extracting subsequences of binary string

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FRANCISCO
FRANCISCO 2013-8-20
评论： FRANCISCO 2013-10-19
as would be the code for the following string have the next subsequences ?
STRING
1(1), 0(2), 1(3), 1(4), 0(5), 0(6), 1(7), 0(8), 0(9), 1(10), 1(11), 1(12), 1(13), 0(14), 0(15), 0(16), 1(17), 1(18), 1(19), 0(20)
SUBSEQUENCES
01: 1(01), 0(02), 1(03), 1(04) -> [1,0,1,1],
02: 1(01), 1(03), 0(05), 1(07) -> [1,1,0,1],
03: 1(01), 1(04), 1(07), 1(10) -> [1,1,1,1],
04: 1(01), 0(05), 0(09), 1(13) -> [1,0,0,1],
05: 1(01), 0(06), 1(11), 0(16) -> [1,0,1,0],
06: 1(01), 1(07), 1(13), 1(19) -> [1,1,1,1],
07: 0(02), 1(03), 1(04), 0(05) -> [0,1,1,0],
08: 0(02), 1(04), 0(06), 0(08) -> [0,1,0,0],
09: 0(02), 0(05), 0(08), 1(11) -> [0,0,0,1],
10: 0(02), 0(06), 1(10), 0(14) -> [0,0,1,0],
11: 0(02), 1(07), 1(12), 1(17) -> [0,1,1,1],
12: 0(02), 0(08), 0(14), 0(20) -> [0,0,0,0],
13: 1(03), 1(04), 0(05), 0(06) -> [1,1,0,0],
14: 1(03), 0(05), 1(07), 0(09) -> [1,0,1,0],
15: 1(03), 0(06), 0(09), 1(12) -> [1,0,0,1],
16: 1(03), 1(07), 1(11), 0(15) -> [1,1,1,0],
17: 1(03), 0(08), 1(13), 1(18) -> [1,0,1,1],
18: 1(04), 0(05), 0(06), 1(07) -> [1,0,0,1],
19: 1(04), 0(06), 0(08), 1(10) -> [1,0,0,1],
20: 1(04), 1(07), 1(10), 1(13) -> [1,1,1,1],
21: 1(04), 0(08), 1(12), 0(16) -> [1,0,1,0],
22: 1(04), 0(09), 0(14), 1(19) -> [1,0,0,1],
23: 0(05), 0(06), 1(07), 0(08) -> [0,0,1,0],
24: 0(05), 1(07), 0(09), 1(11) -> [0,1,0,1],
25: 0(05), 0(08), 1(11), 0(14) -> [0,0,1,0],
26: 0(05), 0(09), 1(13), 1(17) -> [0,0,1,1],
27: 0(05), 1(10), 0(15), 0(20) -> [0,1,0,0],
28: 0(06), 1(07), 0(08), 0(09) -> [0,1,0,0],
29: 0(06), 0(08), 1(10), 1(12) -> [0,0,1,1],
30: 0(06), 0(09), 1(12), 0(15) -> [0,0,1,0],
31: 0(06), 1(10), 0(14), 1(18) -> [0,1,0,1],
32: 1(07), 0(08), 0(09), 1(10) -> [1,0,0,1],
33: 1(07), 0(09), 1(11), 1(13) -> [1,0,1,1],
34: 1(07), 1(10), 1(13), 0(16) -> [1,1,1,0],
35: 1(07), 1(11), 0(15), 1(19) -> [1,1,0,1],
36: 0(08), 0(09), 1(10), 1(11) -> [0,0,1,1],
37: 0(08), 1(10), 1(12), 0(14) -> [0,1,1,0],
38: 0(08), 1(11), 0(14), 1(17) -> [0,1,0,1],
39: 0(08), 1(12), 0(16), 0(20) -> [0,1,0,0],
40: 0(09), 1(10), 1(11), 1(12) -> [0,1,1,1],
41: 0(09), 1(11), 1(13), 0(15) -> [0,1,1,0],
42: 0(09), 1(12), 0(15), 1(18) -> [0,1,0,1],
43: 1(10), 1(11), 1(12), 1(13) -> [1,1,1,1],
44: 1(10), 1(12), 0(14), 0(16) -> [1,1,0,0],
45: 1(10), 1(13), 0(16), 1(19) -> [1,1,0,1],
46: 1(11), 1(12), 1(13), 0(14) -> [1,1,1,0],
47: 1(11), 1(13), 0(15), 1(17) -> [1,1,0,1],
48: 1(11), 0(14), 1(17), 0(20) -> [1,0,1,0],
49: 1(12), 1(13), 0(14), 0(15) -> [1,1,0,0],
50: 1(12), 0(14), 0(16), 1(18) -> [1,0,0,1],
51: 1(13), 0(14), 0(15), 0(16) -> [1,0,0,0],
52: 1(13), 0(15), 1(17), 1(19) -> [1,0,1,1],
53: 0(14), 0(15), 0(16), 1(17) -> [0,0,0,1],
54: 0(14), 0(16), 1(18), 0(20) -> [0,0,1,0],
55: 0(15), 0(16), 1(17), 1(18) -> [0,0,1,1],
56: 0(16), 1(17), 1(18), 1(19) -> [0,1,1,1],
57: 1(17), 1(18), 1(19), 0(20) -> [1,1,1,0],

