finding approximate indices in a monotonically increasing array

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I need to find the indices of an entry in a monotonically increasing array, where the value of this indices is "ref"(as shown below)
I create my increasing array "t" as follow
dt=1/10e9;
N=13600000;
t=((0:N-1)*dt);
ref=1.39459e-05;
I've already tried using the find function but it leads to incorrect results or multiple indices or no answers at all.
can anyone suggest any way I can get this working
  1 comentario
Jan
Jan el 31 de En. de 2013
Please post the find command you have used, when you want us to suggest an improvement.

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Respuesta aceptada

Shashank Prasanna
Shashank Prasanna el 30 de En. de 2013
You can use Nearest Neighbor search. If you have stats toolbox:
f = knnsearch(t',ref)
f =
139460
>> t(f)
ans =
1.39459e-05
IF you don't have the stats toolbox:
>> tri = delaunayn(t');
f = dsearchn(t',tri,ref)
f =
139460
Because you have so many points you have to be patient since it takes time.
  2 comentarios
Sean de Wolski
Sean de Wolski el 31 de En. de 2013
There is so much extra working put into calculating the Delaunay triangulation for all of t when all you need is a simple histogram calculation of one element! I would not recommend this approach.
Foosball?
Shashank Prasanna
Shashank Prasanna el 31 de En. de 2013
I agree, i just like using nearest neighbor search, and it always works. lets go.

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Más respuestas (3)

Sean de Wolski
Sean de Wolski el 30 de En. de 2013
[~,idx] = histc(ref,t);
Use histc() to find the bin containing ref.
  1 comentario
Jan
Jan el 31 de En. de 2013
HISTC is a fast and efficient C-Mex function, which does not create the temporary vector abs(t - ref) and performs a binary search, which is the optimal strategy in theorie. +1

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Jan
Jan el 31 de En. de 2013
dt = 10e-9;
N = 13600000;
t = (0:N-1)*dt;
ref = 1.39459e-05;
[value, index] = min(abs(t - ref));
But without doubt, the binary search of histc is smarter than comparing all values in this brute force approach.

Arsalan
Arsalan el 31 de En. de 2013
Editada: Arsalan el 31 de En. de 2013
Hi Guys, thanks for your suggestions, I never knew about the use of some of this commands before. but I found much more efficient way
dt=1/10e9;
N=13600000;
t=((0:N-1)*dt);
ref=1.39459e-05;
index=round(ref/dt);
my_value=t(index); % should be approximately equal to ref
it works well for values greater then zero, but should work for values less then zero with some manipulation

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