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

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佳丽 周
佳丽 周 2022 年 6 月 8 日
編集済み: Sam Chak 2022 年 6 月 8 日
I want to get EV value at P=50% without real data point at P=0.5 in the lineplot.
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佳丽 周
佳丽 周 2022 年 6 月 8 日
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]

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採用された回答

Torsten
Torsten 2022 年 6 月 8 日
EV = EV(5:14);
P = P(5:14);
EV_05 = interp1(P,EV,0.5)
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佳丽 周
佳丽 周 2022 年 6 月 8 日
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!

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その他の回答 (2 件)

SALAH ALRABEEI
SALAH ALRABEEI 2022 年 6 月 8 日
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
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Sam Chak
Sam Chak 2022 年 6 月 8 日
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?

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Sam Chak
Sam Chak 2022 年 6 月 8 日
No joke, but this is the graphical approach.
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Sam Chak
Sam Chak 2022 年 6 月 8 日
編集済み: Sam Chak 2022 年 6 月 8 日
Cute? Vote if you think it is cute, but I guess you are probably cuter, Ms. Zhou 😊

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