How to find an exact sequence of values?

Hi everyone,

as the title says I'm looking to find an exact sequence of values in a matrix. I have a matrix that I'm turning into a logical array as below. Now I want to be able to say if the consecutive ones in the column are less than X (as in e - see below) they should also be turned into zeroes as well.

I tried to work with find(contain()) but that did nothing. Also ismember() did not turn out the results I was looking for.

Turning matrix into binary representation

for k = 1:1:size(p,1)  
  for l = 1:1:(size(p,2))
    if p(k,l) > z 
      p(k,l) = 1;
    else
      p(k,l) = 0;
    end                  
  end
end

Logical array

  0   0   1   0   0   0   0   0   0   0   0   0   0   0   0   0
   0   1   1   0   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   1   1   1   0   0   0   0   0   0   0   0   0
   0   0   0   1   1   1   1   0   0   0   0   0   0   0   0   0
   0   0   0   1   1   1   1   0   0   0   0   0   0   0   0   0
   0   0   1   0   1   1   1   1   0   0   0   0   0   0   0   0
   0   0   1   0   1   1   1   0   0   0   0   1   0   0   0   0
   0   0   1   0   1   0   0   0   0   0   0   1   0   0   0   0
   0   0   1   0   1   0   0   0   0   0   0   0   0   0   0   0
   0   0   1   0   1   0   0   0   0   0   0   1   0   0   0   0
   0   0   1   0   1   0   0   0   0   0   0   1   1   0   0   0
   0   0   1   0   1   0   0   0   0   0   0   1   1   0   0   0
   0   0   0   0   1   0   0   0   0   0   0   0   1   1   0   0
   0   0   0   0   1   0   0   0   0   0   0   0   1   0   0   0
   0   0   0   0   1   0   0   0   0   0   1   0   0   0   0   1

Size of ones I would like to be able to filter

 e = ones(x,1)

Thank you for your help everyone! It's greatly appreciated.

Cheers,

Peter

[dpb]

Turning matrix into binary representation

p=(p~=0);

I forget whose submittal it is, but there's a nice runs function on FEX that would be ideal...

[Star Strider]

I believe you’re referring to Jan‘s RunLength.

[dpb]

Ayup, SS...I couldn't remember whether the particular one was by Jan or Bruno

BTW, another way would be to convert to char() and regexp() search.

p=char(p+'0').'; % convert to row character string representation

[Pete]

Hey everyone,

thank you for the answers, unfortunately I've been unable to apply RunLength to my problem and if I use regexp() search I cannot enter something along the lines of ones(x,1) and then

x = [1 2 3 4 5];

I'm very new to all of this so I'm probably not doing it right but is there anything else you might be able to recommend?

Thank you for your help!!

[dpb]

Show us what you tried with runlength -- can't imagine it would not work to do the job...and actually, what regexp pattern you used. I'm no whizard on regular expressions but there are those here who are.

[Pete]

Luna's approach worked very well. Thank you for your input everyone!

Answer 1

[Luna]

Try Loren's findpattern2. Here is link:

findpattern

create vector x numbers of ones and run findpattern2.

e = ones(1,x);
for i = 1:size(LogicalArray,2)
   indices  = findpattern2(LogicalArray(:,i),e);
   if ~isempty(indices) % if it finds x elements of this pattern make them zero.
       for k=1:numel(indices)
           LogicalArray(indices(k):indices(k)+x,i) = false; % from indice to pattern length will be zero
       end
   end
end

[Pete]

That worked perfectly! Thank you

[Luna]

Your welcome :)

Answer 2

[Image Analyst]

If you want to use functions already built in to the toolbox, you can use bwareafilt:

binaryImage = logical([...
	  0   0   1   0   0   0   0   0   0   0   0   0   0   0   0   0
   0   1   1   0   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   1   1   1   0   0   0   0   0   0   0   0   0
   0   0   0   1   1   1   1   0   0   0   0   0   0   0   0   0
   0   0   0   1   1   1   1   0   0   0   0   0   0   0   0   0
   0   0   1   0   1   1   1   1   0   0   0   0   0   0   0   0
   0   0   1   0   1   1   1   0   0   0   0   1   0   0   0   0
   0   0   1   0   1   0   0   0   0   0   0   1   0   0   0   0
   0   0   1   0   1   0   0   0   0   0   0   0   0   0   0   0
   0   0   1   0   1   0   0   0   0   0   0   1   0   0   0   0
   0   0   1   0   1   0   0   0   0   0   0   1   1   0   0   0
   0   0   1   0   1   0   0   0   0   0   0   1   1   0   0   0
   0   0   0   0   1   0   0   0   0   0   0   0   1   1   0   0
   0   0   0   0   1   0   0   0   0   0   0   0   1   0   0   0
   0   0   0   0   1   0   0   0   0   0   1   0   0   0   0   1])
x = 4; % Whatever....
% Get rid of stretches of 1's in each row that are less than x long.
for row = 1 : size(binaryImage, 1)
	binaryImage(row, :) = bwareafilt(binaryImage(row, :), [x, inf]);
end

The result is:

binaryImage =

17×16 logical array

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0

0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0

0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0