Extraction a certain part out of Matrix

Hi,

I have a matrix that contains valid and invalid data. The invalid data is already marked by a vector that has either a 1 (if it's a valid data point) or a 0 (if it's invalid):

Col1 Col2 ... Col10 ... 1 ... 1 ... 0 ... 1 ... 1 ... 1 ... 0 ... ...

The ones and zeros in Col10 are distributed irregular and I would like to extract the longest continuous row of ones in Col10.

How do I do that? Any ideas?

Answer 1

[Oleg Komarov]

out = findseq(X,2);
[m,idx] = max(out(:,end));
[r,c] = ind2sub(size(X),out(idx,2))

r and c are the row and column of the first one of the longest contiguos row of ones.

http://www.mathworks.com/matlabcentral/fileexchange/28113-findseq

[Stefan]

findseq - What a handy little function!

Works great.

Is there any way to tell the function what it should look for? In my case it could be that there is more invalid date then valid data. So it might find a longer series of zeros then ones. I could just work with the output of findseq and throw away all zeros but maybe there is another clever way to tell findseq right away what to look for.

Thank you so much already!

[Oleg Komarov]

out in the example in the first column contains the value that is repeated and the last value the number of repetitions (if more than 1). You can sortrows out by the 4th column and see which value comes first, a 0 or a 1.

Answer 2

[Fangjun Jiang]

The code below will find the first longest continuous row of ones.

Data=rand(100,1);
index=Data>0.5;
Data(index)=1;
Data(~index)=0;
DataStr=sprintf('%d',Data)

FoundStr=textscan(DataStr,'%s','delimiter','0','MultipleDelimsAsOne',true);

d=cellfun('length',FoundStr{1});
[dMax,IndMax]=max(d);
Longst=FoundStr{1}{IndMax};
StartIndex=strfind(DataStr,Longst);

DataIndex=StartIndex(1):(StartIndex(1)+dMax-1)

SelectedData=Data(DataIndex)

Answer 3

[Andrei Bobrov]

My favorite version :) all is tired of me

X = rand(20,1)>.35
In = X(:)';
idx = [strfind([0 In],[0 1]);strfind([In 0],[1 0])];
[l,io] = max(diff(idx));
idxout = io;
lgr = l + 1;