how remove tow of three consecutive same values in a vector

1 Ansicht (letzte 30 Tage)
I have two vectors like this
a=[1,2,3,1,1,1,2,3,2,3,1,1,1,1,2,3,3,3,2,2,1],
where * b * is id of * a *
b=[1,2,3,11,12,13,20,21,25,27,31,32,33,34,36,40,41,42,47,48,50]
what i want to achieve is to keep the midle element of three consecutive elements or the even element if te consecuence is more than 3 elements. i want to keep the b[4] or b[11] and b[13] also the b[16]. how to achieve this?? Can any body help?
  9 Kommentare
DuckDuck
DuckDuck am 23 Apr. 2014
Bearbeitet: DuckDuck am 23 Apr. 2014
@Geoff Hayes well i didn't thought about the odd case with more elements, but i thin i would keep the midle of the first three and the 5th, or i will go for every second, so in a bunch of five i would get 2. Thanks for the hint, though. But how about the case when i have 3 consecutive 1s in a, but they are not consecutive in b?
Geoff Hayes
Geoff Hayes am 23 Apr. 2014
I don't understand "..but they are not consecutive in b". I thought that it was only the consecutive integers in a that we were concerned with?

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Akzeptierte Antwort

Azzi Abdelmalek
Azzi Abdelmalek am 23 Apr. 2014
a = [1,2,3,1,1,1,2,3,2,3,1,1,1,1,2,3,3,3,2,2,1];
b = [1,2,3,11,12,13,20,21,25,27,31,32,33,34,36,40,41,42,47,48,50];
c=diff(a)==0;
ii1=strfind([0 c 0],[0 1 1]);
ii2=strfind([0 c 0],[1 1 0])+1;
out=cell2mat(arrayfun(@(x,y) b(x+1:2:y),ii1,ii2,'un',0))
  2 Kommentare
DuckDuck
DuckDuck am 24 Apr. 2014
Bearbeitet: DuckDuck am 24 Apr. 2014
ok but where are the first and last 1?
i need also the lonely 1s and those in pair.
the cyclist
the cyclist am 24 Apr. 2014
You did not mention anything about this in your original question. How were we to know this is what you needed?

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Weitere Antworten (1)

the cyclist
the cyclist am 23 Apr. 2014
This code is just terrible, but I think it does what you want.
a = [1,2,3,1,1,1,2,3,2,3,1,1,1,1,2,3,3,3,2,2,1];
b = [1,2,3,11,12,13,20,21,25,27,31,32,33,34,36,40,41,42,47,48,50];
d = [];
currentRunLength = 1;
for i = 2:numel(a)
if a(i)==a(i-1)
currentRunLength = currentRunLength+1;
if ((currentRunLength==2) && (i~=numel(a)) && a(i)==a(i+1)) || ((currentRunLength>=4) && mod(currentRunLength,2)==0)
d = [d,b(i)];
end
else
currentRunLength = 1;
end
end
  3 Kommentare
the cyclist
the cyclist am 23 Apr. 2014
I think my code covers all cases. Do you have a counter-example?
Geoff Hayes
Geoff Hayes am 23 Apr. 2014
My mistake - I had read your a(i)==a(i-1) as ~= and so figured that a final check would be needed.
That being said, is it only the ids of the duplicates that are to be kept (from b) or should it be the ids of the single elements as well?

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