how remove tow of three consecutive same values in a vector
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DuckDuck
il 23 Apr 2014
Commentato: the cyclist
il 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?
9 Commenti
Geoff Hayes
il 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?
Risposta accettata
Azzi Abdelmalek
il 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 Commenti
the cyclist
il 24 Apr 2014
You did not mention anything about this in your original question. How were we to know this is what you needed?
Più risposte (1)
the cyclist
il 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 Commenti
Geoff Hayes
il 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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