How to search the continuous values with maximal size in a vector ?

How to search continuous values in a vector,meanwhile，the size of the continuous values is the largest? For example,

a=[1 2 3 4 5 6 7 11 13 14 15 18 19]

Here the answer should be

ans=[1 2 3 4 5 6 7]

Help me to find the answer for vector of size n. I want to use this to detect a portion which is linear in a curve. Is there some function in MATLAB which could do this work? Thanks.

Answer 1

[Cedric]

You can go for a solution around the following:

 boundary = find([true, abs(diff(a))~=1, true]) ;
 [blockSize, boundaryStart] = max(diff(boundary)) ;

With this and a defined by

 >> a= [1 2 3 0 5 6 7 11 13 14 15 16 18 19] ;

you get:

 >> boundary                  % Starting ID of blocks of contiguous integers.
 boundary =
     1     4     5     8     9    13    15

 >> blockSize                 % Size of the first largest blocks.
 blockSize =
     4

 >> boundaryStart             % Starting ID of first largest block.
 boundaryStart =
     5

which you can use e.g. as follows:

 >> blockRange = boundary(boundaryStart) + (0:blockSize-1) ;
 >> a(blockRange)
 ans =
    13    14    15    16

Note that this method can be easily updated so you get all blocks that have the largest size.. it would be something like (not tested):

 bDiff = diff(boundary) ;
 boundaryStarts = find(bDiff==max(bDiff)) ;

and then you can iterate through elements of boundaryStarts to get all relevant blocks.

Answer 2

[Alessandro Renna]

I wrote this function:

function y=continuous(a)
n=length(a);
for i=1:n-1
    if diff(a(i:i+1))==1
        z(i)=a(i);
        z(i+1)=a(i+1);
    end
end
y=z(z~=0);

In this way, if a is [1 2 3 4 5 6 7 11 13 14 15 18 19], the result y will be only the continuous part of a, [1 2 3 4 5 6 7 13 14 15 18 19].

[sandberry]

Hi,thanks,for the continuous set you have got, I have added some line, so that the size of the continuous values is the largest.it goes just like this:

function y=continuous(a) n=length(a); for i=1:n-1 if diff(a(i:i+1))==1 z(i) = a(i); z(i+1) = a(i+1); end end z = z(z~=0); ind = find(diff(z)>1); ind = [0 ind length(z)]; s = cell(1,length(ind)-1); for i=1:length(ind)-1 s{i}= z(ind(i)+1:ind(i+1)); len(i)=length(s{i}); end y = s{find(max(len))};

In this way, if a is [1 2 3 4 5 6 7 11 13 14 15 18 19], the result y will be[1 2 3 4 5 6 7].

But regretful,I feel the method is not so well--kind of complicated for such a simple problem. Any good suggestions?