How to find x value for certain y value of a lineplot in matlab

I want to get EV value at P=50% without real data point at P=0.5 in the lineplot.

[佳丽 周]

EV=[-15 -14 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15]

P=[0.0312500000000000,0,0.0312500000000000,0.0156250000000000,0.00625000000000000,0.0208333333333333,0.0186011904761905,0.0429687500000000,0.0282118055555556,0.0312500000000000,0.0539772727272727,0.0758838383838384,0.130080856643357,0.158545100732601,0.256227106227106,0.439955357142857,0.648752289377290,0.705156822344322,0.740785256410257,0.809659090909091,0.869444444444444,0.903125000000000,0.912946428571429,0.949218750000000,0.971726190476191,0.989583333333333,0.987500000000000,0.992187500000000,0.968750000000000,1,1]

Answer 1

[Torsten]

EV = EV(5:14);
P = P(5:14);
EV_05 = interp1(P,EV,0.5)

[佳丽 周]

I still get NaN result in my data,thank you anyway!

[Torsten]

Can't be true according to your graph.

I assumed EV are the x-axis values and P are the y-axis values.

And I count 17 values each.

[佳丽 周]

I am quite confused now :(

[Torsten]

EV=[-15 -14 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15];
P=[0.0312500000000000,0,0.0312500000000000,0.0156250000000000,0.00625000000000000,0.0208333333333333,0.0186011904761905,0.0429687500000000,0.0282118055555556,0.0312500000000000,0.0539772727272727,0.0758838383838384,0.130080856643357,0.158545100732601,0.256227106227106,0.439955357142857,0.648752289377290,0.705156822344322,0.740785256410257,0.809659090909091,0.869444444444444,0.903125000000000,0.912946428571429,0.949218750000000,0.971726190476191,0.989583333333333,0.987500000000000,0.992187500000000,0.968750000000000,1,1];
EV = EV(10:20);
P = P(10:20);
EV_05 = interp1(P,EV,0.5,'linear')

EV_05 = 0.2876

[佳丽 周]

Thank you !!!

[Torsten]

Explanation: You have to choose an interval for P where the values are monotonically increasing. Otherwise, you'll get NaN from the interpolation. Since this is not the case for your complete P vector, I only took the values from P(10) to P(20).

[佳丽 周]

Before posting online, I tried to fit a linear equation according to the nearest two data points. But I didn't get what I want proprely. May be chose a larger interval will be better fit my data.Thanks for your good explanation!

Answer 2

[SALAH ALRABEEI]

You can find the the EV from the index of p value.

Example;

a=[4,2,3,5,6,7]

a = 1×6

4 2 3 5 6 7

b = a.^2 

b = 1×6

16 4 9 25 36 49

Now, to find what is a at b= 9!

inx = find(b==9);
a(inx)

ans = 3

[佳丽 周]

Thanks,but it doesn't work for my data.In your example,I want to find specific "a" value when b' doesn't exist in real ”b“.

[Sam Chak]

Hi @佳丽 周

If the specific data and governing equation are unavailable, then you can only rely on Interpolation and Approximation Theory to find out the value.

More importantly, can you furnish the 17 visible data points (EV, P) for one-step further investigation of your NaN result?

Answer 3

[Sam Chak]

Hi @佳丽 周

No joke, but this is the graphical approach.

[佳丽 周]

Okk! I got what your meaning.Interpolation work for my data finally.Your graph is pretty straightforward and cute.Thanks for your assitance

[Sam Chak]

Cute? Vote if you think it is cute, but I guess you are probably cuter, Ms. Zhou 😊