Find "intersection" of data array

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Christian
Christian el 3 de Sept. de 2018
Comentada: Christian el 5 de Sept. de 2018
Hello everybody,
I would like to find the "intersections" of different data arrays. The arrays look similar this:
x=[0 1.2 2.5 3.4 4.6 5.4 6.8 7.3 8.6 9.8 7.1 8.5 9.9 10.1 11.4]
y=[100 90 80 70 65 60 55 50 45 40 40 42 44 52 62]
I would like to cut off the part under the intersection and redefine the data set. So it would be enough to find the two connection points (in this example (8.6;45) and (9.9;44)). Any ideas, how to do this? I have ran out of ideas and really need new ideas/inspiration from you guys!
Cheers, Christian
  6 comentarios
Walter Roberson
Walter Roberson el 4 de Sept. de 2018
What would be the expected result for a case such as
9
13
10
11 12
where the data "doubles back" but does not intersect itself?
Christian
Christian el 4 de Sept. de 2018
@Walter Roberson
all of my data-sets intersect, so I won't face a case as you have described. But thanks to "Stephen Cobeldick" I found a perfect solution for my Problem -> check the answer below!

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Respuesta aceptada

Stephen23
Stephen23 el 4 de Sept. de 2018
Editada: Stephen23 el 4 de Sept. de 2018
Download and use selfintersect:
I tried it on your example values:
>> x = [0,1.2,2.5,3.4,4.6,5.4,6.8,7.3,8.6,9.8,7.1,8.5,9.9,10.1,11.4];
>> y = [100,90,80,70,65,60,55,50,45,40,40,42,44,52,62];
>> [~,~,s] = selfintersect(x,y)
s =
9 12
>> xnew = x([1:s(1),1+s(2):end])
xnew =
0.00000 1.20000 2.50000 3.40000 4.60000 5.40000 6.80000 7.30000 8.60000 9.90000 10.10000 11.40000
>> ynew = y([1:s(1),1+s(2):end])
ynew =
100 90 80 70 65 60 55 50 45 44 52 62
>> plot(x,y,'-*',xnew,ynew,'-*')
If you have multiple intersects you will have to use a loop (or some alternative) to process each row of the output matrix s.
  3 comentarios
Stephen23
Stephen23 el 4 de Sept. de 2018
@Christian: I hope that it helps. If you find selfintersects useful, then please also go and give it a rating at the page where you downloaded it from. I am sure that its author will appreciate that.
Christian
Christian el 5 de Sept. de 2018
Done! :)

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