SUBSTRINGS CREATION WITH DIFFERENT ORDER

2013 11 21
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更新时间：2013 11 21
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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

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Azzi Abdelmalek
Azzi Abdelmalek 2013-11-21
编辑：Azzi Abdelmalek 2013-11-21

0 个投票

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)

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FRANCISCO
FRANCISCO 2013-11-21
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
Azzi Abdelmalek 2013-11-21
编辑：Azzi Abdelmalek 2013-11-21
I don't know what pattern you are referring to.Post what is missing with the above example
FRANCISCO
FRANCISCO 2013-11-21
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
Azzi Abdelmalek 2013-11-21
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]

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更多回答（1 个）

Andrei Bobrov
Andrei Bobrov 2013-11-21
编辑：Andrei Bobrov 2013-11-21

0 个投票

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

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