how remove tow of three consecutive same values in a vector

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?

[Azzi Abdelmalek]

This is not clear

[DuckDuck]

why is not clear? b is the id of a from another array. i have consecutive same values in a, and looking at b, you can see that those values in the array were also consecutive. i need only one of them, the midle one, when they are in 3 or every second when more than 3.

[the cyclist]

@effess, you say you want the middle element when there is a group of three. But then you say you want b[4]. That's the first element of the group of three, not the middle. b[5] is the middle one. Which do you want?

[DuckDuck]

excuse me, i wrote it b[4] because i started the array from 0, like in programming.

[the cyclist]

"like in programming" (hehe).

Just remember that MATLAB is 1-based index, not 0-based.

[Geoff Hayes]

@effess, MATLAB uses one-based array indexing so the first element in b is accessed by b(1) .

A relatively simple solution would be to just loop through all elements of a and count the number of consecutive integers that are identical starting at index i say. Upon a change in the integer, at index j say, where a(j)~=a(j-1) then you know that all integers in a from index i to index j-1 are identical.

If the count is one (i.e. (j-1)-i+1==1) then you need only copy/keep b(i). If the count is three, then just keep the middle element b(i+1), and if the count is even, then keep every other element, b(i+1:2:j-1).

What I'm not sure of, is how to handle the case of (for example) 5 consecutive integers. Do you keep the middle of the first three, and then the last: so b(3) and *b(5) (if identical from 1 to 5) or do something else? How do you expect to handle the case of 7 consecutive integers? Or 9?

[the cyclist]

Note that taking the middle element (which is the 2nd element) of a group of 3 is a special case of the general rule of taking the even-numbered elements of groups of 3 or more.

[DuckDuck]

@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]

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?

Answer 1

[Azzi Abdelmalek]

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))

[DuckDuck]

 ok but where are the first and last 1?
i need also the lonely 1s and those in pair.

[the cyclist]

You did not mention anything about this in your original question. How were we to know this is what you needed?

Answer 2

[the cyclist]

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

[Geoff Hayes]

Don't forget to capture, outside of the for loop, the case where consecutive identical integers are at the end of vector a i.e. currentRunLength>1.

[the cyclist]

I think my code covers all cases. Do you have a counter-example?

[Geoff Hayes]

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?