Finding the first occurance using interp1

Hello. I have some data (red curve) and I'm trying to find the X value at which Y = 0.2.
I have used
Y20 = interp1(Y,X,0.2,'linear')
which works well, but finds the last occurrance.
How can I find the first occurrance (i.e. around x=9)
Thanks

 Réponse acceptée

Matt J
Matt J le 27 Fév 2018
Modifié(e) : Matt J le 27 Fév 2018

0 votes

Use only the first two data X,Y data points in the interpolation.

6 commentaires

Jason
Jason le 27 Fév 2018
Hi, but its not always the 1st two data points. sometime my real data is very similar to the black curve.
Then go through your data from the beginning, note when the Y-value is first greater or equal to 0.2 (say at position I in the Y-array) and then call interp1 as
Y20 = interp1(Y(1:I),X(1:I),0.2,'linear')
Best wishes
Torsten.
Jason
Jason le 27 Fév 2018
Modifié(e) : Jason le 27 Fév 2018
Like this:
[~,I] = min(abs(Y - 0.2))
i=find(Y(1:end-1)>=0.2 & Y(2:end)<=0.2,1);
Y20=interp1(Y(i:i+1),X(i:i+1),0.2);
Torsten
Torsten le 27 Fév 2018
It could happen that Y is increasing, couldn't it ?
Matt J
Matt J le 27 Fév 2018
Modifié(e) : Matt J le 27 Fév 2018
Not according to the posted figure, but even if it could, I think the extension is an exercise I'll leave for the OP.

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Plus de réponses (1)

Sean de Wolski
Sean de Wolski le 27 Fév 2018
Modifié(e) : Sean de Wolski le 27 Fév 2018
Use cummax and cummin to find the the first set of points that cross 0.2. Then interp just them.
x = 1:10
y = sin(x)
plot(x,y)
yval = 0.2;
idx = find(cummin(y)<0.2 & cummax(y)>0.2, 1, 'first')
interp1(y([idx-1 idx]), x([idx-1, idx]), 0.2)

1 commentaire

Jason
Jason le 27 Fév 2018
Thankyou for your answer. Im sorry I can't accept both. Matt came first.

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