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Repeated event in a array

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Mate 2u
Mate 2u el 4 de Mayo de 2012
Hi there, I have a vector of size 10,000,000 x 1. The vector has inputs of 1 0 and -1 only. I want to find repeated events.
For instance 1 1 0 1 1 -1 -1 1 , I want to test if this exact same elements have appeared again in the data.
If so I would like a graph to represent the whole data plus the pattern in a colour.
Looking forward to a answer. Thanks.
  2 comentarios
Jan
Jan el 4 de Mayo de 2012
The question is not clear.
Your input is in = [1 1 0 1 1 -1 -1 1];
What is the wanted output? What are "repeated events"? What kind of graph do you want? For a suggestion how to draw a pattern in color, we have to knwo, what a "pattern" is in your case.
Mate 2u
Mate 2u el 4 de Mayo de 2012
Ok here is a example... A = [ 1 0 1 0 1 1 1 -1 -1 1 0 1 0 1 0 -1 -1]
And if I want to search for 1010, it will plot a graph and whereever there is 1 0 1 0 in order it will come out coloured in that part of the graph

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Daniel Shub
Daniel Shub el 4 de Mayo de 2012
Loren talks about findings patterns in arrays.
Processing your example
A = [1 0 1 0 1 1 1 -1 -1 1 0 1 0 1 0 -1 -1];
x = [1 0 1 0];
with Loren's method would give
[1 10 12]
This can then be plotted with
stem(A);
hold on;
arrayfun(@(n)stem((0:3)+n, A((0:3)+n),'r'), [1, 10, 12]);
  3 comentarios
Daniel Shub
Daniel Shub el 4 de Mayo de 2012
yes.
Mate 2u
Mate 2u el 4 de Mayo de 2012
Is there a way where I can only plot the matched elements + the next 20 elements of the full array?
So for example, forget plotting everything, just plot what was found and then do seperate plots for each but with each one continuing for a certain period?

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Jan
Jan el 4 de Mayo de 2012
Are you looking for neighboring elements with equal values? Then use diff. Are you looking for a specific patter, use strfind(data, [1 1 0 1 1 -1 -1 1]).
  2 comentarios
Mate 2u
Mate 2u el 4 de Mayo de 2012
You are right, but is there a way to plot it so that the repeated parts are in different colour?
Mate 2u
Mate 2u el 4 de Mayo de 2012
Thank you Jan! You were very helpful! Sorry if I was not clear enough.

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