like function or method of finiding x1 at a given y_target
x2 = xline(7.2,'-');

采纳的回答

Matt J
Matt J 2025-7-10
编辑:Matt J 2025-7-10
One way:
x = [50 43 35 22 15 12 11 10 8.5 6.5 5.75 5 4 3 2.75 2.5 2.25 2 1.67 1.37 1.2 1 0.7 0.55 0.3 0.25];
y = [0.3 0.35 0.43 0.52 0.6 0.67 0.79 0.91 1.02 1.1 1.2 1.35 1.23 1.23 1.20 1.17 1.15 1.14 1.07 1.02 0.95 0.83 0.73 0.31 0.21 0.18];
x=flip(x); y=flip(y)*0.75;
y_target = 0.8;
[ymax,i]=max(y);
xmax=x(i);
fun=@(xq) abs(interp1(x,y,xq) -y_target );
x1=fminbnd( fun , x(1), xmax)
x1 = 1.6500
x2=fminbnd( fun , xmax, x(end)),
x2 = 7.3333
semilogx(x,y); xline([x1,x2]); yline(y_target);

更多回答(1 个)

Matt J
Matt J 2025-7-11
编辑:Matt J 2025-7-11
Here's yet another approach that uses linexlines2D, downloadable from,
xy = [[50 43 35 22 15 12 11 10 8.5 6.5 5.75 5 4 3 2.75 2.5 2.25 2 1.67 1.37 1.2 1 0.7 0.55 0.3 0.25];
0.75*[0.3 0.35 0.43 0.52 0.6 0.67 0.79 0.91 1.02 1.1 1.2 1.35 1.23 1.23 1.20 1.17 1.15 1.14 1.07 1.02 0.95 0.83 0.73 0.31 0.21 0.18]];
y_target = 0.8;
I=linexlines2D( xy(:,1:end-1), xy(:,2:end), [0,1,-y_target]); %intersections
I(:,all(isnan(I),1))=[];
semilogx(xy(1,:) , xy(2,:)); yline(y_target); hold on
plot(I(1,:), I(2,:),'or'); hold off

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