Find X-value given Y

Hello

I've been doing some research regarding this but I can't seem to find a solution for when I want to find the X-value given Y.

I have the following data:

>> X

ans =

            20          50         100         200         500        1000        2000        5000       10000       20000

>> Y

ans =

     -0.1755   -0.1755   -0.9151   -2.8534   -8.4043  -13.1515  -20.0000  -27.5350  -33.9794  -40.0000

Where the X is plotted on a logarithmic scale. I tried interp1 with X and Y switched but that dosent work since its not strictly monotonic increasing.

Does anyone have a solution on this? Say I want to find the value of X given Y=-5

Answer 1

[Star Strider]

The ‘x’ and ‘y’ vectors were not the same sizes, so I added 25 to ‘x’, then added a small amount to the duplicated element in ‘y’ and did the interpolation:

x = [25  50         100         200         500        1000        2000        5000       10000       20000];
y = [ -0.1755   -0.1755   -0.9151   -2.8534   -8.4043  -13.1515  -20.0000  -27.5350  -33.9794  -40.0000];
dy = diff([0 y]);
dyix = find(dy == 0);
y(dyix) = y(dyix-1)+1E-8;

xint = interp1(y, x, -10);              % Find The Value Of ‘x’ Corresponding To y=-10’

[Adaptine]

Works like a charm. Do you mind explaining what's happening there?

[Star Strider]

Don’t mind at all. This is MATLAB Answers!

The code first takes the difference in the ‘y’ vector (in ‘dy’), padding it with a leading 0 so the indices will line up. It next finds the index of the zero element (in ‘dyix’), and then adds a very small value to the corresponding ‘y’ element, so the vector is now monotonically increasing. That problem now solved, interp1 is happy.

This can be extended to other situations with multiple duplicates, but I didn’t generalise to that here because you only have one duplicate.

[Adaptine]

Thank you very much. So the duplicate y-value is the reason interp1 didnt like swapped x and y?

[Star Strider]

My pleasure.

Exactly. The interp1 function has no idea how to interpret two values of the dependent variable for one value of the independent variable. (Nor would I, without making some assumptions that interp1 wisely avoids.) By clearing up that ambiguity (so that the independent variable is monotonically increasing and the duplication in it no longer exists), interp1 proceeeds. (It also needs its first two arguments to have the same size in one dimension.)

[the cyclist]

Note that his X and Y vectors did actually have equal numbers of elements. They scroll off the right-hand side of the window, though.

Answer 2

[the cyclist]

Here is a terribly kludgy way to do it, but it gets the job done. It sorts Y, and then in the case of equal entries in Y, it will add a tiny offset. (This will work even if there are long series of consecutive equal entries.)

X = [20          50         100         200         500        1000        2000        5000       10000       20000];
Y = [-0.1755   -0.1755   -0.9151   -2.8534   -8.4043  -13.1515  -20.0000  -27.5350  -33.9794  -40.0000];

[sortedY sortingIndex] = sort(Y);
sortedX = X(sortingIndex);

for ny = 2:numel(sortedY);
    if sortedY(ny)==sortedY(ny-1)
        sortedY(ny) = sortedY(ny) + eps(sortedY(ny));
    end
end

interpolated_X = interp1(sortedY,sortedX,-5)

[Sean Eruppakkattu]

Nice solution for preventing consecutive equal entries, though maybe I would use for the sake of intuitivity the following code to order the X and Y arrays:

A = [X,Y]; % matrix
res = sortrows(A,2); % matrix is ordered with respect with the Y-column

with also significant one-order speed improvement. Then the for-loop would simple refer to the column vector of the matrix.