How to find x value for a known y in array plot?

I have to draw a plot based on a long matrix, which I did, after that I got the maximum of y value (max(y)) but I need the x value for the known y value y=0.8*max(y) THIS y value is not included in the matrix, so I can't use the Index of the maximum to get my x

I did found a similar question https://de.mathworks.com/matlabcentral/answers/258556-how-to-i-find-a-x-value-from-a-given-y

and tried to apply the same code x(y==0.8*max(y)) but I get a 0x1 empty column vector.

Can anyone help me out? thanks in advance

Answer 1

[Star Strider]

Possibly:

idx = find(diff(sign(y - 0.8*max(y))));

xvals = x(idx)

That should find the closest indices to all of them.

[amoda]

Hello Star Strider,

thank you for suggestion, I got a vector of 89 Indices and another with 89 corresponding x-values, but to be honest I am not sure how should I understand that, which one is the closest one to my desired x-value.

[Star Strider]

Without the data, it is difficult to determine that. It will be necessary to define a specific x-value to determine which is closest.

Example —

x = linspace(0, 20, 500);                               % Create Data
y = sin(2*pi*x) + cos(3*pi*x) - (x-10).^2/100;          % Create Data

idx = find(diff(sign(y  - 0.8*max(y))));
xvals = x(idx)

xvals = 1×12

5.2505 5.3307 7.2144 7.3747 9.2184 9.3788 11.2224 11.3828 13.2265 13.3467 15.2305 15.3106

x_desired = 10;                                         % Define Target X-Value
[xabsdist,xidx] = min(abs(xvals - x_desired));
% idx(xidx)
% x(idx(xidx))
% y(idx(xidx))
fprintf(1,'Target Y-Value: %.3f\nClosest X-Value: %.3f\nCorresponding Y-Value: %.3f\nY-Distance |Desired-Actual|: %.3f\n', 0.8*max(y), x(idx(xidx)), y(idx(xidx)), abs(0.8*max(y)-y(idx(xidx))))

Target Y-Value: 1.517
Closest X-Value: 9.379
Corresponding Y-Value: 1.596
Y-Distance |Desired-Actual|: 0.080

figure
plot(x, y, 'DisplayName','Data')
hold on
plot(xvals, y(idx), 'ks', 'DisplayName','Closest Indices')
plot(x(idx(xidx)), y(idx(xidx)), 'rs', 'MarkerFaceColor','r', 'DisplayName','Closest X-Value')
yline(0.8*max(y), '--g', 'DisplayName','0.8\times y_{max} Reference')
xline(x_desired, '--r', 'DisplayName','Desired X-Value')
hold off
grid
legend('Location','best')

This uses synthetic data, however it should work for your actual data.

Make appropriate changes to get the result you want. (It is also possible to interpolate to get the exact x- and y-values rather than just using the indices.)

EDIT — (11 Aug 2022 at 1:00)

Added interpolation example.

Example —

tgt = 0.8*max(y);
L = numel(x);
for k = 1:numel(idx)
    idxrng = max(1,idx(k)-1) : min(L,idx(k)+1);
    xv(k) = interp1(y(idxrng), x(idxrng), tgt);
end

[xvabsdist,xvidx] = min(abs(xv - x_desired));
fprintf(1,'Closest X-Value: %.3f\nDistance |xv - x-desired|: %.3f\n', xv(xvidx),xvabsdist)

Closest X-Value: 9.386
Distance |xv - x-desired|: 0.614

figure
plot(x, y, 'DisplayName','Data')
hold on
plot(xv, ones(size(xv))*tgt, 'ks', 'DisplayName','Exact X-Values')
plot(xv(xvidx), tgt, 'rs', 'MarkerFaceColor','r', 'DisplayName','Closest X-Value')

yline(0.8*max(y), '--g', 'DisplayName','0.8\times y_{max} Reference')
xline(x_desired, '--r', 'DisplayName','Desired X-Value')
hold off
grid
legend('Location','best')

.

Answer 2

[David Hill]

[~,idx]=min(abs(y-.8*max(y)));%.8*max(y) is not part of the y-array, find the closest y-idx to that value
x(idx)

[amoda]

Hey David,

thanks a lot for your answer, unfortunately your code delivers no satisfiying result, because it gives me back, the corresponding X-value of max(y) which I used to get my y=0.8*max(y)

Answer 3

[amoda]

meanwhile I tried to use the interpolation function Interp1, hoping it could deliver the best result, but I was wrong, I think it didnt work because I have only vectors and no functions.

Answer 4

[Bruno Luong]

*

x = linspace(-3,3);

Assuming your data are monotonic on the neighborhood of the max

y = exp(-x.^2);

[maxy, imax] = max(y);

yt = 0.8*maxy;

% left side
i1 = find(y<yt & 1:length(y)<imax, 1, 'last');
if isempty(i1)
    error('no x found on the left side')
end
xl = interp1(y([i1 i1+1]), x([i1 i1+1]), yt)

xl = -0.4729

% right side
i2 = find(y<yt & 1:length(y)>=imax, 1, 'first');
if isempty(i2)
    error('no x found on the right side')
end
xr = interp1(y([i2-1 i2]), x([i2-1 i2]), yt)

xr = 0.4729

Answer 5

[amoda]

thank you all for your constructive answers, they all led me to find finally the most acceptable solution in my opinion:

yMax=max(y)

[idx,idy]=find(y==yMax)

Xcoresp=x(idx,idy)

y80=0.80*y

then I created linear vector which include my y80 value and apply the interpolation function

x1=linspace(x(1,1),Xcoresp,1000)

y1=linspace(y(1,1),yMax,1000)

xDesired=interp1(y1,x1,y80,'nearest')