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

조회 수: 13(최근 30일)
amoda
amoda 2022년 8월 7일
답변: amoda 2022년 8월 14일
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
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

채택된 답변

Star Strider
Star Strider 2022년 8월 7일
Possibly:
idx = find(diff(sign(y - 0.8*max(y))));
xvals = x(idx)
That should find the closest indices to all of them.
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Star Strider
Star Strider 2022년 8월 10일
편집: Star Strider 2022년 8월 11일
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')
.

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추가 답변(4개)

David Hill
David Hill 2022년 8월 7일
[~,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)
  댓글 수: 1
amoda
amoda 2022년 8월 10일
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)

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amoda
amoda 2022년 8월 10일
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.

Bruno Luong
Bruno Luong 2022년 8월 10일
편집: Bruno Luong 2022년 8월 10일
*
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

amoda
amoda 2022년 8월 14일
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')

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