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

 Akzeptierte Antwort

Matt J
Matt J am 27 Feb. 2018
Bearbeitet: Matt J am 27 Feb. 2018

0 Stimmen

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

6 Kommentare

Jason
Jason am 27 Feb. 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 am 27 Feb. 2018
Bearbeitet: Jason am 27 Feb. 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 am 27 Feb. 2018
It could happen that Y is increasing, couldn't it ?
Matt J
Matt J am 27 Feb. 2018
Bearbeitet: Matt J am 27 Feb. 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.

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (1)

Sean de Wolski
Sean de Wolski am 27 Feb. 2018
Bearbeitet: Sean de Wolski am 27 Feb. 2018

0 Stimmen

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 Kommentar

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

Melden Sie sich an, um zu kommentieren.

Gefragt:

am 27 Feb. 2018

Bearbeitet:

am 27 Feb. 2018

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by