SUBSTRINGS CREATION WITH DIFFERENT ORDER

I wish to estimate substrings of length 4-13, a primary secuence currently about 150,000 binary numbers. How to create the substrings be as follows:

let's assume we have the following sequence (numbers in parentheses indicate the order):

s = 0 (1) 1 (2) 1 (3) 0 (4) 1 (5) 1 (6) 0 (7) 1 (8) 0 (9) 0 (10) 0 (11) 1 (12)

substrings of length 4 would be:

0 (1) 1 (2) 1 (3) 0 (4)

1 (2) 1 (3) 0 (4) 1 (5)

1 (3) 0 (4) 1 (5) 1 (6)

0 (4) 1 (5) 1 (6) 0 (7)

1 (5) 1 (6) 0 (7) 1 (8)

1 (6) 0 (7) 1 (8) 0 (9)

0 (7) 1 (8) 0 (9) 0 (10)

1 (8) 0 (9) 0 (10) 0 (11)

0 (9) 0 (10) 0 (11) 1 (12) .

Now would do the same with the substrings of length 5 After substrings of length 6 ..... up length 13

Many thanks

Answer 1

[Azzi Abdelmalek]

s =[ 0  1  1  0  1  1  0  1  0  0 0 1 ];
n=6;
m=numel(s)-n+1;
A=zeros(m,n);
idx=cell2mat(arrayfun(@(x) x:x+n-1,(1:m)','un',0));
out=s(idx)

[FRANCISCO]

Thank you very much, a question that I have forgotten; as could also add a counter, I count the number of occurrences for each pattern? Because I will use a much longer sequence, it is likely that patterns repeat, and I like to have a column to show me the number of occurrences of each pattern. How would it?

[Azzi Abdelmalek]

I don't know what pattern you are referring to.Post what is missing with the above example

[FRANCISCO]

okei, for example we have the following sequence:

  s = [0 0 0 0 0 0 0 0 0 1];

I apply the code you created to create substrings of length 4:

if true

    % code
n=4;

m=numel(s)-n+1;

A=zeros(m,n);

idx=cell2mat(arrayfun(@(x) x:x+n-1,(1:m)','un',0));

out=s(idx)

end

and the result of "out" is:

      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

we see that the pattern [0 0 0 0] has been repeated 6 times. How would that "out" out:

[0 0 0 0] | 6

[0 0 0 1] | 1

Thank you very much.

[Azzi Abdelmalek]

s =[ 0  1  1  0  1  1  0  1  0  0 0 1 ];
n=4;
m=numel(s)-n+1;
A=zeros(m,n);
idx=cell2mat(arrayfun(@(x) x:x+n-1,(1:m)','un',0))
out=s(idx)
[a,b,c]=unique(out,'rows','stable')
freq=accumarray(c,ones(size(c)))
[a freq]

Answer 2

[Andrei Bobrov]

s =  [0  1  1  0  1  1  0  1  0  0  0  1 ];
n = numel(s);
k=4:n;
out = cell(n-3,1);
for jj = 1:numel(k)
    p = s(hankel(1:k(jj),k(jj):n)');
    [p2,~,ii] = unique(p,'rows');
    out{jj} = [p2, accumarray(ii,1)];
end