how remove tow of three consecutive same values in a vector

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DuckDuck
DuckDuck 2014-4-23
评论： the cyclist 2014-4-24
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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DuckDuck
DuckDuck 2014-4-23
编辑：DuckDuck 2014-4-23
@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 2014-4-23
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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采纳的回答

Azzi Abdelmalek
Azzi Abdelmalek 2014-4-23
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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DuckDuck
DuckDuck 2014-4-24
编辑：DuckDuck 2014-4-24
ok but where are the first and last 1?
i need also the lonely 1s and those in pair.
the cyclist
the cyclist 2014-4-24
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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更多回答（1 个）

the cyclist
the cyclist 2014-4-23
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
Geoff Hayes 2014-4-23
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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