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采纳的回答

Andrei Bobrov
Andrei Bobrov 2013-8-21
编辑：Andrei Bobrov 2013-8-21
N = 20;
n = 4;
A = hankel(1:N-n+1,N-n+1:N);
k = 0:n-1;
idx = [];
for ii = 1:size(A,1)
p = A(ii,:);
while p(end,end) + k(end) <= N
p = [p;p(end,:)+k];
end
idx=[idx;p];
end
or
N = 20;
n = 4;
A = hankel(1:N-n+1,N-n+1:N);
k = 0:n-1;
c = ceil((N - A(:,end) + 1)/k(end));
i2 = cumsum(c);
i1 = i2 - c + 1;
idx = zeros(i2(end),n);
for jj = 1:N-n+1
idx(i1(jj):i2(jj),:) = bsxfun(@plus,A(jj,:),(0:c(jj)-1)'*k);
end
ADD
s = [1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0];
[j1,j2,j2] = unique(s(idx),'rows')
out = [j1, histc(j2,1:max(j2))/i2(end)]; % This row corrected
  8 个评论
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FRANCISCO
FRANCISCO 2013-8-21
sorry, I have not understood the code. This it does is calculate the number of times to repeat each subsequence?. It calculates the sub but if calculated occurrences each subsequence?. it?

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更多回答（2 个）

Roger Stafford
Roger Stafford 2013-8-20
编辑：Roger Stafford 2013-8-21
n = 20;
d = 4;
c = zeros(sum([1,floor((d:n-1)/(d-1))]),d); % Allocate space for c
j = 0;
for k = 1:n-d+1
r = 1;
while k+r*(d-1) <= n
j = j+1;
c(j,:) = k:r:k+r*(d-1);
r = r+1;
end
end
The c array will be a 57 x 4 matrix of subsequence indices taken from 1:20.
c =
1 2 3 4
1 3 5 7
1 4 7 10
.....
17 18 19 20
If you replace the line "c(j,:) = k:r:k+r*(d-1);" by
c(j,:) = s(k:r:k+r*(d-1));
where s is your string, this will generate the subsequence of binary strings you are (apparently) asking for.
  3 个评论
显示 1更早的评论
FRANCISCO
FRANCISCO 2013-8-21
thank you very much, that command should now be used to calculate the number of times to repeat each subsequence? is to calculate the probability by dividing the number of occurrences of that subsequence by the total number of subsequences. But I'm not sure which command used to count the number of occurrences of each subsequence
FRANCISCO
FRANCISCO 2013-10-19
One question, as I can do with structure for you automatically calculate subsequences of length 4-20? ie, d = 4:20 but applying for so I said why not have the same dimension:
if true
% code
for d=4:20
c(d)=zeros(sum([1,floor((d:n-1)/(d-1))]),d);
j=0;
for k=1:n-d+1
r=1;
while k+r*(d-1)<=n
j=j+1;
c(j,:)=s(k:r:k+r*(d-1));% s es la cadena binaria / me da las subsecuencias
r=r+1;
end
end
end
end

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Roger Stafford
Roger Stafford 2013-8-22
编辑：Roger Stafford 2013-8-22
Here is a slightly shorter version:
n = 20;
d = 4;
f2 = cumsum([0,floor((n-1:-1:d-1)/(d-1))]);
f1 = f2(1:end-1)+1;
f2 = f2(2:end);
c = repmat(0:d-1,f2(end),1);
for k = 1:length(f1)
c(f1(k),:) = c(f1(k),:) + k;
c(f1(k):f2(k),:) = cumsum(c(f1(k):f2(k),:),1);
end

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