How to find x values from y value in "fit" function?

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Kanan Yagublu
Kanan Yagublu 2022-5-10
评论： Kanan Yagublu 2022-5-11
Hi, I have some data and with this data I create a fit function:
f = fit(x,y,'smoothingspline');
plot(f,x,y);
I need to find X values in two points like in the picture.
I tried to use :
z = x(y==y_i);
But what I get is:
0×1 empty double column vector
How can I solve this problem?
Thanks in advance

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采纳的回答

David Hill
David Hill 2022-5-10
yval=-62;
eqn=@(x)feval(f,x)-yval
X1=fzero(eqn,4.8);
X2=fzero(eqn,5.3);
  2 个评论
Kanan Yagublu
Kanan Yagublu 2022-5-11
Thank you for your answer, but how did you get 4.8 and 5.3? Are they approximation from the graph?
Because when I change the data set the code doesn't work
Kanan Yagublu
Kanan Yagublu 2022-5-11
@David Hill and how can I modify it so it will not depend on 4.8 and 5.3 ?

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更多回答（1 个）

Steven Lord
Steven Lord 2022-5-10
Ran in:
Set up a sample polynomial fit.
x = randn(10, 1);
y = (x-1).*(x+1); % polynomial is y = x^2-1 = (x-1)*(x+1)
p = fit(x, y, 'poly2')
p =
Linear model Poly2: p(x) = p1*x^2 + p2*x + p3 Coefficients (with 95% confidence bounds): p1 = 1 (1, 1) p2 = 6.259e-17 (-2.697e-16, 3.949e-16) p3 = -1 (-1, -1)
Find the points where p takes on the value y = 3.
plusOneSolution = fzero(@(x) p(x)-3, 1)
plusOneSolution = 2
minusOneSolution = fzero(@(x) p(x)-3, -1)
minusOneSolution = -2
Check that evaluating the fit at those two points gives us the value y = 3.
check = p([plusOneSolution, minusOneSolution])
check = 2×1
3.0000 3.0000

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