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

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sandberry
sandberry 2013-4-27
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.

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采纳的回答

Cedric
Cedric 2013-4-28
编辑：Cedric 2013-4-28
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.

更多回答（1 个）

Alessandro Renna
Alessandro Renna 2013-4-27
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].
  1 个评论
sandberry
sandberry 2013-4-28
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?

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