# how remove tow of three consecutive same values in a vector

1 Ansicht (letzte 30 Tage)
DuckDuck am 23 Apr. 2014
Kommentiert: the cyclist am 24 Apr. 2014
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?
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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 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))
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the cyclist am 24 Apr. 2014

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### Weitere Antworten (1)

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
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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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