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

If I want to find x value for y==0.5, how can I find that? We only have 9 points.

[Dyuman Joshi]

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]

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.

Answer 1

[John D'Errico]

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

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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:

https://www.mathworks.com/matlabcentral/fileexchange/11837-fast-and-robust-curve-intersections?s_tid=ta_fx_results

intersections is fast and robust. It can handle cases where the curve has an infinite slope for example, where interp1 will fail miserably.

[wuwei han]

Thank you very much! I have already known that.

Answer 2

[Star Strider]

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

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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')

.

[wuwei han]

Thank you very much!

[Star Strider]

As always, my pleasure!

[wuwei han]

Sorry, I found that that green point is not on the line.

[Star Strider]

It is on the red line, and it is at 0.5.

[wuwei han]

-0.79244

0.79244

But this should be the true result.

[wuwei han]

Thanks a lot. The function 'intersections' solves my question perfectly.

[Star Strider]

I did not previously see that you had supplied the ‘x’ vector in the plot call. That is now corrected.

[wuwei han]

Thank you for your answer!>.<

[Star Strider]

My pleasure!

Answer 3

[Yash]

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!

[wuwei han]

Thank you very much. I have known there may be no way to find 'x' by 'y' directly.