Arrays and finding a chain of ones

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Rasmus
Rasmus el 13 de Abr. de 2014
Respondida: Image Analyst el 13 de Abr. de 2014
Okay I have an array which gives me 566 ones and zeros in total. The code i am using is the following.
x0=[zeros(1,276) ones(1,290)];
x0(randperm(566));
So what i need to figure out is how to find the longest chain of ones in this array. Any who has a good solution for this?

Respuesta aceptada

Azzi Abdelmalek
Azzi Abdelmalek el 13 de Abr. de 2014
Editada: Azzi Abdelmalek el 13 de Abr. de 2014
x0=[1 1 1 0 0 1 0 1 1 1 1 1 0 0 1]
x=[0 x0 0]
idx1=strfind(x,[1 0])-1
idx0=strfind(x,[0 1])
[max_length,ii]=max(idx1-idx0+1)
index1=idx0(ii) % The chain containing max_length ones begins at index1
  3 comentarios
Rasmus
Rasmus el 13 de Abr. de 2014
it seems like this code looks for a chain with a defined max length? but what if i don't want to define a length, and just want to find the longest chain, without knowing how long it is?
Image Analyst
Image Analyst el 13 de Abr. de 2014
Almost didn't read this because I saw it was accepted but it looked like a trivial thing to do with the Image Processing Toolbox. It's like 3 lines of code. Let me know if you want to see it.

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Más respuestas (1)

Image Analyst
Image Analyst el 13 de Abr. de 2014
Well, for the benefit of anyone who does have the Image Processing Toolbox and wants to know how to find the starting and ending indexes of the longest stretch of "1"s in the array, here is the code:
% Create sample data
x0 = [1 1 0 0 0 1 0 1 1 1 1 1 1 0 0 1 0 1 1 1 0]
% Make measurements of lengths of connected "1" regions.
measurements = regionprops(logical(x0), 'Area', 'PixelIdxList');
% Sort them to find the longest one.
[sortedAreas, sortIndexes] = sort([measurements.Area], 'Descend')
% Get the starting and ending indexes of the largest one.
startingIndex = measurements(sortIndexes(1)).PixelIdxList(1)
endingIndex = measurements(sortIndexes(1)).PixelIdxList(end)
In the command window:
x0 =
1 1 0 0 0 1 0 1 1 1 1 1 1 0 0 1 0 1 1 1 0
sortedAreas =
6 3 2 1 1
sortIndexes =
3 5 1 2 4
startingIndex =
8
endingIndex =
13

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