Could someone please explain me this code?

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x = linspace(min([x1,x2],[],2), max([x1,x2],[],2))
y1 = interp1(x1, y1, x, 'pchip','extrap')
y2 = interp1(x2, y2, x, 'pchip','extrap')
index = find(diff(sign(y1-y2)))
for k = 1:numel(index)
indexrange = max(1,index(k)-2) : min(numel(x),index(k)+2);
xi(k) = interp1(y1(indexrange)-y2(indexrange), x(indexrange), 0);
yi(k) = interp1(x(indexrange), y1(indexrange), xi(k));
end
Xc = [xi; yi]
Could you please explain me all the steps of this code?

Accepted Answer

Star Strider
Star Strider on 17 Oct 2021
That looks like something I wrote!
It would be helpful to know the context (I don’t remember where I posted it), however it appears that ‘x1’ and ‘x2’ are vectors spanning different ranges. The ‘x’ assignment produces a vector that spans both ranges, The ‘y1’ and ‘y2’ assignments interpolate and extrapolate the original ‘y1’ and ‘y2’ to the new ‘x’, and the ‘index’ assignment returns the approximate intersection indices of ‘y1’ and ‘y2’. The loop then interpolates first to find the accurate ‘x’ value of the intersection (as ‘xi’), and then uses that value to interpolate the accurate ‘y’ value of the intersection (as ‘yi’). The ‘indexrange’ vector are the indices of both vectors for the interpolation, and is constructed so that the first value is never less than 1 and the last value is never greater than the length of the vector. The ‘Xc’ matrix concatenates them so that ‘xi’ is the first row and ‘yi’ is the second row. This simply makes it easier to view the result of the loop.
I usually comment-document my code, so my apologies if I forgot to in that instance.
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More Answers (1)

Anitha Limann
Anitha Limann on 17 Oct 2021
That is very true!
This is written by you. You had comments there but i was quite confused with the for loop here. and why we have to intrapolate and extrapolate x and y. Also i am not familiar with numel. That is why i wanted to get a clarification on the code.
I am really thankful for your help at the time and today again!

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