Identifying x value at y on an xy plot

I have a fairly simple problem that I imagine has a simple solution. I have a plot in which the x-axis represents a location (0-8) and the y-axis is the probability of a certain action occuring at that location. I need to identify the x value when there is a 50% chance of the action happening. Example data below.

x = [0,1,2,3,4,5,6,7,8];
y = [0,0,0,.25,.25,.25,.75,1,1];
plot(x,y);
line([0,8],[.5,.5]);

I just need to identify the x value when y is 0.5. Is there a simple solution I am missing here? Thank you!

Answer 1

[Star Strider]

A common problem interpolating y-values that are not unique is how to make them appropriate for interpolation. One way is to simply choose a small range of data in the vicinity of the desired interpolation point.

Try this:

x = [0,1,2,3,4,5,6,7,8];
y = [0,0,0,.25,.25,.25,.75,1,1];

idx = find(y <= 0.5, 1, 'last');                        % Select Small Range Of Data Based On Desired Interpolation Point
idxrng = [0 1]+idx;

Lxq = interp1(y(idxrng), x(idxrng), 0.5, 'linear');     % Interpolate Over Small Range Of Data
Pxq = interp1(y(idxrng), x(idxrng), 0.5, 'spline');     % Interpolate Over Small Range Of Data

figure
plot(x,y)
hold on
plot(Lxq, 0.5, 'p')
hold off
line([0,8],[.5,.5]);

Here the two interpolation menthods give the same result. That may not always be the situation, so specify an appropriate method.

Experiment to get different results.

[Hannah Douglas]

This solution works really well for me, thank you! So far the linear and spline interpolation produce the same value, but I'll keep them both in for comparison. Thanks

[Star Strider]

As always, my pleasure!

The linear and spline methods produce the same result in this example. However in different situations they would produce significantly different results, the reason I recommend always specifying a method to get the desired result. (The linear method is the default with interp1.)

Answer 2

[dpb]

P=0.5;                                    % P to find
iy=find(y>P,1);                           % first point past P
xp=interp1(y(iy-1:iy),x(iy-1:iy),P);      % interpolate between prior point and point

You'll have problem where your proability points are identical between observations, though.