Finding a minumum range of sequential values

Further to a question that was asked a few days ago (which Matt Fig answered to the satisfaction of the author - see link below):

http://www.mathworks.com/matlabcentral/answers/9393-basic-question-how-to-find-range-of-index-values

In particular, I am searching for a sequence of a minimum of 5 ones in order to perform another function on these elements. That is to say that I would like to perform the second function on any sequence where 5 or more ones may occur. Currently my code looks like:

S3 = findstr(B, ones(1, 5));

But when performing the second action, should the sequence happen to be greater than 5, I find that only the first 5 values identified with the code above are addressed. For example, I have a run of 10 ones, where only the first five values are are found and subsequently computed. Instead, I would like to compute values for all 10 elements appearing in the sequence.

Any ideas on how I might be able to get around this?

The data is a single row vector (1 X 14620); A range of integers where min value = 1 and max value = 6578

e.g

B = [17 4 1 3 2 3 2 1 1 1 1 1 1 1 1 1 1 2 .........12 1 15 372 3 1 1 1 2 42 93 1 105 5 1 1 1 1 78 1 2]

Answer 1

[Matt Fig]

What else does your data look like? Is it ones and zeros? Integers only? A single row vector, or matrix? Give all the details you can...

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EDIT With more details added...

x = B==1;  % Sequences of 1s, change this line to find other values.
t = [1 diff(x)]~=0;
x = x(t);
t = find([t 1]);
n = diff(t);
idx = (x==1 & n>=5);  % Seq. length >= 5, change this for other lengths.

n = n(idx)    % Length of each sequence of 1s.
idx = t(idx)  % Starting index of each sequence.

[bugguts99]

The data is a single row vector (1 X 14620); A range of integers where min value = 1 and max value = 6578.

[bugguts99]

Thanks Matt. Works perfectly.

Answer 2

[Image Analyst]

If you have the Image Processing Toolbox, it's trivial. Just call bwlabel and regionprops (asking for Area and PixelIdxList), find those with area >= 5, and you're done in only about 3 lines of code, with the indexes of all 1's in a run of 5 or more.

[bugguts99]

Also works perfectly when

x = B == 1;

Thanks for the tips!

Answer 3

[Walter Roberson]

See the techniques described in this other recent Question

By the way:

endpos = find(B(S3(1):end)~=1,1);

if endpos is empty then everything to the end of B is 1's.

Answer 4

[Andrei Bobrov]

x = B == 1;
idx1int = [strfind([0 x],[0 1]); strfind(x,[1 0])];
idxout = idx1int(:,diff(idx1int,1)+1>=5);
nout = diff(idxout,1)+1;