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How to get x value for a given y value in a interp1 figure?

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If I want to find x value for y==0.5, how can I find that? We only have 9 points.
  2 comentarios
Dyuman Joshi
Dyuman Joshi el 28 de Mzo. de 2024
Use the interp1 function as you have mentioned.
What seems to be the problem?
Refer to its documentation for information regarding accepted syntaxes.
wuwei han
wuwei han el 28 de Mzo. de 2024
Sorry, I described wrong situation.
vv=[0.190934428093434,0.277000826121106,0.477820361464529,0.703789686451856,1,0.703789686451856,0.477820361464529,0.277000826121106,0.190934428093434]
plot(-40/3/8:10/3/8:40/3/8,vv, 'r*-')
like this, how can I find x for y==0.5?
Thanks to your answer.

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Respuesta aceptada

John D'Errico
John D'Errico el 28 de Mzo. de 2024
Editada: John D'Errico el 28 de Mzo. de 2024
You need to understand there is no unique solution. So asking for THE value of x has no answer, since there are two possible solutions.
vv=[0.190934428093434,0.277000826121106,0.477820361464529,0.703789686451856,1,0.703789686451856,0.477820361464529,0.277000826121106,0.190934428093434]
vv = 1×9
0.1909 0.2770 0.4778 0.7038 1.0000 0.7038 0.4778 0.2770 0.1909
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
plot(-40/3/8:10/3/8:40/3/8,vv, 'r*-')
Can you solve the problem? Well, yes. It not not that difficult, as long as you accept the idea of multiple non-unique solutions, and you use a tool that can do the job correctly. One such tool is intersections, as found on the file exchange (written by Doug Schwarz.)
You asked for the x-value of those points where the curve crosses y==0.5.
[xint,yint] = intersections(-40/3/8:10/3/8:40/3/8,vv,[-2 2],[.5 .5])
xint =
-0.79244
0.79244
yint =
0.5
0.5
You can find intersections for free download here:
intersections is fast and robust. It can handle cases where the curve has an infinite slope for example, where interp1 will fail miserably.

Más respuestas (2)

Star Strider
Star Strider el 28 de Mzo. de 2024
Editada: Star Strider el 28 de Mzo. de 2024
To get both of them —
vv=[0.190934428093434,0.277000826121106,0.477820361464529,0.703789686451856,1,0.703789686451856,0.477820361464529,0.277000826121106,0.190934428093434];
L = numel(vv)
L = 9
xv = -40/3/8:10/3/8:40/3/8;
dv = 0.5;
zxi = find(diff(sign(vv - dv)));
for k = 1:numel(zxi)
idxrng = max(1, zxi(k)-1) : min(L,zxi(k)+1);
xp(k) = interp1(vv(idxrng), xv(idxrng), dv);
end
format long
xp
xp = 1×2
-0.792436118381859 0.792436118381859
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
figure
plot(xv, vv, '*-r')
hold on
plot(xp, ones(size(xp))*dv, 'sg', 'MarkerSize', 10)%, 'MarkerFaceColor','g')
plot(xp, ones(size(xp))*dv, '+k', 'MarkerSize',10)
hold off
yline(0.5, '--k')
.
  9 comentarios
wuwei han
wuwei han el 28 de Mzo. de 2024
Thank you for your answer!>.<

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Yash
Yash el 28 de Mzo. de 2024
Hi Wuwei,
You can use the "interp1" function to find the values from the interpolation.
Given below is an example:
x_ref = [-1.7 -1.2 -0.7 -0.4 0 0.4 0.7 1.2 1.7];
y_ref = [0.2 0.3 0.5 0.7 1 0.7 0.5 0.3 0.2];
x_test = [0.5];
y_test = interp1(x_ref, y_ref, x_test)
y_test = 0.6333
plot(x_ref, y_ref, 'r', x_test, y_test, 'ko')
Since "x" is not a function of "y" here (one value of "y" leads to multiple values of "x"), you cannot directly use "interp1" for "x" vs "y". Either use half of the points in that case, or use some other techniques to make "x" a function of "y".
You can refer to the following documentation of the interp1 function for more details: https://www.mathworks.com/help/matlab/ref/interp1.html
Hope this helps!
  1 comentario
wuwei han
wuwei han el 28 de Mzo. de 2024
Thank you very much. I have known there may be no way to find 'x' by 'y' directly.

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